Ch.17

Chapter 17

397

Practice Exercises 17.1 Ba3(PO4)2(s) 3Ba2+(aq) + 2PO4

3+(aq) Ksp = [Ba2+]3[PO4

3+]2 17.2 (a) Ksp = [Ba2+][C2O4

2–] (b) Ksp = [Ag+]2[SO42–]

17.3 TlI(s) Tl+(aq) + I–(aq)

Ksp = [Tl+][I–]

mol TlI = 5.9 × 10–3 g 1 mol TlI331.3 g TlI

⎛ ⎞⎜ ⎟⎝ ⎠

= 1.78 × 10–5 mol TlI

[Tl+] = [I–] = 51.78 10 mol

1 L

−× = 1.78 × 10–5 M

Ksp = (1.8 × 10–5)(1.8 × 10–5) = 3.2 × 10–10

17.4 PbF2(s) Pb2+(aq) + 2F–(aq) 22+

spK = Pb F−⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

( ) ( )( )23 3spK = 2.15 10 2 2.15 10 =− −× × 3.98 × 10–8

17.5 Ag2SO4(s) 2Ag+(aq) + SO4

2–(aq)

Mol Na2SO4 added = (28.4 g Na2SO4) 2 4

2 4

1 mol Na SO142.04 g Na SO⎛ ⎞⎜ ⎟⎝ ⎠

= 0.200 mol Na2SO4

Concentration of SO42– =

240.200 mol SO

1 L solution

− = 0.200 M SO4

2–

[Ag+] [SO4

2–] I – 0.200 C + (4.3 × 10–3) × 2 + 4.3 × 10–3

E + 8.6 × 10–3 0.200 + 4.3 × 10–3

Ksp = [Ag+]2[SO4

2–] Ksp (8.6 ×10–3)2(0.204) = 1.5 × 10–5

17.6 CoCO3(s) Co2+(aq) + CO3

2–(aq)

[Co2+] [CO32–]

I – 0.10 C +1.0 × 10–9 + 1.0 × 10–9 E +1.0 × 10–9 0.10 + 1.0 × 10–9

Substituting the above values for equilibrium concentrations into the expression for Ksp gives: Ksp = [Co2+][CO3

2–] = (1.0 × 10–9)(0.10 + 1.0 × 10–9) = 1.0 × 10–10

Chapter 17

398

17.7 PbF2(s) Pb2+(aq) + 2F–(aq)

[Pb2+] [F–] I 0.10 – C + 3.1 × 10–4 + 2(3.1 × 10–4) E 0.10 + (3.1 × 10–4) + 6.2 × 10–4

Substituting the above values for equilibrium concentrations into the expression for Ksp gives: Ksp = [Pb2+][F–]2 = [0.10 + 3.1 × 10–4][6.2 × 10–4]2 Now (0.10 + 3.1 × 10–4) is also ≈ 0.10: Hence, Ksp = (0.10)(6.2 × 10–4)2 = 3.9 × 10–8

17.8 Ag3PO4(s) 3Ag+(aq) + PO4

3+(aq) Ksp = [Ag+]3[PO4

3+] = 2.8 × 10–18

[Ag+] [PO43+]

I – – C + 3x + x E + 3x + x

Ksp = [3x]3[x] = 2.8 × 10–18 2.8 × 10–18 = 27x4 x4 = 1.04 × 10–19

x = 4 191.04 10−× x = 1.79 × 10–5 Thus the solubility is 1.8 × 10–5 M Ag3PO4.

17.9 (a) AgBr(s) Ag+(aq) + Br–(aq) Ksp = [Ag+][Br–] = 5.0 × 10–13

[Ag+] [Br–] I – – C + x + x E + x + x

Substituting the above values for equilibrium concentrations into the expression for Ksp gives: Ksp = 5.0 × 10–13 = [Ag+][Br–] = (x)(x)

13 7x 5.0 10 7.1 10− −= × = × Thus the solubility is 7.1 × 10–7 M AgBr.

(b) Ag2CO3(s) 2Ag+(aq) + CO3

2–(aq) Ksp = [Ag+]2[CO32–] = 8.1 × 10–12

[Ag+] [CO3

2–] I – – C + 2x + x E + 2x + x

Substituting the above values for equilibrium concentrations into the expression for Ksp gives: Ksp = 8.1 x 10–12 = [Ag+]2[CO3

2–] = (2x)2(x) and 4x3 = 8.1 × 10–12

Chapter 17

399

( )12 43x 8.1 10 / 4 1.3 10− −= × = ×

Thus the molar solubility is 1.3 × 10–4 M Ag2CO3.

17.10 AgI(s) Ag+(aq) + I–(aq) Ksp = [Ag+][I–] = 8.3 × 10–17 The initial concentration of I– is 2 × 0.20 M from the CaI2.

[Ag+] [I–] I – 0.40 C + x + x E + x 0.40 + x

Substituting the above values for equilibrium concentrations into the expression for Ksp gives: Ksp = 8.3 × 10–17 = [Ag+][I–] = (x)(0.40 + x) We know that the value of Ksp is very small, and it suggests the simplifying assumption that (0.40 + x) ≈ 0.40: Hence, 8.3 × 10–17 = (0.40)x, and x = 2.1 × 10–16. The assumption that (0.40 + x) ≈ 0.40 is seen to be valid indeed. Thus 2.1 × 10–16 M of AgI will dissolve in a 0.20 M CaI2 solution. In pure water, Ksp = 8.3 × 10–17 = [Ag+][I–] = (x)(x) x = [AgI(aq)] = 9.1 × 10–9 M (much more soluble)

17.11 Fe(OH)3(s) Fe3+(aq) + 3OH–(aq) Ksp = [Fe3+][OH–]3 = 1.6 × 10–39

[Fe3+] [OH–] I – 0.050 C + x + 3x E + x 0.050 + 3x

Substituting the above values for equilibrium concentrations into the expression for Ksp gives:

Ksp = 1.6 × 10–39 = [Fe3+][OH–]3 = (x)[0.050 + 3x]3 We try to simplify by making the approximation that (0.050 + 3x) ≈ 0.050:

1.6 × 10–39 = (x)(0.050)3 or x = 1.3 × 10–35 Clearly the assumption that (0.050 + 3x) ≈ 0.050 is justified. Thus 1.3 × 10–35 M of Fe(OH)3 will dissolve in a 0.050 M sodium hydroxide solution.

17.12 The expression for Ksp is Ksp = [Ca2+][SO4

2–] = 2.4 × 10–5 and the ion product for this solution would be: [Ca2+][SO4

2–] = (2.5 × 10–3)(3.0 × 10–2) = 7.5 × 10–5 Since the ion product is larger than the value of Ksp, a precipitate will form.

17.13 The solubility product constant is Ksp = [Ag+]2[CrO4

2–] = 1.2 × 10–12 and the ion product for this solution would be: [Ag+]2[CrO4

2–] = (4.8 × 10–5)2(3.4 × 10–4) = 7.8 × 10–13 Since the ion product is smaller than the value of Ksp, no precipitate will form.

Chapter 17

400

17.14 We expect PbSO4(s) since nitrates are soluble.

Because two solutions are to be mixed together, there will be a dilution of the concentrations of the various ions, and the diluted ion concentrations must be used. In general, on dilution, the following relationship is found for the concentrations of the initial solution (Mi) and the concentration of the final solution (Mf): MiVi = MfVf Thus the final or diluted concentrations are:

( )2+ 3 4100.0 mLPb = 1.0 10 M = 5.0 10 M200.0 mL

− −⎛ ⎞⎡ ⎤ × ×⎜ ⎟⎣ ⎦ ⎝ ⎠

( )2 3 34

100.0 mLSO = 2.0 10 M = 1.0 10 M200.0 mL

− − −⎛ ⎞⎡ ⎤ × ×⎜ ⎟⎣ ⎦ ⎝ ⎠

The value of the ion product for the final (diluted) solution is: [Pb2+][SO4

2–] = (5.0 × 10–4)(1.0 × 10–3) = 5.0 × 10–7 Since this is smaller than the value of Ksp (6.3 × 10–7), a precipitate of PbSO4 is not expected.

17.15 We expect a precipitate of PbCl2 since nitrates are soluble.

We proceed as in Practice Exercise 14. MiVi = MfVf

( )2+ 50.0 mLPb = 0.10 M = 0.071 M70.0 mL

⎛ ⎞⎡ ⎤ ⎜ ⎟⎣ ⎦ ⎝ ⎠

( ) 20.0 mLCl = 0.040 M = 0.011 M70.0 mL

− ⎛ ⎞⎡ ⎤ ⎜ ⎟⎣ ⎦ ⎝ ⎠

The value of the ion product for such a solution would be: [Pb2+][Cl–]2 = (7.1 × 10–2)(1.1 × 10–2)2 = 8.6 × 10–6 Since the ion product is smaller than Ksp (1.7 × 10–5), a precipitate of PbCl2 is not expected.

17.16 CoS will precipitate if the H+ concentration is too low. Solving for Q and then comparing Q to Kspa, we can

determine whether or not CoS will precipitate. 2

12spa 2

[Co ][H S]K = 5 10

[H ]

+−

+= ×

232

2 4 2[Co ][H S] [0.005][0.10]Q = 5 10

[H ] [3.16 10 ]

+

+ −= = ×

×

Q > Kspa Since Q is greater than Kspa, then the reaction will move to reactants and CoS solid will form.

17.17 Consulting Table 17.2, we find that Fe2+ is much more soluble in acid than Hg2+. We want to make the H+ concentration large enough to prevent FeS from precipitating, but small enough that HgS does precipitate. First, we calculate the highest pH at which FeS will remain soluble, by using Kspa for FeS. (Recall that a saturated solution of H2S = 0.10 M.)

222

spa 2 2[Fe ][H S] [0.010][0.10]K = 6 10

[H ] [H ]

+

+ += = ×

[H+] = 0.0013 M pH = –log [H+] = 2.9

Chapter 17

401

Since Fe2+ is much more soluble in acid than Hg2+ we already know that this pH will precipitate HgS, but we can check it by using Kspa for HgS:

2322

spa 2 2[Hg ][H S] [0.010][0.10]K = 2 10

[H ] [H ]

+−

+ += = ×

[H+] = 2.2 × 1014 M (This concentration is impossibly high, but it tells us that this much acid would be required to dissolve HgS at these concentrations.)

17.18 BaC2O4(s) Ba2+(aq) + C2O4

2–(aq) Ksp = 1.2 × 10–7 = [Ba2+][C2O42–]

[Ba2+] = 0.050 M 1.2 × 10–7 = (0.050)[C2O4

2–] [C2O4

2–] = 2.4 × 10–6 M H2C2O4 H+ + HC2O4

– Ka1 = 5.6 × 10–2 HC2O4

– H+ + C2O42– Ka2 = 5.4 × 10–5

H2C2O4 2H+ + C2O42– Ka = (5.4 × 10–5) × (5.6 × 10–2) = 3.0 × 10–6

[H2C2O4] = 0.10 [C2O4

2–] = 2.4 × 10–6 M

Ka = 3.0 × 10–6 = [ ]

2 22 4

2 2 4

H C O

H C O

+ −⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ =

( )( )

2 6H 2.4 10

0.10

+ −⎡ ⎤ ×⎣ ⎦

Since the amount of oxalate formed is so small, the concentration of oxalic acid is essentially unchanged. [H+] = 3.55 × 10–1 This is the minimum concentration of H+ that will prevent the formation of BaC2O4 precipitate.

17.19 Follow the exact procedure outlined in Example 17.10. Ksp = [Ca2+][CO3

2–]= 4.5 × 10–9 Ksp = [Ni2+][CO3

2–] = 1.3 × 10–7 NiCO3 is more soluble and will precipitate when:

[CO32–] =

7sp 62

K 1.3 10 1.3 10 0.10[Ni ]+×

= = ×

CaCO3 will precipitate when:

9sp2 8

3 2

K 4.5 10 [CO ] 4.5 10 0.10[Ca ]

−+

×= = = ×

CaCO3 will precipitate and NiCO3 will not precipitate if [CO3

2–] > 4.5 × 10–8 and [CO32–] < 1.3 × 10–6.

Now, using the equation in example 17.10 we get:

2 172

3

0.030[H ] (2.4 10 )[CO ]

+ −−

⎛ ⎞⎜ ⎟= ×⎜ ⎟⎝ ⎠

NiCO3 will precipitate if:

2 17 136

0.030[H ] (2.4 10 ) 5.5 101.3 10

+ − −−

⎛ ⎞= × = ×⎜ ⎟

×⎝ ⎠

[H+] = 7.4 × 10–7 pH = 6.13

Chapter 17

402

CaCO3 will precipitate: 2 17 11

8

6

0.030[H ] (2.4 10 ) 1.6 104.5 10

[H ] 4.0 10 pH 5.40

+ − −−

+ −

⎛ ⎞= × = ×⎜ ⎟

×⎝ ⎠

= × =

So CaCO3 will precipitate and NiCO3 will not if the pH is maintained between pH = 5.40 and pH = 6.13 17.20 The overall equilibrium is AgCl(s) + 2NH3(aq) Ag(NH3)2

+(aq) + Cl–(aq)

+3 2

c 23

Ag(NH ) ClK =

NH

−⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

⎡ ⎤⎣ ⎦

In order to obtain a value for Kc for this reaction, we need to use the expressions for Ksp of AgCl(s) and the

Kform of Ag(NH3)2+:

+ 10sp

+3 2 7

form 23

+3 2 3

c sp form 23

K = Ag Cl = 1.8 10

Ag(NH )K = = 1.6 10

Ag NH

Ag(NH ) ClK = K K = 2.9 10

NH

− −

+

−−

⎡ ⎤ ⎡ ⎤ ×⎣ ⎦ ⎣ ⎦⎡ ⎤⎣ ⎦ ×⎡ ⎤⎡ ⎤⎣ ⎦⎣ ⎦

⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦× = ×

⎡ ⎤⎣ ⎦

Now we may use an equilibrium table for the reaction in question:

[NH3] [Ag(NH3)2+] [Cl–]

I 0.10 – – C – 2x + x + x E 0.10 – 2x x x

Substituting these values into the mass action expression gives:

3c 2

(x)(x)K = 2.9 10 = (0.10 2x)

−×−

Take the square root of both sides to get (x)0.054 = (0.10 2x)−

Solving for x we get, x = 4.9 × 10–3 M. The molar solubility of AgCl in 0.10 M NH3 is therefore 4.9 × 10–3 M.

In order to determine the solubility in pure water, we simply look at Ksp AgCl(s) Ag+ (aq) + Cl–(aq) Ksp = [Ag+][Cl–] = 1.8 × 10–10 At equilibrium; [Ag+] = [Cl–] = 1.3 × 10–5. Hence the molar solubility of AgCl in 0.10 M NH3 is about 380

times greater than in pure water.

Chapter 17

403

17.21 We will use the information gathered for the last problem. Specifically, AgCl(s) + 2NH3(aq) Ag(NH3)2

+(aq) + Cl–(aq) +

3 2 3c 2

3

Ag(NH ) ClK = = 2.9 10

NH

−−

⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ ×

⎡ ⎤⎣ ⎦

If we completely dissolve 0.20 mol of AgCl, the equilibrium [Cl–] and [Ag(NH3)2

+] will be 0.20 M in a one liter container. The question asks, therefore, what amount of NH3 must be initially present so that the equilibrium concentration of Cl– is 0.20 M?

[NH3] [Ag(NH3)2

+] [Cl–] I Z – – C – 2x + x + x E Z – 2x x x

3c 2

(x)(x)K = 2.9 10 = (Z 2x)

Take the square root of both sides to get;x 0.200.054

Z 2x Z 0.40

−×−

= =− −

We have substituted the known value of x. Solving for Z we get, Z = 4.1 M Consequently, we would need to add 4.1 moles of NH3 to a one liter container of 0.20 M AgCl in order to

completely dissolve the AgCl. Review Questions 17.1 The ion product is the quantity which results from the mass action expression and it is a product of the ion

concentrations. The ion product constant, also called the solubility product constant, is equal to the product of the ion concentrations for a saturated solution of a sparingly soluble substance.

17.2 The addition of a common ion to a saturated solution lowers the solubility of a sparingly soluble ionic salt.

According to Le Châtelier’s principle, addition of an ion to a saturated solution will shift the equilibrium so as to absorb as much of the added ion as possible. This results in the precipitation of the sparingly soluble salt. In the case of AgCl, Ksp = 1.8 × 10–10, addition of NaCl to a saturated solution containing Ag+ and Cl– will result in the precipitation of AgCl(s). The common ion in this case is Cl–. It is present in the solution due to the NaCl as well as the AgCl.

17.3 Kc = ( )

3 22 34

3 4 2

Ba PO

Ba PO

+ −⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦⎡ ⎤⎣ ⎦

The denominator of the mass action expression is a constant since the concentration of substance within a pure solid is constant. Consequently, we define a new constant which is the product of Kc and the concentration of the pure solid.

3 22+ 3

sp 4K = Ba PO −⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

17.4 When the value of the ion product is greater than Ksp, a precipitate will form.

Chapter 17

404

17.5 Ksp values assume 100% dissociation of the ions, but in solution, the ions are not fully dissociated, but instead ion pairs are often formed.

17.6 When sodium acetate is added to a solution of acetic acid, the equilibrium is disrupted and more acetic acid

is formed. This lowers the amount of hydronium ions in solution thus raising the pH. 17.7 Na2S(s) 2Na+(aq) + HS–(aq) + OH–(aq) 17.8 The oxide ion reacts with the water to produce hydroxide ion: O2– + H2O 2OH– 17.9 As the pH of an oxalic acid solution is decreased, the hydrogen ion concentration is increased, this shifts

the equilibrium to the undissociated acid, thus increasing the H2C2O4 concentration. The opposite is true as the pH of the solution is increased.

17.10 By combining the three dissociation expressions for phosphoric acid, we come up with a single expression

that relates the hydrogen ion concentration and the phosphate ion concentration: H3PO4(aq) 3H+(aq) + PO4

3–(aq) K = Ka1

× Ka2 × Ka3

17.11 (a) CoS(s) Co2+(aq) + HS–(aq) + OH–(aq) Ksp = [CO2+][HS–][OH–]

(b) CoS(s) + 2H+(aq) Co2+(aq) + H2S(aq) [ ]2+

2spa 2+

Co H SK =

H

⎡ ⎤⎣ ⎦

⎡ ⎤⎣ ⎦

17.12 NH3(aq) + H2O NH4

+(aq) + OH–(aq) The addition of NH4Cl to the mixture of Mg(OH)2 and water causes the Mg(OH)2 to dissolve because the NH4

+ reacts with the OH– and shifts the equilibrium towards the dissolution of the solid. 17.13 AgCl(s) Ag+(aq) + Cl–(aq) Ag+(aq) + 2NH3(aq) Ag(NH3)2

+(aq) Silver chloride is an insoluble solid. However, any Ag+ ions present react with added NH3 to form the

Ag(NH3)2+ complex ion. According to Le Châtelier’s Principle, as NH3 is added to a solution containing

Ag+ ions, the complex ion forms using up the Ag+ ions. This disrupts the equilibrium and forces AgCl to dissolve.

Upon addition of HNO3, a strong acid, H+ reacts with NH3 to form NH4

+. This disrupts the complex equilibria and causes the Ag(NH3)2

+ to dissociate and form free Ag+ ions. Once again, the [Ag+] reaches a value sufficient in the presence of Cl– to precipitate AgCl in accordance with Le Châtelier’s Principle.

17.14 AgCl and AgBr are both insoluble compounds. However, AgBr is less soluble than AgCl. When solid

AgCl is added to an aqueous solution of MgBr2, some of the AgCl dissociates. The Ag+ ion reacts with the aqueous Br– to form insoluble AgBr. This disrupts the AgCl/Ag+ equilibrium and additional AgCl dissociates. With sufficient stirring, and perhaps a little heat, all of the AgCl will dissolve and AgBr will precipitate.

Chapter 17

405

17.15 PbCl2(s) Pb2+(aq) + 2Cl–(aq) Pb2+(aq) + 3Cl–(aq) PbCl3

–(aq)

( )( )

( )( )

33

form 3 32+ 2

33form 32

mol PbClvolumePbCl

K = = Pb Cl mol Pb mol Cl

volume volume

mol PbClK = volume

mol Pb mol Cl

−−

− + −

−

+ −

⎛ ⎞⎜ ⎟⎡ ⎤ ⎜ ⎟

⎣ ⎦ ⎝ ⎠

⎡ ⎤ ⎡ ⎤ ⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎣ ⎦ ⎣ ⎦ ⎝ ⎠⎝ ⎠

×

Notice that the above expression is the product of a ratio of mole amounts and a volume3 term. The

constant Kform does not change on dilution, but the volume term is changed by dilution. This means that the ratio of moles term in the above expression must change on dilution, in order to hold the product constant. If the volume is doubled, the ratio of moles would have to become smaller by a factor of 8 (= 23) in order for the entire argument to have a constant value, i.e, in order for Kform to remain constant on dilution.

This means that the concentrations of Pb2+ and Cl– must increase as the solution containing the complex ion

is diluted. Eventually the ion product for PbCl2 will exceed the value of Ksp for PbCl2, resulting in its precipitation.

Review Problems 17.16 (a) CaF2(s) Ca2+ + 2F– Ksp = [Ca2+][F–]2 (b) Ag2CO3(s) 2Ag+ + CO3

2– Ksp = [Ag+]2[CO32–]

(c) PbSO4(s) Pb2+ + SO42– Ksp = [Pb2+][SO4

2–] (d) Fe(OH)3(s) Fe3+ + 3OH– Ksp = [Fe3+][OH–]3 (e) PbI2(s) Pb2+ + 2I– Ksp = [Pb2+][I–]2 (f) Cu(OH)2(s) Cu2+ + 2OH– Ksp = [Cu2+][OH–]2 17.17 (a) AgI(s) Ag+ + I– Ksp = [Ag+][I–] (b) Ag3PO4(s) 3Ag+ + PO4

3– Ksp = [Ag+]3[PO43–]

(c) PbCrO4(s) Pb2+ + CrO42– Ksp = [Pb2+][CrO4

2–] (d) Al(OH)3(s) Al3+ + 3OH– Ksp = [Al3+][OH–]3 (e) ZnCO3(s) Zn2+ + CO3

2– Ksp = [Zn2+][CO32–]

(f) Zn(OH)2(s) Zn2+ + 2OH– Ksp = [Zn2+][OH–]2 17.18 BaSO3(s) Ba2+ + SO3

2– Ksp = [Ba2+][SO32–]

Ksp = (0.10)(8.0 × 10–6) = 8.0 × 10–7 In this problem, all of the Ba2+ comes from the BaCl2. 17.19 To solve this problem we need to realize that the concentration of the solution is equal to the number of

moles of solid recovered divided by the volume of the solution, i.e.,

[ ] 4 44

4

0.649 g CaCrO 1 mole CaCrO 1000 mLCaCrO = = 0.0267 M156 mL 156.1 g CaCrO 1 L

⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠⎝ ⎠

The equilibrium for this problem is CaCrO4(s) Ca2+ + CrO42–

Ksp = [Ca2+][CrO42–] = (0.0267)2 = 7.13 × 10–4

Chapter 17

406

17.20 PbCl2 Pb2+ + 2Cl– Ksp = [Pb2+][Cl–]2 At equilibrium [Pb2+] = 0.016 M and [Cl–] = 0.032 M so Ksp = (0.016)(0.032)2 = 1.6 × 10–5 17.21 BaF2 Ba2+(aq) + 2F–(aq) Ksp = [Ba2+][F–]-2

First find the concentration of the Ba2+ and F– that was in solution and then find the value for Ksp. Using the amount of BaF2 recovered; determine the number of moles of each ion, then find the concentration of each ion.

mole BaF2 = 0.132 g BaF2 2

2

1 mol BaF175.32 g BaF⎛ ⎞⎜ ⎟⎝ ⎠

= 7.53 × 10–4 mol BaF2

[Ba2+] = 4 2+

2

2

7.53 10 mol BaF 1 mol Ba 1000 mL100 mL solution 1 mol BaF 1 L

−⎛ ⎞⎛ ⎞⎛ ⎞×⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠

= 7.53 × 10–3 M Ba2+

[F–] = 4

2

2

7.53 10 mol BaF 2 mol F 1000 mL100 mL solution 1 mol BaF 1 L

− −⎛ ⎞⎛ ⎞⎛ ⎞×⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠

= 1.51 × 10–2 M F–

Ksp = [Ba2+][F–] = (7.53 × 10–3)(1.51 × 10–2)2 = 1.7 × 10–6 17.22 Ag3PO4(s) 3Ag+ + PO4

3– Ksp = [Ag+]3[PO43–]

Ksp = [3(1.8 × 10–5)]3[1.8 × 10–5] = 2.8 × 10–18 17.23 Ba3(PO4)2(s) 3Ba2+ + 2PO4

3– Ksp = [Ba2+]3[PO43–]2

Ksp = [3(1.4 × 10–8)]3[2(1.4 × 10–8)]2 = 5.8 × 10–38

17.24 ( ) 44 4

4

1 mole BaSOmol BaSO = 0.00245 g BaSO

233.3906 g BaSO⎛ ⎞⎜ ⎟⎝ ⎠

= 1.05 × 10–5 mol

[Ba2+] = [SO42–] = 1.05 × 10–5 M

Ksp = [Ba2+][SO4

2–] = (1.05 × 10–5)2 = 1.10 × 10–10

17.25 ( ) 32 3 22 3 2 2 3 2 2 3 2

2 3 2

1 mol AgC H Omol AgC H O = 0.800 g AgC H O = 4.79 10 mol AgC H O

166.9 g AgC H O−⎛ ⎞

×⎜ ⎟⎝ ⎠

[AgC2H3O2] = 3

2 3 24.79 10 mol AgC H O0.100 L solution

−× = 4.79 × 10–2 M

One mole of Ag+ and one mole of C2H3O2 will be produced for every mole of AgC2H3O2. Therefore, Ksp = [Ag+][C2H3O2

–] = (4.79 × 10–2)2 = 2.29 × 10–3. 17.26 For every mole of CO3

2– produced, 2 moles of Ag+ will be produced. Let x = [CO32–] at equilibrium and

[Ag+] = 2x at equilibrium. Ksp = [Ag+]2[CO32–] = 8.1 × 10–12 = (2x)2(x) = 4x3. Solving we find x = 1.3 ×

10–4. Thus, the molar solubility of Ag2CO3 is 1.3 × 10–4 moles/L. 17.27 For every mole of Pb2+ produced, 2 moles of F– will be produced.

Let x = [Pb2+] at equilibrium and [F–] = 2x at equilibrium. Ksp

= [ Pb2+][F–]2 = 3.6 × 10–8 = (x)(2x)2 = 4x3. Solving we find x = 2.1 × 10–3 M. Thus, the molar solubility of PbF2 is 2.1 × 10–3 moles/L.

Chapter 17

407

17.28 PbBr2(s) Pb2+ + 2Br– Ksp = [Pb2+][Br–]2

[Pb2+] [Br–] I – – C + x + 2x E + x + 2x

Ksp = (x)(2x)2 = 4x3 = 2.1 × 10–6, 6

33 2.1 10x 8.1 104

−−×

= = × M

17.29 Ag2CrO4(s) 2Ag+ + CrO4

2– Ksp = [Ag+]2[CrO42–]

[Ag+] [CrO4

2–] I – – C + 2x + x

E + 2x + x

Ksp = (2x)2(x) = 4x3 = 1.2 × 10–12, 12

53 1.2 10x 6.7 104

−−×

= = × M

17.30 First determine the molar solubility of the MX salt. Let x = [M+] = [X–], Ksp = [M+][X–] = (x)(x) = 3.2 × 10–10 x = 1.8 × 10–5 M. This is the equilibrium concentration of the two ions. For the MX3 salt, let x = equilibrium concentration of M3+, [X–] = 3x. Ksp = [M+][X–]3 = (x)(3x)3 = 27x4. The value of x in this expression is the value determined in the first part

of this problem. So, Ksp = (27)(1.8 × 10–5)4 = 2.8 × 10–18 17.31 First determine the molar solubility of the M2X3 salt. M2X3 (s) 2M3+ (aq) + 3X2–(aq) Ksp = [M3+]2[X2–]3

[M3+] [X2–] I – – C + 2x + 3x E + 2x + 3x

Ksp = (2x)2(3x)3= 2.2 × 10–20 = 108x5 x = 4.6 × 10–5 M The molar solubility of this compound is 4.6 × 10–5 moles/L We want the molar solubility of the M2X compound to be twice the value just calculated or 9.2 × 10–5

moles/L. We need to solve the equilibrium expression: M2X (s) 2M+ (aq) + X2–(aq) Ksp = [M+]2[X2–]

[M+] [X2–] I – – C + 2x + x E + 2x + x

Ksp = (2x)2(x) = 4x3 where x = 9.2 × 10–5 M So, Ksp = 4(9.2 × 10–5)3 = 3.1 × 10–12

Chapter 17

408

17.32 To solve this problem, determine the molar solubility for each compound. LiF: let x = [Li+] = [F–] Ksp = [Li+][F–] = x2 = 1.7 × 10–3 x = 4.1 × 10–2 moles/L = molar solubility of LiF. BaF2: let x = [Ba2+]; [F–]2 = 2x Ksp = [Ba2+][F–]2 = (x)(2x)2 = 1.7 × 10–6 4x3 = 1.7 × 10–6, and x = 7.5 × 10–3 M = molar solubility of BaF2. Because the molar solubility of LiF is greater than the molar solubility of BaF2, LiF is more soluble. 17.33 To solve this problem, first determine the molar solubility for each compound. AgCN: let x = [Ag+] = [CN–] Ksp = [Ag+][CN–] = x2 = 2.2 × 10–16 x = 1.5 × 10–8 moles/L = molar solubility of AgCN. The number of grams of AgCN that will dissolve in

100 mL is;

8

71.5 10 moles 133.9 gg (100 mL) 2.0 10 g1000 mL 1 mole

−−⎛ ⎞× ⎛ ⎞= = ×⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

Zn(CN)2: let x = [Zn2+], [CN–] = 2x; Ksp = [Zn2+][CN–]2 = (x)(2x)2 = 3 × 10–16 4x3 = 3 × 10–16, and x = 4.2 × 10–6 M = molar solubility of Zn(CN)2. The number of grams of Zn(CN)2

that will dissolve in 100 mL is;

6

54.2 10 moles 117.4 gg (100 mL) 4.9 10 g1000 mL 1 mole

−−⎛ ⎞× ⎛ ⎞= = ×⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

More Zn(CN)2 will dissolve in 100 mL of water so it has the larger solubility. 17.34 CaSO4(s) Ca2+(aq) + SO4

2–(aq) Ksp = [Ca2+][SO42–] = 2.4 × 10–5

let x = [Ca2+] = [SO42–] Ksp = x2 = 2.4 × 10–5 and x = 4.9 × 10–3 M.

The molar solubility of CaSO4 is 4.9 × 10–3 moles/L. 17.35 CaCO3(s) Ca2+ + CO3

2– Ksp = [Ca2+][CO32–] = 4.5 × 10–9

let x = [Ca2+] = [CO32–]

x2 = 4.5 × 10–9; x = 6.7 × 10–5 M The solubility of CaCO3 is 6.7 × 10–5 M. So, 6.7 × 10–5 moles of CaCO3 dissolves in 1 L of H2O. If 100

mL of water are available, 6.7 × 10–6 moles of CaCO3 will dissolve. Converting this to #g/100 mL we multiply by the molar mass of 100 g/mole and determine that 6.7 × 10–4 g will dissolve in 100 mL of water.

17.36 Mg(OH)2 Mg2+ + 2OH– Ksp = [Mg2+][OH–]2 = 7.1 × 10–12

The concentration of OH– is determined from the pH: pOH = 14 – 12.50 = 1.50 [OH–] = 0.032 M [Mg2+] = x [OH–] = 0.032 M Ksp = x(0.032)2 = 7.1 × 10–12 x = 6.9 × 10–9 M The molar solubility of Mg(OH)2 is 6.9 × 10–9 M in a solution with a pH of 12.50.

Chapter 17

409

17.37 Al(OH)3(s) Al3+ + 3OH– Ksp = [Al3+][OH–]3 = 3 × 10–34 The concentration of OH– is determined from the pH: pOH = 14 – 9.50 = 4.50 [OH–] = 3.2 × 10–5 M [Al3+] = x [OH–] = 3.2 × 10–5 Ksp = x(3.2 × 10–5)3 = 3 × 10–34 x = 9.2 × 10–21 The molar solubility of Al(OH)3 is 9.2 × 10–21 M in a solution with a pH of 9.50.

17.38 Ag2CrO4 (s) 2Ag+ (aq) + CrO4

2–(aq) Ksp = [Ag+]2[CrO42–] = 1.2 × 10–12

(a) [Ag+] [CrO4

2–] I 0.200 – C 0.200 + 2x + x E 0.200 + 2x + x

Ksp = (0.200+2x)2(x) Assume that x << 0.200 1.2 × 10–12 = (0.200)2(x) x = 3.0 × 10–11 The molar solubility is 3.0 × 10–11 moles/L (b)

[Ag+] [CrO42–]

I – 0.200 C +2x 0.200 + x E +2x 0.200 + x

Ksp = (2x)2(0.200+x) Assume that x << 0.200 1.2 × 10–12 = (2x)2(0.200) x = 1.2 × 10–6 The molar solubility is 1.2 × 10–6 moles/L. 17.39 MgOH2 (s) Mg2+(aq) + 2OH–(aq) Ksp = [Mg2+][OH–]2 = 7.1 × 10–12

(a) [Mg2+] [OH–] I – 0.20 C + x 0.20 + 2x E + x 0.20 + 2x

Ksp = (x)(0.20 + 2x)2 Assume 2x << 0.20 Ksp = (x)(0.20)2 = 7.1 × 10–12 x = 1.8 × 10–10 M The assumption is valid and the molar solubility of Mg(OH)2 in 0.20 M NaOH is 1.8 × 10–10 moles/L

(b)

[Mg2+] [OH–] I 0.20 – C 0.20 + x + 2x E 0.20 + x + 2x

Ksp = (0.20 + x)(2x)2 Assume x << 0.20 Ksp = (0.20)(2x)2 = 7.1 × 10–12 x = 3.0 × 10–6 M The assumption is valid and the molar solubility of Mg(OH)2 in 0.20 M MgSO4 is 3.0 × 10–6 moles/L

Chapter 17

410

17.40 Fe(OH)2(s) Fe2+(aq) + 2 OH–(aq) 22

spK = Fe OH+ −⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

mol OH– = 2.20 g NaOH(1 mol/40.01 g NaOH) = 0.0550 mol NaOH [OH–] = mol OH–/L solution = 0.0550 mol/0.250 L = 0.22 M

[Fe2+] [OH–] I – 0.22 C + x + 2x E x 0.22 + 2x

We assume that x << 0.22, so that 0.22 + 2x ≈ 0.22, then we enter the equilibrium values of the above table

into the Ksp expression:

22

spK = Fe OH+ −⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

7.9 × 10–16 = x(0.22)2 x = molar solubility = 1.6 × 10–14 M Next, we must determine how many moles of Fe(OH)2 are formed in the reaction. This is a limiting reactant problem. The number of moles of OH– is 0.0550 (see above). The number of moles of Fe2+ is (0.250 L)(0.10 mol/L) = 0.025 mol From the balanced equation at the top, we need two OH– for every one Fe2+. This would be 2(0.025 mol) = 0.050 mol OH–. Looking at the molar quantities above, we have more than enough OH– so, Fe2+ is our limiting reactant: 0.025 mol Fe(OH)2 will form in 0.25 L solution. If dissolved, this would be a concentration of 0.025 mol/0.25 L = 0.10 M. But from above, the maximum molar solubility of is 1.6 × 10–14 M. This means that remainder of Fe(OH)2 in excess of this value precipitates: 0.10 – 1.6 × 10–14 ≈ 0.10 M. This works out to 0.25 L(0.10 mol/L) = 0.025 mol Fe(OH)2(89.8 g/mol) = 2.2 g solid Fe(OH)2 (essentially all of it). The remaining OH–, 0.005 mol, gives a concentration of OH– of 0.005 mol OH

0.250 L

− = 0.02 M OH–

7.9 × 10–16 = [Fe2+][0.02]2 [Fe2+] = 2.0 × 10–12 M

17.41 Ni(OH)2(s) Ni2+(aq) + 2OH–(aq) mol OH– = 1.75 g NaOH(1 mol/40.01 g NaOH) = 0.0437 mol NaOH First, assume that all of the ions are in solution and no precipitate has formed. At that point the

concentrations are:

[OH–] = 0.0437 mol OH0.250 L

− = 0.175 M OH–

[Ni2+] = 0.10 M

Chapter 17

411

Next, determine the limiting reactant: 2+

2+0.10 mol Ni 2 mol OH

1 L soln 1 mol Ni

−⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

= 0.20 M OH–

Only 0.175 M OH– is available, therefore OH– is the limiting reagent. Using the limiting reagent, assume that all of the OH– has precipitated

[OH–] = 0.0 M

[Ni2+] = 0.10 M Ni2+ –2+0.175 mol OH 1 mol Ni

1 L soln 2 mol OH

−

−

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

= 0.10 M Ni2+ – 0.088 M Ni2+ = 0.012 M Ni2+ From here, calculate the equilibrium concentrations:

[Ni2+] [OH–] I 0.012 – C + x + 2x E 0.012 + x 2x

We assume that x << 0.012, so that 0.012 + 2x ≈ 0.012, then we enter the equilibrium values of the above table into the Ksp expression:

22

spK = Ni OH+ −⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

6 × 10–16 = 0.012(x)2 x = 2.2 × 10–7 M The amount of Ni(OH)2 formed can be calculated from the amount of OH– added to the solution since essentially all of the OH– was precipitated:

g Ni(OH)2 = 0.0437 mol OH– 2 2

2

1 mol Ni(OH) 92.71 g Ni(OH)1 mol Ni(OH)2 mol OH−

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠ = 2.03 g Ni(OH)2

To determine the pH of the final solution, the [OH–] needs to be found: From the ICE table, the equilibrium concentration of OH– = 2x

x = 2.2 × 10–7 M 2x = 4.4 × 10–7 M = [OH–] We need to add in the concentration of OH–,1 × 10–7 M, from the dissociation of water: [OH–] = 4.4 × 10–7 M + 1 × 10–7 = 5.4 × 10–7 M pOH = –log[OH–] = –log[5.4 × 10–7] = 6.27 pH = 14 – pOH = 14 – 6.27 = 7.73

17.42 (a) CuCl(s) Cu+(aq) + Cl–(aq) Ksp = [Cu+][Cl–] = 1.9 × 10–7

[Cu+] [Cl–] I – – C + x + x E + x + x

Ksp = x2 = 1.9 × 10–7 ∴ x = molar solubility = 4.4 × 10–4 M (b) CuCl(s) Cu+(aq) + Cl–(aq) Ksp = [Cu+][Cl–] = 1.9 × 10–7

[Cu+] [Cl–] I – 0.0200 C + x + x E + x 0.0200 + x

Ksp = (x)(0.0200+x) = 1.9 × 10–7 Assume that x << 0.0200 ∴ x = molar solubility = 9.5 × 10–6 M

Chapter 17

412

(c) CuCl(s) Cu+(aq) + Cl–(aq) Ksp = [Cu+][Cl–] = 1.9 × 10–7

[Cu+] [Cl–] I – 0.200 C + x + x E + x 0.200 + x

Ksp = (x)(0.200+x) = 1.9 × 10–7 Assume that x << 0.200 ∴ x = molar solubility = 9.5 × 10–7 M (d) CuCl(s) Cu+(aq) + Cl–(aq) Ksp = [Cu+][Cl–] = 1.9 × 10–7 Note that the Cl– concentration equals (2)(0.150 M) since two moles of Cl– are produced for every mole of

CaCl2.

[Cu+] [Cl–] I – 0.300 C + x + x E + x 0.300 + x

Ksp = (x)(0.300+x) = 1.9 × 10–7 Assume that x << 0.300 ∴ x = molar solubility = 6.3 × 10–7 M 17.43 AuCl3(s) Au3+ + 3Cl– Ksp = [Au3+][Cl–]3 = 3.2 × 10–25 (a) let x = [Au3+]; then [Cl–] = 3x Ksp = (x)(3x)3 = 27x4

x = 25

4 3.2 1027

−× = 3.3 × 10–7 M

The molar solubility of AuCl3 is 3.3 × 10–7 M in H2O. (b) [Au3+] = x; [Cl–] = 0.010 + 3x Ksp = (x)(0.010 + 3x)3: Assume 3x << 0.010 Ksp = (x)(0.010)3 x = 3.2 × 10–19 M The molar solubility of AuCl3 is 3.2 × 10–19 M in 0.010 M HCl. (c) [Au3+] = x; [Cl–] = 0.020 + 3x Ksp = (x)(0.020 + 3x)3: Assume 3x << 0.020 Ksp = (x)(0.020)3 x = 4.0 × 10–20 M The molar solubility of AuCl3 is 4.0 × 10–20 M in 0.010 M MgCl2. (d) [Au3+] = 0.010 + x; [Cl–] = 3x Ksp = (0.010 + x)(3x)3: Assume x << 0.010 Ksp = (0.010)(3x)3

x = 25

3 3.2 100.27

−× = 1.1 × 10–8 M

The molar solubility of AuCl3 is 1.1 × 10–8 M in 0.010 M Au(NO3)3.

Chapter 17

413

17.44 Fe(OH)2(s) Fe2+(aq) + 2 OH–(aq) 22

spK = Fe OH+ −⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ = 7.9 × 10–16

pH = 9.50 pOH = 14.00 – pH = 4.50 [OH–] = 10–4.50 = 3.16 × 10–5 M

[Fe2+] [OH–] I – 3.16 × 10–5 C + x + 2x E x (3.16 × 10–5) + 2x

Since Ksp for iron(II) hydroxide is so small, we can safely assume that 2x << 3.16 × 10–5, so that (3.16 × 10–5) + 2x ≈ 3.16 × 10–5, then we enter the equilibrium values of the

above table into the Ksp expression:

22

spK = Fe OH+ −⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

7.9 × 10–16 = x(3.16 × 10–5)2 x = molar solubility = 7.9 × 10–7 M

17.45 (a) Ca(OH)2(s) Ca2+(aq) + 2OH–(aq) Ksp = [Ca2+][OH–]2 = 6.5 × 10–5

let 2x = [OH–], [Ca2+] = x + 0.10 Ksp = (0.10 + x)(2x)2 = 6.5 × 10–5 By successive approximations, x = 0.012

The molar solubility of Ca(OH)2 in 0.10 M CaCl2 is 0.012 moles/L. (b) Ca(OH)2(s) Ca2+(aq) + 2OH–(aq) Ksp = [Ca2+][OH–]2 = 6.5 × 10–5 let x = [Ca2+], [OH–] = 2x + 0.10 Ksp = (x)(2x + 0.10)2 = 6.5 × 10–5

By successive approximations, x = 0.0053

The molar solubility of Ca(OH)2 in 0.010 M NaOH is 0.053 moles/L. 17.46 PbCl2(s) Pb2+ + 2Cl– Ksp = [Pb2+][Cl–]2 = 1.7 × 10–5

[Cl–] = 0.10 M [Pb2+][Cl–]2 =[Pb2+][0.10]2 = 1.7 × 10–5 [Pb2+] = 1.7 × 10–3 M

17.47 PbBr2 will be less soluble in a solution of 0.10 M NaBr because the values of Q and Ksp for lead bromide

are dependent upon the [Br–]2. 17.48 HC2H3O2 + H2O H3O+ + C2H3O2

– +

53 2 3 2a

2 3 2

[H O ][C H O ]K = = 1.8 10

HC H O

−−×

⎡ ⎤⎣ ⎦

First, we calculate the % ionization in water (no added sodium acetate):

[HC2H3O2] [H3O+] [C2H3O2–]

I 0.10 – – C 0.10–x +x +x E 0.10 –x x +x

Assume x << 0.10

Chapter 17

414

[ ]

5 3a 3

[x][x]K = = 1.8 10 x 1.3 10 [H O ]0.10

− +× = × =

% ionization = ([H+]/[total acid]) × 100 = (1.3 × 10–3/0.10) × 100 = 1.3 %

Now we calculate the % ionization using 0.050 mol/0.500L = 0.10 M for the concentration of sodium acetate:

[HC2H3O2] [H3O+] [C2H3O2–]

I 0.10 – 0.10 C 0.10–x +x +x E 0.10 –x x 0.10 + x

Assume x << 0.10

[ ]

5 -5a 3

[x][0.10]K = = 1.8 10 x 1.8 10 [H O ]0.10

− +× = × =


So the % ionization decreased by (1.3 – 0.018) = 1.3 %, using correct significant figures. (This does not mean it has no dissociation, but that the dissociation is very small compared to 1.3%.) Using the [H+] values above, the pH initially was: –log [H+] = –log [1.3 × 10–3] = 2.89 After addition of the sodium acetate it was: –log [H+] = –log [1.8 × 10–5] = 4.74 The pH changes by (4.7 – 2.9) = +1.85 pH units

17.49 HC2H3O2 + H2O H3O+ + C2H3O2

– +

53 2 3 2a

2 3 2

[H O ][C H O ]K = = 1.8 10

HC H O

−−×

⎡ ⎤⎣ ⎦

First, we calculate the % ionization in water (no added HCl):

[HC2H3O2] [H3O+] [C2H3O2–]

I 0.10 – – C 0.10 – x + x + x E 0.10 – x x + x

Assume x << 0.10

[ ]

5 3a 3

[x][x]K = = 1.8 10 x 1.3 10 [H O ]0.10

− − +× = × =


Now we calculate the % ionization using 0.025 mol/0.500 L = 0.050 M for the concentration of H3O+:

[HC2H3O2] [H3O+] [C2H3O2–]

I 0.10 0.050 – C 0.10 – x + x + x E 0.10 – x 0.050 + x x

Assume x << 0.050

Chapter 17

415

[ ]

5 5a 2 3 2

[x][0.050]K 1.8 10 x 3.6 10 [C H O ]0.10

− − −= = × = × =

% ionization = ([C2H3O2

–]/[total acid]) × 100 = (3.6 × 10–5/0.10) × 100 = 0.036 %

So the % ionization decreased by (1.3 – 0.036) = 1.3%, using correct significant figures. (This does not mean it has no dissociation, but that the dissociation is very small compared to 1.3%.) Using the [H+] values above, the pH initially was: –log [H+] = –log [1.3 × 10–3] = 2.89 After addition of the hydrogen chloride it was: –log [H+] = –log [0.050] = 1.30 The pH changes by (1.30 – 2.89) = –1.59 pH units

17.50 AgCl(s) Ag+ + Cl– +

spK = Ag Cl−⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ = 1.8 × 10–10

AgI(s) Ag+ + I– +spK = Ag I−⎡ ⎤ ⎡ ⎤

⎣ ⎦ ⎣ ⎦ = 8.3 × 10–17

When AgNO3 is added to the solution, AgI will precipitate before any AgCl does due to the lower

solubility of AgI. In order to answer the question, i.e., what is the [I–] when AgCl first precipitates, we need to find the minimum concentration of Ag+ that must be added to precipitate AgCl.

Let x = [Ag+]; Ksp = (x)(0.050) = 1.8 × 10–10; x = 3.6 × 10–9 M When the AgCl starts to precipitate, the solution will have a [Ag+] of 3.6 × 10–9 M. Now we ask, what is

the [I–] if [Ag+] = 3.6 × 10–9 M? So, +

spK = Ag I−⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ = (3.6 × 10–9)(x) = 8.3 × 10–17; x = 2.3 × 10–8 M = [I–]

17.51 This problem is similar to Problem 17.50, except that the Ksp constants are closer in value. We first determine

the minimum amount of SO42– that must be added to initiate the precipitation of CaSO4. CaSO4 will precipitate

after SrSO4 due to its larger value for Ksp: Ksp(CaSO4) = 2.4 × 10–5 and Ksp(SrSO4) = 3.2 × 10–7 (see Table 17.1) (a) Let x = 2+Ca⎡ ⎤

⎣ ⎦ ; 2+ 2sp 4K = Ca SO −⎡ ⎤ ⎡ ⎤

⎣ ⎦ ⎣ ⎦ = (0.15)(x) = 2.4 × 10–5

x = 24SO −⎡ ⎤

⎣ ⎦ = 1.6 × 10–4 M

When the 24SO −⎡ ⎤

⎣ ⎦ = 1.6 × 10–4 the CaSO4 will start to precipitate. Now we ask, what is the

2+Sr⎡ ⎤⎣ ⎦ if 2

4SO −⎡ ⎤⎣ ⎦ = 1.6 × 10–4 M?

SrSO4 (s) Sr2+ (aq) + SO42–(aq) 2+ 2

sp 4K = Sr SO −⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

[Sr2+] [SO4

2–] I – 1.6 × 10–4 C + x 1.6 × 10–4 + x E + x 1.6 × 10–4 + x

2+ 2

sp 4K = Sr SO −⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ = (x)(1.6 × 10–4 + x) = 3.2 × 10–7

For this problem, we must solve the quadratic equation and we determine that x = 4.9 × 10–4 M. Thus, the [Sr2+] = 4.9 × 10–4 M when the CaSO4 starts to precipitate.

Chapter 17

416

(b) Initially the solution had a concentration of 0.15 M. The solution now has a [Sr2+] = 4.9 × 10–4. So, the percentage of Sr2+ precipitated is;

-40.15 4.9 10 100% = 99.7 %0.15

− ××

17.52 In order for a precipitate to form, the value of the reaction quotient, Q, must be greater than the value of

Ksp. For PbCl2, Ksp = 1.7 × 10–5 (see Table 17.1).

22+Q = Pb Cl−⎡ ⎤ ⎡ ⎤

⎣ ⎦ ⎣ ⎦ = (0.0150)(0.0120)2 = 2.16 × 10–6. Since Q < Ksp, no precipitate will form.

17.53 In order for a precipitate to form, the value of the reaction quotient, Q, must be greater than the value of

Ksp. For AgC2H3O2, Ksp = 2.3 × 10–3. +

2 3 2Q = Ag C H O −⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ = (0.015)(0.50) = 7.5 × 10–3. Since Q > Ksp, a precipitate will form. (Note: The

concentration of C2H3O2– is twice the concentration of Ca(C2H3O2)2 since one mole of Ca(C2H3O2)2

produces two moles of C2H3O2–).

17.54 The precipitate that may form is PbBr2(s). To determine if a precipitate will form, a value for the reaction

quotient, Q, must be calculated: Q = [Pb2+][Br–]2. In performing this calculation, the dilution of the ions must be considered:

[Pb2+] = [Br–] = 0.00500 M. Q = [0.00500][0.00500]2 = 1.3 × 10–7

If Q > Ksp, a precipitate will form. Ksp for PbBr2(s) is 2.1 × 10–6. Therefore, a precipitate will not form. Hence, the concentrations calculated are the diluted concentrations. Since no precipitate forms, the concentrations are not equilibrium values.

17.55 Ksp = [Mn2+][OH–]2 = 1.6 × 10–13

Ksp = [Cd2+][OH–]2 = 5.0 × 10–15 Mn(OH)2 is more soluble, so we need to determine the hydroxide ion concentration when it begins to precipitate. Ksp = [Mn2+][OH–]2 = 1.6 × 10–13 = (0.10)(x)2 x = 1.3 × 10–6 M OH–

Use this value for the [OH–] to solve for the concentration of the [Cd2+] left in solution: Ksp = [Cd2+][OH–]2 = 5.0 × 10–15 = (x)[1.3 × 10–6]2 x = 3.1 × 10–3 M Cd2+

17.56 To solve this problem, determine the value for Q and apply LeChâtelier’s Principle. (a) 2+Pb⎡ ⎤

⎣ ⎦ = (50.0 mL)(0.0100 moles/L)/(100.0 mL) = 5.00 × 10–3

Br−⎡ ⎤⎣ ⎦ = (50.0 mL)(0.0100 moles/L)/(100.0 mL) = 5.00 × 10–3

22+Q = Pb Br−⎡ ⎤ ⎡ ⎤

⎣ ⎦ ⎣ ⎦ = (5.00 × 10–3)(5.00 × 10–3)2 = 1.25 × 10–7

For PbBr2, Ksp = 2.1 × 10–6 Since Q < Ksp, no precipitate will form. (b) 2+Pb⎡ ⎤

⎣ ⎦ = (50.0 mL)(0.0100 moles/L)/(100.0 mL) = 5.00 × 10–3

Br−⎡ ⎤⎣ ⎦ = (50.0 mL)(0.100 moles/L)/(100.0 mL) = 5.00 × 10–2

22+Q = Pb Br−⎡ ⎤ ⎡ ⎤

⎣ ⎦ ⎣ ⎦ = (5.00 × 10–3)(5.00 × 10–2)2 = 1.25 × 10–5

For PbBr2, Ksp = 2.1 × 10–6 Since Q > Ksp, a precipitate will form.

Chapter 17

417

17.57 In order for a precipitate to form, the value of the reaction quotient, Q, must be greater than the value of Ksp. For AgC2H3O2, Ksp = 2.3 × 10–3.

+Ag⎡ ⎤⎣ ⎦ = (22.0 mL)(0.100 M)/(67.0 mL) = 3.28 × 10–2 M

2 3 2C H O −⎡ ⎤⎣ ⎦ = (45.0 mL)(0.0260 M)/(67.0 mL) = 1.75 × 10–2 M

+2 3 2Q = Ag C H O −⎡ ⎤ ⎡ ⎤

⎣ ⎦ ⎣ ⎦ = (3.28 × 10–2)(1.75 × 10–2) = 5.74 × 10–4.

Since Q < Ksp, no precipitate will form. 17.58 Cu(OH)2(s) Cu2+(aq) + 2 OH–(aq)

22spK = Cu OH+ −⎡ ⎤ ⎡ ⎤

⎣ ⎦ ⎣ ⎦

[ ]2-204.8 10 = 0.10 OH−⎡ ⎤× ⎣ ⎦

[OH–] = 6.9 × 10–10 M pOH = –log[OH–] = –log[6.9 × 10–10] = 9.2 pH = 14.00 –pOH = 4.8

4.8 is the pH below which all the Cu(OH)2 will be soluble.

Mn(OH)2(s) Mn2+(aq) + 2 OH–(aq) 22

spK = Mn OH+ −⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

[ ]2-131.6 10 = 0.10 OH−⎡ ⎤× ⎣ ⎦

[OH−] = 1.3 × 10–6 M pOH = –log[OH−] = –log[1.3 × 10–6] = 5.9 pH = 14.00 – pOH = 8.1

8.1 is the pH below which all the Mn(OH)2 will be soluble.

Therefore, from pH = 4.8–8.1 Mn(OH)2 will be soluble, but some Cu(OH)2 will precipitate out of solution. 17.59 The following reactions are possible:

CaC2O4(s) Ca2+(aq) + C2O42–(aq) Ksp = [Ca2+][C2O4

2–] = 2.3 × 10–9 MgC2O4(s) Mg2+(aq) + C2O4

–2(aq) Ksp = [Mg2+][C2O42–] = 8.6 × 10–5

H2C2O4(aq) H+(aq) + HC2O4–(aq)

[ ]1

–2 4

a2 2 4

H HC OK

H C O

+⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦= = 5.6 × 10–2

HC2O4–(aq) H+(aq) + C2O4

2–(aq) 2

2–2 4

a –2 4

H C OK

HC O

+⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦=⎡ ⎤⎣ ⎦

= 5.4 × 10–5

H2C2O4(aq) 2H+(aq) + C2O42–(aq)

[ ]

2 2–2 4

O2 2 4

H C OK

H C O

+⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦= = 3.0 × 10–6

Assume that the concentration of H2C2O4 at the end of the process is 0.10 M First, calculate the concentration of C2O4

2–, at which the MgC2O4 will precipitate Ksp = [Mg2+][C2O4

2–] = 8.6 × 10–5 [Mg2+] = 0.10 M (0.10)(x) = 8.6 × 10–5 x = 8.6 × 10–4 M = [C2O4

2–]

Chapter 17

418

As long as the concentration of C2O42– is kept below 8.6 × 10–4 M, the Mg2+ will remain in solution

In order for the concentration of oxalate to remain below = 8.6 × 10–4 M, the pH will be:

H2C2O4(aq) 2H+(aq) + C2O42–(aq)

[ ]

2 2–2 4

O2 2 4

H C OK

H C O

+⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦= = 3.0 × 10–6

[ ]

2 –4

O

H 8.6 10K

0.10

+⎡ ⎤ ⎡ ⎤×⎣ ⎦ ⎣ ⎦= = 3.0 × 10–6

[H+] = 0.0188 M pH = 1.73

By keeping the pH < 1.73, the MgC2O4 will not precipitate. To precipitate the CaC2O4, a similar calculation needs to be done:

Ksp = [Ca2+][C2O42–] = 2.3 × 10–9

[Ca2+] = 0.10 M (0.10)(x) = 2.3 × 10–9 x = 2.3 × 10–8 M = [C2O4

2–] As long as the concentration of C2O4

2– is kept above 2.3 × 10–8 M, the Ca2+ will precipitate In order for the concentration of oxalate to remain above = 2.3 × 10–8 M, the pH must be:

H2C2O4(aq) 2H+(aq) + C2O42–(aq)

[ ]2

2 2–2 4

a2 2 4

H C OK

H C O

+⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦= = 3.0 × 10–6

[ ]

2 –8H 2.3 10K

0.10

+⎡ ⎤ ⎡ ⎤×⎣ ⎦ ⎣ ⎦= = 3.0 × 10–6

[H+] = 13.0 M pH = –1.12

17.60 The less soluble substance is PbS. We need to determine the minimum [H+] at which CoS will precipitate.

[ ]2+

2spa 2 + 2+

Co H S (0.010)(0.1)K = = = 0.5 (from Table 17.2)[H ]H

⎡ ⎤⎣ ⎦

⎡ ⎤⎣ ⎦

+ (0.010)(0.1)[H ] = = 0.0450.5

pH = –log[H+] = 1.35. At a pH lower than 1.35, PbS will precipitate and CoS will not. At larger values of pH, both PbS and CoS will precipitate.

17.61 The less soluble substance is SnS so we will determine the maximum amount of H+ that is permitted before

MnS starts to precipitate.

[ ]2+

2 7spa 2 + 2+

Mn H S (0.010)(0.1)K = = = 3 10 (from Table 17.2)[H ]H

⎡ ⎤⎣ ⎦ ×

⎡ ⎤⎣ ⎦

+ 67

(0.010)(0.1)[H ] = = 6 103 10

−××

pH = –log[H+] = 5.2. At pH values equal to or less than 5.2, MnS will not precipitate.

Chapter 17

419

17.62 (a) Cu2+(aq) + 4Cl–(aq) CuCl42–(aq)

24

form 42+

CuClK =

Cu Cl

−

−

⎡ ⎤⎣ ⎦

⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

(b) Ag+(aq) + 2I–(aq) AgI2–(aq)

2form 2+

AgIK =

Ag I

−

−

⎡ ⎤⎣ ⎦

⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

(c) Cr3+(aq) + 6NH3(aq) Cr(NH3)63+(aq)

33 6

form 63+3

Cr(NH )K =

Cr NH

+⎡ ⎤⎣ ⎦⎡ ⎤⎡ ⎤⎣ ⎦⎣ ⎦

17.63 (a) Ag+(aq) + 2S2O32–(aq) Ag(S2O3)2

3–(aq) 3

2 3 2form 2+ 2

2 3

Ag(S O )K =

Ag S O

−

−

⎡ ⎤⎣ ⎦

⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

(b) Zn2+(aq) + 4NH3(aq) Zn(NH3)42+(aq)

23 4

form 42+3

Zn(NH )K =

Zn NH

+⎡ ⎤⎣ ⎦⎡ ⎤⎡ ⎤⎣ ⎦⎣ ⎦

(c) Sn4+(aq) + 3S2–(aq) SnS32–(aq)

23

form 34+ 2

SnSK =

Sn S

−

−

⎡ ⎤⎣ ⎦

⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

17.64 (a) Co3+(aq) + 6NH3(aq) Co(NH3)63+(aq)

33 6

form 63+3

Co(NH )K =

Co NH

+⎡ ⎤⎣ ⎦⎡ ⎤⎡ ⎤⎣ ⎦⎣ ⎦

(b) Hg2+(aq) + 4I–(aq) HgI42–(aq)

24

form 42+

HgIK =

Hg I

−

−

⎡ ⎤⎣ ⎦

⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

(c) Fe2+(aq) + 6CN–(aq) Fe(CN)64–(aq)

46

form 62+

Fe(CN)K =

Fe CN

−

−

⎡ ⎤⎣ ⎦

⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

17.65 (a) Hg2+(aq) + 4NH3(aq) Hg(NH3)42+(aq)

23 4

form 42+3

Hg(NH )K =

Hg NH

+⎡ ⎤⎣ ⎦⎡ ⎤⎡ ⎤⎣ ⎦⎣ ⎦

(b) Sn4+(aq) + 6F–(aq) SnF62–(aq)

26

form 64+

SnFK =

Sn F

−

−

⎡ ⎤⎣ ⎦

⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

Chapter 17

420

(c) Fe3+(aq) + 6CN–(aq) Fe(CN)63–(aq)

36

form 63+

Fe(CN)K =

Fe CN

−

−

⎡ ⎤⎣ ⎦

⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

17.66 There are two events in this net process: one is the formation of a complex ion (an equilibrium which has

an appropriate value for Kform), and the other is the dissolving of Fe(OH)3, which is governed by Ksp for the solid.

Fe(OH)3(s) Fe3+(aq) + 3OH–(aq) 33+ 39

spK = Fe OH = 1.6 10− −⎡ ⎤ ⎡ ⎤ ×⎣ ⎦ ⎣ ⎦

Fe3+(aq) + 6CN–(aq) Fe(CN)63–(aq)

36 31

form 63+

Fe(CN)K = = 1.0 10

Fe CN

−

−

⎡ ⎤⎣ ⎦ ×

⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

The net process is: Fe(OH)3(s) + 6CN–(aq) Fe(CN)6

3–(aq) + 3OH–(aq) The equilibrium constant for this process should be:

336

c 6

Fe(CN) OHK =

CN

− −

−

⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

⎡ ⎤⎣ ⎦

The numerical value for the above Kc is equal to the product of Ksp for Fe(OH)3(s) and Kform for Fe(CN)6

3–, as can be seen by multiplying the mass action expressions for these two equilibria:

Kc = Kform × Ksp = 1.6 × 10–8 Because Kform is so very large, we can assume that all of the dissolved iron ion is present in solution as the

complex, thus: [Fe(CN)63–] = 0.11 mol/1.2 L = 0.092 M. Also the reaction stoichiometry shows that each

iron ion that dissolves gives 3 OH– ions in solution, and we have: [OH–] = 0.092 × 3 = 0.28 M. We substitute these values into the Kc expression and rearrange to get:

[ ]33

66

c

Fe(CN) OHCN =

K

− −⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦−

36

8(0.092)(0.28) =

1.6 10−×

Thus we arrive at the concentration of cyanide ion that is required in order to satisfy the mass action

requirements of the equilibrium: [CN–] = 7.1 mol L–1. Since this concentration of CN– must be present in 1.2 L, the number of moles of cyanide that are required is: 7.1 mol L–1 × 1.2 L = 8.5 mol CN–.

Additionally, a certain amount of cyanide is needed to form the complex ion. The stoichiometry requires

six times as much cyanide ion as iron ion. This is 0.11 moles × 6 = 0.66 mol. This brings the total required cyanide to (8.5 + 0.66) = 9.2 mol.

9.2 mol × 49.0 g/mol = 450 g NaCN are required.

Chapter 17

421

17.67 AgBr(s) Ag+ + Br– Ksp = 5.0 × 10–13 Ag+ + 2S2O3

2– Ag(S2O3)23– Kf = 2.0 × 1013

AgBr(s) + 2S2O32– Ag(S2O3)2

3– + Br– Kc = Ksp*Kf = 10

[S2O3

2–] [Ag(S2O3)23–] [Br–]

I 1.20 – – C – 2x + x + x E 1.20 – 2x + x + x

Note: Since the AgBr(s) has a constant concentration, it may be neglected.

32 3 2

c 2 22 3

2

c 2

[Ag(S O ) ][Br ]K = = 10

[S O ]

xK = = 10(1.20 2x)

− −

−

−

To solve this equation, take the square root of both sides and then solve for x. x = 0.518 M = [Ag(S2O3)2

3–] Since 1 mole of AgBr produces 1 mole of Ag(S2O3)2

3–, we can determine the number of grams of AgBr that will dissolve in 125 mL.

g AgBr = (0.125 L)(0.518 moles/L)(187.77 g/mole) = 12.2 g AgBr 17.68 Kc = Ksp × Kform = (1.7 × 10–5)(2.5 × 101) = 4.3 × 10–4 17.69 Kc = Ksp × Kform = (1.2 × 10–16)(5.3 × 1018) = 6.4 × 102

17.70 The applicable equilibria are as follows:

AgI(s) Ag+(aq) + I–(aq) + 17spK = Ag I = 8.3 10− −⎡ ⎤ ⎡ ⎤ ×⎣ ⎦ ⎣ ⎦

Ag+(aq) + 2CN–(aq) Ag(CN)2–(aq)

( )2 18form 2+

Ag CNK = = 5.3 10

Ag CN

−

−

⎡ ⎤⎢ ⎥⎣ ⎦ ×⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

The two equations above may be combined and Kc found as follows:

AgI(s) + 2CN–(aq) Ag(CN)2–(aq) + I–(aq)

( )2c 2

Ag CN IK = =

CN

− −

−

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎣ ⎦⎣ ⎦

⎡ ⎤⎣ ⎦

Ksp × Kform = 4.4 × 102

We begin with 0.010 M CN–. This is reduced by some amount (2x) as it reacts with the silver ions, and [AgI2

–] is increased by x:

[CN–] [Ag(CN)2–] I–

I 0.010 – – C – 2x + x + x E 0.010 – 2x x x

Chapter 17

422

Now we insert the equilibrium values into the above equation:

( )2c 2

Ag CN IK = =

CN

− −

−

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎣ ⎦⎣ ⎦

⎡ ⎤⎣ ⎦

4.4 × 102

[ ][ ][ ]

c 2x x

K = =0.010 2x−

4.4 × 102

Take the square root of both sides and solve for x: x = 4.9 × 10–3

This value represents the change in concentration of I– which, from the balanced equation, equals the change in concentration of AgI(s). The molar solubility of AgI in 0.010 M KCN is 4.9 × 10–3 M.

17.71 In case (a), the formation constant is relatively small indicating that the complex is not very stable. At the

same time, the extremely small value for Ksp indicates that the ML2 solid is very stable. Consequently, the solution will contain very little M2+.

In case (b), the solubility of ML2 is even smaller than in case (a). However, the large value for the

formation constant indicates that any M2+ ions in solution will react with any ligand present to form the complex ion. As a result, more of the ML2 solid will dissolve increasing the amount of M2+ in solution.

17.72 Recall that Kinst = 1/ Kform.

Zn(OH)2 (s) Zn2+(aq) + 2OH–(aq) 22+ -16

spK = Zn OH = 3.0 10−⎡ ⎤ ⎡ ⎤ ×⎣ ⎦ ⎣ ⎦

Zn2+(aq) + 4NH3 (aq) Zn(NH3)4

2+(aq) 2

3 4form 42+

3

Zn(NH )K = = ?

Zn NH

+⎡ ⎤⎣ ⎦⎡ ⎤⎡ ⎤⎣ ⎦⎣ ⎦

Combined, this is:

Zn(OH)2 (s) + 4NH3 (aq) Zn(NH3)4

2+(aq) + 2OH–(aq) 22 -

3 4c 4

3

Zn(NH ) OHK =

NH

+⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

⎡ ⎤⎣ ⎦

[NH3] [Zn(NH3)4

2+] [OH–] I 1.0 – – C – 4x + x + 2x E 1.0 – 4x x 2x

[ ][ ][ ]

2

c 4x 2x

K = 1.0 - 4x

he problem gives the molar solubility of Zn(OH)2 as 5.7 × 10–3 M. This means in one liter of 1.0 M NH3, x = 5.7 × 10–3 moles. Substituting this value in for x, we get Kc = 8.1 × 10–7.

Kc = Ksp × Kform 8.1 × 10–7 = 3.0 × 10–16 × Kform

Kform = 2.7 × 109 Kinst = 1/Kform Kinst = 1/(2.7 × 109) = 3.7 × 10–10

Chapter 17

423

17.73 Cu(OH)2 (s) Cu2+(aq) + 2OH–(aq) 22+ -20

spK = Cu OH = 4.8 10−⎡ ⎤ ⎡ ⎤ ×⎣ ⎦ ⎣ ⎦

Cu2+(aq) + 4NH3 (aq) Cu(NH3)4

2+(aq) Combined, this is:

Cu(OH)2 (s) + 4NH3 (aq) Cu(NH3)4

2+(aq) + 2OH–(aq)

22 -3 4

c 43

Cu(NH ) OHK =

NH

+⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

⎡ ⎤⎣ ⎦ = 5.3 × 10–7

[NH3] [Cu(NH3)4

2+] [OH–] I 2.0 – – C – 4x + x + 2x E 2.0 – 4x x 2x

[ ][ ][ ]

2

c 4x 2x

K = 2.0 4x−

= 5.3 × 10–7

Solve for x using successive approximations. x = 1.1 × 10–2 [Cu(NH3)4

2+] = 1.2 × 10–2 Since all of the Cu2+ comes from the Cu(OH)2, the molar solubility of Cu(OH)2 is 1.1 × 10–2 M

17.74 The applicable equilibria are as follows:

AgI(s) Ag+(aq) + I–(aq) + -17spK = Ag I = 8.3 10−⎡ ⎤ ⎡ ⎤ ×⎣ ⎦ ⎣ ⎦

Ag+(aq) + 2I–(aq) AgI2–(aq)

2 11form 2+

AgIK = = 1 10

Ag I

−

−

⎡ ⎤⎣ ⎦ ×

⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

When a solution of AgI2– is diluted, all of the concentrations of the species in Kform above decrease.

However, the decrease of [I–] has more effect on equilibrium because its expression is squared. Hence, the denominator is decreased more than the numerator in the reaction quotient, Q. The system reacts according to Le Châtelier’s Principle, by moving to the left (toward reactants) to increase the value of [I–].

As the system moves to the left, more Ag+ is created, which has an effect on the first equilibrium above. Again, Le Châtelier’s Principle causes the reaction to move to the left to re-establish equilibrium, which produces AgI(s) precipitate.


AgI(s) + I–(aq) AgI2–(aq)

2c

AgIK = =

I

−

−

⎡ ⎤⎣ ⎦⎡ ⎤⎣ ⎦

Ksp × Kform = 8.3 × 10–6

To answer the second question, we make a table and fill in what we know. We begin with 1.0 M I–. This is reduced by some amount (x) as it reacts with the silver ions, and [AgI2

–] is increased by the same amount:

[I–] [AgI2–]

I 1.0 – C – x + x E 1.0 – x + x

Chapter 17

424

Now we insert the equilibrium values into the above equation:

2c

AgIK = =

I

−

−

⎡ ⎤⎣ ⎦⎡ ⎤⎣ ⎦

8.3 × 10–6

[ ][ ]c

xK = =

1.0 x−8.3 × 10–6

x = 8.3 × 10–6 This value represents the change in concentration of I– which, from the balanced equation, equals the change in concentration of AgI(s). The given volume is 100 mL, which allows us to find the amount of AgI reacting:

0.100 L(8.3 × 10–6 mol/L) = 8.3 × 10–7 mol AgI

8.3 × 10–7 mol AgI(234.8 g/mol) = 1.9 × 10–4 g AgI 17.75 The applicable equilibria are as follows:

AgI(s) Ag+(aq) + I–(aq) + 17spK = Ag I = 8.3 10− −⎡ ⎤ ⎡ ⎤ ×⎣ ⎦ ⎣ ⎦

Ag+(aq) + 2I–(aq) AgI2–(aq)

( )2 11form 2+

Ag IK = = 1 10

Ag I

−

−

⎡ ⎤⎢ ⎥⎣ ⎦ ×⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦


AgI(s) + I–(aq) Ag(I)2–(aq)

( )2c

Ag IK = =

I

−

−

⎡ ⎤⎢ ⎥⎣ ⎦

⎡ ⎤⎣ ⎦

Ksp × Kform = 8.3 × 10–6

If all of the AgI dissolves, it will be in the form of Ag(I)2–, therefore the concentration of Ag(I)2

– is:

[Ag(I)2–] =

( )20.020 mol Ag I

0.100 L solution

−

= 0.200 M Ag(I)2–

[CN–] = 60.200

8.3 10−× =2.4 × 104 M

The amount of KI that must be added is: (2.4 × 104 M)(0.100 L) = 2400 mol KI g KI = (2400 mol KI)(166 g/mol) = 3.98 × 103 g KI

Additional Exercises 17.76 We must first calculate the solubility in terms of # mols/L, i.e.,

( )3 42

2

1 mol Mg(OH)gmol 7.05 10 = 1.21 10L L 58.32 g Mg(OH)− −⎛ ⎞

= × ×⎜ ⎟⎝ ⎠

M

Next, use this to establish the individual ion concentrations based on the equilibrium: Mg(OH)2 Mg2+ + 2OH–

2+ 4Mg = 1.21 10−⎡ ⎤ ×⎣ ⎦ M

4OH = 2.42 10− −⎡ ⎤ ×⎣ ⎦ M

Finally, calculate Ksp using the standard expression:

( )( )2 22+ 4 4 12spK = Mg OH = 1.21 10 2.42 10 = 7.09 10− − − −⎡ ⎤ ⎡ ⎤ × × ×⎣ ⎦ ⎣ ⎦

Chapter 17

425

17.77 To solve this problem, recognize that for a solution having a density of 1.00 g mL–1, 1 ppm = 1 mg L–1. Therefore, the initial hard water solution has a concentration of 278 mg Ca2+ / 1 L

solution. Converting to molar concentration:

2 2 2 2

2 2

3 2

mol Ca 278 mg Ca 1 g Ca 1 mol Ca = L solution 1 L solution 1000 mg Ca 40.078 g Ca

= 6.94 10 M Ca

+ + + +

+ +

− +

⎛ ⎞⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠

×

The concentration of CO3

2– is:

2 23 2 3 2 3 3

2 3 2 3

3 23

mol CO 1.00 g Na CO 1 mol Na CO 1 mol CO =

L solution 1 L solution 105.99 g Na CO 1 mol Na CO

= 9.43 10 M CO

− −

− −

⎛ ⎞⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠

×

Comparing the concentrations of Ca2+ and CO3

2–, we observe that Ca2+ is the limiting reactant. Because of the small value of Ksp, we can assume that CaCO3 will precipitate using all of the available Ca2+ and leaving 9.43 × 10–3 – 6.94 × 10–3 = 2.49 × 10–3 M CO3

2–. The question now becomes, how much Ca2+ will be present in a solution having a [CO3

2–] = 2.49 × 10–3 M? Use the solubility product constant for Ksp to answer this question.

9 2+ 2

sp 3K = 4.5 10 = Ca CO− −⎡ ⎤ ⎡ ⎤× ⎣ ⎦ ⎣ ⎦

9sp2+ 632

3

K 4.5 10Ca = = = 1.8 10 M2.49 10CO

−−

−−

×⎡ ⎤ ×⎣ ⎦ ⎡ ⎤ ×⎣ ⎦

Converting back to units of ppm (mg L–1) we get:

6 2+ 2+2+

2+

2 2+

1.8 10 mol Ca 40.08 g Ca 1000 mgppm Ca = 1 L solution 1 g1 mol Ca

= 7.2 10 ppm Ca

−

−

⎛ ⎞⎛ ⎞⎛ ⎞×⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠

×

17.78 First, we must calculate the mass of CaSO4 dissolved. The volume is a

cylinder, with V = hπr2: h = 0.50 in (2.54 cm/1 inch) = 1.27 cm r = 1/2 diameter = 0.50 cm

Therefore: V = 1.0 cm3, and mass = 1.0 cm3(0.97 g/cm3) = 0.97 g

However, because it is a hydrate (CaSO4·2H2O), plaster is only (136.2/172.2 = 0.79) 79% calcium sulfate, therefore the true mass of CaSO4 is: Mass = 0.97(.79) = 0.77 g CaSO4 In moles, this is 0.77 g CaSO4 = (1 mol/136.2g) = 0.0056 mol CaSO4 Now we must calculate the volume of water necessary to dissolve 0.77 g of CaSO4. We start by finding its molar solubility:

2+ 2sp 4K = Ca SO −⎡ ⎤ ⎡ ⎤

⎣ ⎦ ⎣ ⎦ = 2.4 × 10–5

[ ][ ]spK = x x = 2.4 × 10–5 x = 4.9 × 10–3 mol/L

Chapter 17

426

So the volume of water needed is: 0.0056 mol(1 L/4.9 × 10–3 mol) = 1.2 L Finally, we find the amount of time needed to produce this much water: 1.2 L (1 day/2.00 L) = 0.60 days, or about 14 hours.

17.79 Initially, both Ag+ and HC2H3O2 are at 1.0 M concentrations. These values will be used to determine if the

AgC2H3O2 will precipitate. First, determine the concentration of the acetate ion from the equilibrium:

HC2H3O2 H+ + C2H2O2–

Ka = –

2 3 2

2 3 2

H C H O

HC H O

+⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

⎡ ⎤⎣ ⎦ = 1.8 × 10–5

[HC2H3O2] [H3O+] [C2H3O2–]

I 1.0 – – C – x + x + x E 1.0 – x x x

Assume x << 1.0

[ ]5 -3

a 2 3 2[x][x]K = = 1.8 10 x 4.2 10 [C H O ]1.0

− −× = × =

Next, using the concentration of the acetate ion, determine whether or not a precipitate will form. AgC2H3O2(s) Ag+ + C2H3O2

– Ksp = [Ag+][C2H3O2

–] Q = [Ag+][C2H3O2

–] before equilibrium is established Q = (1.0 M)(4.2 × 103 M) = 4.2× 103 Q > Ksp therefore a precipitate will form.

17.80 Step 1: Determine the [OH–] from the NH3 reaction with water:

NH4+ + OH– NH3 + H2O

–4

b3

–45

3

NH OHK =

NH

NH OH1.8 10 =

NH

+

+−

⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

⎡ ⎤⎣ ⎦⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦×

⎡ ⎤⎣ ⎦

[NH3] [OH–] [NH4+]

I 0.10 – – C – x + x + x E 0.10 – x x x

Assume x << 0.10 [ ][ ][ ]

5 x x1.8 10 =

0.10−×

x = 1.3 × 10–3 = [OH–] Step 2: Find the concentration of Mg2+ at the given concentration of OH–.

The concentration of OH– from the NH3 is 1.3 × 10–3, there is an additional amount of OH– from the equilibrium of the Mg(OH)2, which makes the calculation: Ksp = 7.1 × 10–12 = [Mg2+][OH–]2

7.1 × 10–12 = (x)(1.3 × 10–3 + 2x)2

Chapter 17

427

The additional amount of OH– can be ignored since it will be less than 1.3× 10–3: Using the molar solubility of Mg(OH)2 in distilled water: 7.1 × 10–12 = [Mg2+][OH–]2 s = [Mg2+] and 2s = [OH–] 4s3 = 7.1 × 10–12 s = 1.2 × 10–4 The solubility of Mg(OH)2 in distilled water is less than the amount of OH– supplied by the ammonia so we a re justified in ignoring its contribution. We may now solve for x

x = 4.2 × 10–6 M = [Mg2+] The molar solubility of Mg(OH)2 = 4.2 × 10–6 M

17.81 (a) Mg(OH)2(s) Mg2+ + 2OH– NH4

+ + OH– NH3 + H2O Mg(OH)2(s) + 2NH4

+(aq) Mg2+(aq) + 2H2O + 2NH3(aq) (b) We want all of the Mg(OH)2 to go into solution. The NH4

+ reacts with any OH– produced in the dissociation of Mg(OH)2 thereby shifting the equilibrium to the right. Using the Ksp value for Mg(OH)2, we may find the hydroxide ion concentration under these conditions:

Mg(OH)2(s) Mg2+ + 2OH–

[ ]

22+sp

212

6

K = Mg OH

7.1 10 = 0.10 OH

OH 8.4 10

−

− −

− −

⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

⎡ ⎤× ⎣ ⎦⎡ ⎤ = ×⎣ ⎦

Now we can use this value in the following, simultaneous equilibrium: NH4

+(aq) + OH–(aq) H2O + NH3 (aq)

Kc = 1/KbNH3 = 1/1.8 × 10–5 = 5.6 × 104 [ ]3 4

c -64 4

NH 0.20K 5.6 10

OH NH 8.4 10 NH− + +

⎡ ⎤⎣ ⎦= = = ×⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤×⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

(We know that [NH3] = 0.20 M because in the equation below 2 moles of ammonia are formed for every one mole of magnesium ion:

Mg(OH)2(s) + 2NH4+(aq) Mg2+(aq) + 2H2O + 2NH3

(aq))

Solving for [NH4+], we get 0.43 M.

So the total [NH3] + [NH4

+] = 0.20 + 0.43 = 0.63 M

One must therefore add 0.63 mol NH4Cl to a liter of solution.

(c) The resulting solution will contain 0.20 mol of NH3. Solve the weak base equilibrium problem for NH3. The pH = 11.28.

17.82 Start by determining the [Cl–] if the [Pb2+] = 0.0050 M.

5sp 2

2

K 1.7 10Cl 5.8 100.0050Pb

−− −

+

×⎡ ⎤ = = = ×⎣ ⎦ ⎡ ⎤⎣ ⎦

Then determine the concentration of chloride in a saturated solution, Ksp = 4x3 where x = [Pb2+] and [Cl–] =

2x, sp 2 23K

x 1.6 10 M, Cl 3.2 104

− − −⎡ ⎤= = × = ×⎣ ⎦ M

Chapter 17

428

The volume of 0.10 M HCl which needs to be added is: M1V1 = M2V2 (x mL)(0.10 M HCl) = (3.2 × 10–2 M HCl)(x + 100 mL) 0.1 x = 3.2× 10–2x + 3.2 6.8 × 10–2 x = 3.2 x = 47 mL 47 mL of 0.10 M HCl need to be added.

17.83 In this problem, we have two simultaneous equilibria occurring:

Mn(OH)2(s) Mn2+(aq) + 2OH–(aq) 22+

spK = Mn OH−⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ = 1.6 × 10–13

Fe2+(aq) + 2OH–(aq) Fe(OH)2(s) c 22 -

1K = Fe OH+⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

= 1/(7.9 × 10–16) = 1.3 × 1015

The second equilibrium represents the opposite equation from that of Ksp. Therefore, its value is 1/Ksp for Fe(OH)2.

Combined, and omitting spectator ions, this is:

Mn(OH)2(s) + Fe2+(aq) Mn2+(aq) + Fe(OH)2(s)

( )( )2

13 15c sp (Mn) c (Fe)2

MnK = K K 1.6 10 1.3 10 208

Fe

+−

+

⎡ ⎤⎣ ⎦ = ⋅ = × × =⎡ ⎤⎣ ⎦

[Fe2+] [Mn2+] I 0.100 – C – x + x E 0.100 – x x

2

c 2

MnK =

Fe

+

+

⎡ ⎤⎣ ⎦⎡ ⎤⎣ ⎦

[ ][ ]

x208 =

0.100-x

20.8 – 208x = x 20.8 = 209x x = 0.0995

Therefore, [Fe2+] = 0.100 – x = 0.001 M and [Mn2+] = 0.0995 M

Since Ksp for Fe(OH)2 and Mn(OH)2 are so small, we assume there is almost no free hydroxide ion present and therefore the pH would remain neutral, around 7.

17.84 The reaction for this problem is the formation of Ag(NH3)2+:

Ag+(aq) + 2NH3(aq) Ag(NH3)2+ 3 2 7

form 2+3

Ag(NH )K = = 1.6 10

Ag NH

+⎡ ⎤⎣ ⎦ ×⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

We can rearrange this equation and substitute the values from the text to determine the [Ag+]:

( )

( )( )

33 2+ 10

2 27form 3

Ag(NH ) 2.8 10Ag = = = 1.8 10

1.6 10 1K NH

+ −−

⎡ ⎤ ×⎣ ⎦⎡ ⎤ ×⎣ ⎦ ×⎡ ⎤⎣ ⎦ M

Chapter 17

429

17.85 FeS(s) + 2H+(aq) Fe2+(aq) + H2S(aq) [ ]2+

2spa 2+

Fe H SK =

H

⎡ ⎤⎣ ⎦

⎡ ⎤⎣ ⎦

[H+] [Fe2+] [H2S] I 8 – – C 8 – 2x + x + x E 8 – 2x + x + x

[ ]2+2

spa 2 2+

Fe H S (x)(x)K = = = 600 (from Table 17.2)(8 2x)H

xtake the square root of both sides to get; = 24.5(8 2x)

⎡ ⎤⎣ ⎦

−⎡ ⎤⎣ ⎦

−

Solving gives x = 3.92 M. FeS is very soluble in 8 M acid. 17.86 [Ag+] = 0.200 M

[H+] = 0.10 M First, the concentration of acetate ion needs to be determined at the point that the silver acetate precipitates:

AgC2H3O2(s) Ag+ + C2H3O2–

Ksp = [Ag+][C2H3O2–] = 2.3 × 10–3

Let x = [C2H3O2–]

2.3 × 10–3 = (0.200)(x) x = 1.2 × 10–2 M = [C2H3O2

–] When NaC2H3O2 is added to the solution, the C2H3O2

– will react with the H+ from the nitric acid to form HC2H3O2. This will give a concentration of 0.10 M HC2H3O2. This will then come to equilibrium Now, find the concentration of HC2H3O2 using the Ka for acetic acid:

HC2H3O2 H+ + C2H3O2–

[HC2H3O2](0.200 L) + [C2H3O2–](0.200 L) = mole NaC2H3O2 that needs to be added.

[HC2H3O2] + [H+] = 0.10 M which is from the nitric acid [HC2H3O2] = 0.10 M – [H+]

2 3 2a

2 3 2

H C H OK

HC H O

+ −⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦=

⎡ ⎤⎣ ⎦

Let x = [H+]

[ ][ ]

–2-5

x 1.2 101.8 10

0.10 – x

⎡ ⎤×⎣ ⎦× =

x = 1.5 × 10–4 M = [H+] [HC2H3O2] = 0.10 M – [H+] = 0.10 – 1.5 × 10–4 M = 9.99 × 10–4 M The amount of NaC2H3O2 to be added is: [HC2H3O2](0.200 L) + [C2H3O2

–](0.200 L) = mole NaC2H3O2 [9.99 × 10–4 M](0.200 L) + [1.2 × 10–2 M](0.200 L) = 2.6 × 10–3 mole NaC2H3O2 The number of grams to be added is

g NaC2H3O2 = 2.6 × 10–3 mole NaC2H3O2 2 3 2

2 3 2

82.03 g NaC H O1 mol NaC H O

⎛ ⎞⎜ ⎟⎝ ⎠

= 0.21 g NaC2H3O2

Chapter 17

430

17.87 (a) The number of moles of the two reactants are: 0.12 M Ag+ × 0.050 L = 6.0 × 10–3 moles Ag+ 0.048 M Cl– × 0.050 L = 2.4 × 10–3 moles Cl– The precipitation of AgCl proceeds according to the following stoichiometry: Ag+ + Cl– AgCl(s). If we assume that the product is completely insoluble, then 2.4 × 10–3

moles of AgCl will be formed because Cl– is the limiting reagent (see above.)

( )3 143.3 g AgClg AgCl = 2.4 10 mol AgCl = 0.35 g AgCl1 mol AgCl

− ⎛ ⎞× ⎜ ⎟

⎝ ⎠

(b) The silver ion concentration may be determined by calculating the amount of excess silver added

to the solution: [Ag+] = (6.0 × 10–3 moles – 2.4 × 10–3 moles)/0.100 L = 3.6 × 10–2 M The concentrations of nitrate and sodium ions are easily calculated since they are spectators in this

reaction: [NO3

–] = (0.12 M)(50.0 mL)/(100.0 mL) = 6.0 × 10–2 M [Na+] = (0.048 M)(50.0 mL)/(100.0 mL) = 2.4 × 10–2 M In order to determine the chloride ion concentration, we need to solve the equilibrium expression.

Specifically, we need to ask what is the chloride ion concentration in a saturated solution of AgCl that has a [Ag+] = 3.6 × 10–2 M.

AgCl(s) Ag+ + Cl– Ksp = 1.8 × 10–10

[Ag+] [Cl–] I 0.036 – C + x + x E 0.036 + x + x

+

spK = Ag Cl−⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ = (0.036+x)(x) = 1.8 × 10–10

x = 5.0 × 10–9 M if we assume that x<<0.036 Therefore, [Cl–] = 5.0 × 10–9 M. (c) The percentage of the silver that has precipitated is:

(2.4 × 10–3 moles)/(6.0 × 10–3 moles) × 100% = 40% 17.88 Let x = mols of PbI2 that dissolve per liter; Let y = mols of PbBr2 that dissolve per liter. Then, at equilibrium, we have [Pb2+] = x + y, [I–] = 2x and [Br–] = 2y We know: PbBr2(s) Pb2+ + 2Br– Ksp = 2.1 × 10–6 = [Pb2+][Br–]2 PbI2(s) Pb2+ + 2I– Ksp = 7.9 × 10–9 = [Pb2+][I–]2

Chapter 17

431

Substituting we get: 2.1 × 10–6 = (x + y)(2y)2 and 7.9 × 10–9 = (x + y)(2x)2 Solving for x and y we find: x = 4.85 × 10–4 y = 7.91 × 10–3 Thus, [Pb2+] = 8.40 × 10–3 M, [I–] = 9.70 × 10–4 M and [Br–] = 1.58 × 10–2 M Note: [Br–] > [I–] because PbBr2 is more soluble than PbI2. 17.89 A saturated solution of La2(CO3)3 satisfies the following equilibrium expression:

2 334 3+ 2

sp 3K = 4.0 10 = La CO− −⎡ ⎤ ⎡ ⎤× ⎣ ⎦ ⎣ ⎦

If [La3+] = 0.010 M, then the carbonate concentration of a saturated solution is:

( )

34sp2 1033 3 2 23+

K 4.0 10CO = = = 1.6 100.010La

−− −×⎡ ⎤ ×⎣ ⎦ ⎡ ⎤

⎣ ⎦

We do the same calculation for PbCO3: 14 2+ 2

sp 3K = 7.4 10 = Pb CO− −⎡ ⎤ ⎡ ⎤× ⎣ ⎦ ⎣ ⎦

If [Pb2+] = 0.010 M, then the carbonate concentration of a saturated solution is:

14sp2 12

3 2+

K 7.4 10CO = = = 7.4 10 M0.010Pb

−− −×⎡ ⎤ ×⎣ ⎦ ⎡ ⎤

⎣ ⎦

Therefore, at a carbonate ion concentration between 7.4 × 10–12 M and 1.6 × 10–10 M, PbCO3 will precipitate, but La2(CO3)3 will not precipitate. The upper limit for the carbonate ion concentration is therefore 1.6 × 10–10 M.

The equilibrium we need to look at now is: H2CO3(aq) 2H+(aq) + CO3

2–(aq) The Ka for this reaction is the product of Ka1

and Ka2 for carbonic acid. From Table 18.1 we see that Ka1

= 4.5 × 10–7 and Ka2

= 4.7 × 10–11. So the equilibrium expression and value for the reaction of interest is:

1 2

2+ 23 17

a a a2 3

H COK = = K K = 2.1 10

H CO

−−

⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ × ×

⎡ ⎤⎣ ⎦

This equation is rearranged and the values above and the values given in the problem are substituted in order to determine the pH range over which PbCO3 will selectively precipitate:

( )( )17 2

a 2 3+2 2

3 3

2.1 10 3.3 10K H COH = =

CO CO

− −

− −

× ×⎡ ⎤⎣ ⎦⎡ ⎤⎣ ⎦ ⎡ ⎤ ⎡ ⎤

⎣ ⎦ ⎣ ⎦

If we substitute 2 123CO = 7.4 10 M− −⎡ ⎤ ×⎣ ⎦ we determine + 4H = 3.1 10 M−⎡ ⎤ ×⎣ ⎦ and the pH = 3.51.

Substituting 2 103CO = 1.6 10 M− −⎡ ⎤ ×⎣ ⎦ , + 5H = 6.6 10 M−⎡ ⎤ ×⎣ ⎦ and

pH = 4.18. Consequently, if [H+] = 6.6 × 10–5 M (pH = 4.18), La2(CO3)3 will not precipitate but PbCO3 will precipitate.

At pH = 3.51 and below, neither carbonate will precipitate.

Chapter 17

432

17.90 Three reactions are occurring in the solution:

1) Cu2+(aq) + 4NH3–(aq) Cu(NH3)4

2+–(aq) Kform = 1.1 × 1013 = 3 442

3

Cu(NH )

Cu NH+

⎡ ⎤⎣ ⎦⎡ ⎤⎡ ⎤⎣ ⎦⎣ ⎦

2) NH3(aq) + H2O NH4+(aq) + OH–(aq) Kb = 1.8 × 10–5 =

–4

3

NH OH

NH

+⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

⎡ ⎤⎣ ⎦

3) Cu(OH)2(s) Cu2+ + 2OH– Ksp = 4.8 × 10–20 = 22 –Cu OH+⎡ ⎤ ⎡ ⎤

⎣ ⎦ ⎣ ⎦

Starting with the first reaction, using the Kform of Cu(NH3)42+, calculate the concentration of NH3.

The concentration of Cu2+ before any reaction has occurred is 0.050 M, and the concentration of NH3 before any reaction has occurred is 0.50 M. Assume that all of the Cu2+ has reacted and then the reaction has come to equilibrium. Therefore, the initial concentration of Cu2+ is 0 and the initial concentration of NH3 is 0.50 M – (4 × 0.05 M)

[Cu2+] [NH3] Cu(NH3)42+

I – 0.30 0.050 M C + x + 4x – x E + 2x 0.30 + 4x 0.050 – x

Kform = 1.1 × 1013 = 3 442

3

Cu(NH )

Cu NH+

⎡ ⎤⎣ ⎦⎡ ⎤⎡ ⎤⎣ ⎦⎣ ⎦

Kform = 1.1 × 1013 =[ ]

[ ][ ]40.050 – x

2x 0.30 4x+ assume x << 0.050

x = 2.8 × 10–13 [NH3] = 0.30 [Cu2+] = 5.6 × 10–13

Next, determine the [OH–] from the reaction of NH3 with H2O

NH3(aq) + H2O NH4+(aq) + OH–(aq) Kb = 1.8 × 10–5 =

–4

3

NH OH

NH

+⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

⎡ ⎤⎣ ⎦

[NH3] [NH4

+] OH–

I 0.30 – – C – x + x +x E 0.30 – x x x

Kb = 1.8 × 10–5 =–

4

3

NH OH

NH

+⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

⎡ ⎤⎣ ⎦

Kb = 1.8 × 10–5 =[ ][ ]

[ ]x x

0.30 – x assume x << 0.30

x = 2.3 × 10–3 = [OH–]

Finally, with the concentration of OH– and the concentration of Cu2+, determine if any precipitate has formed:

Cu(OH)2 Cu2+ + 2OH– Ksp = 4.8 × 10–20 = 22 –Cu OH+⎡ ⎤ ⎡ ⎤

⎣ ⎦ ⎣ ⎦

[OH–] = 2.3 × 10–3 [Cu2+] = 5.6 × 10–13

Chapter 17

433

Q = 22 –Cu OH+⎡ ⎤ ⎡ ⎤

⎣ ⎦ ⎣ ⎦ = (5.6 × 10–13)(2.3 × 10–3)2 = 3.0 × 10–18

Q > Ksp therefore a precipitate forms. Assume that all of the Cu2+ has reacted with the OH– and then the solution returns to equilibrium [Cu2+] [OH–] I – 2.3 × 10–3 C + x + 2x E + x 2.3 × 10–3 + 2x

Ksp = 4.8 × 10–20 = 22 –Cu OH+⎡ ⎤ ⎡ ⎤

⎣ ⎦ ⎣ ⎦

Ksp = 4.8 × 10–20 = [ ]2–3x 2.3 10 2x⎡ ⎤× +⎣ ⎦ x << 2.3 × 10–3

x = 9.1 × 10–15 M = [Cu2+] 17.91 In order to answer this question, we need the [OH–] at equilibrium. Mg(OH)2 is a sparingly soluble

compound. According to Table 17.1, Ksp = 7.1 × 10–12.

MgOH2(s) Mg2+(aq) + 2OH–(aq) 22+

spK = Mg OH−⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

[Mg2+] [OH–] I – – C + x + 2x E + x + 2x

Ksp = (x)(2x)2 = 4x3 = 7.1 × 10–12 x = 1.2 × 10–4 M, [OH–] = 2x = 2.4 × 10–4 M: pOH = –log[OH–] = 3.62 The pH = 14.00 – pOH = 10.38. 17.92 Al(OH)3 has such an exceedingly small Ksp, there is almost no dissociation in

pure water. Therefore, the pH would be expected to be about 7. In doing the calculations, this is borne out:

33+

spK = Al OH−⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

[ ][ ]3343 10 x 3x−× = x = 1.8 × 10–9

[OH–] = 5.5 × 10–9

[OH–] = (1 × 10–7) + (5.5 × 10–9) = 1.055 × 10–7 pOH = 6.98 pH = 7.02

17.93 Fe(OH)3(s) Fe3+ + 3OH– Ksp = [Fe3+][OH–]3 = 1.6 × 10–39

[Fe3+] [OH–] I – 1.0 × 10–7 C + x + 3x E + x 1.0 × 10–7 + 3x

Assume 3x << 1.0 × 10–7, Ksp = (x)(1.0 × 10–7)3, solving for x we get, x = 1.6 × 10–18 M. Thus, 1.6 × 10–18

mol of Fe(OH)3 dissolve in 1 L of water.

Chapter 17

434

17.94 At its simplest, this is only a Ksp problem. The concentration of Mn2+ is (0.400 L)(0.10 M Mn2+)/(0.500 L) = 0.080 M Mn2+

22+

spK = Mn OH−⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

[ ]2-131.6 10 0.080 OH−⎡ ⎤× = ⎣ ⎦

-6OH 1.4 10−⎡ ⎤ = ×⎣ ⎦

When [OH–] = 1.4 × 10–6 M, Mn(OH)2 precipitates. 0.100 L(2.0 mol/L) = 0.20 mol NH3 are added to 400 mL solution which would make an initial

concentration of 0.20 mol/0.500 L = 0.40 M NH3. The following equilibrium is set up: H2O + NH3 NH4

+(aq) + OH–(aq) -5bK = 1.8 10×

The problem tells us that all of the Sn2+ is precipitated as Sn(OH)2: Sn2+(aq) + 2OH–(aq) Sn(OH)2(s)

The concentration of the Sn2+ is (0.400 L)(0.10 M Sn2+)/(0.500 L) = 0.080 M Sn2+, just before it precipitates. The Sn2+ immediately uses the first 2(0.080 M) = 0.16 M OH– which is produced from the reaction of ammonia with water above, using 0.16 M NH3. This effectively brings our initial concentration of NH3 to 0.24 M.

Now we determine how much NH3 will produce [OH–] = 1.3 × 10–6 M.

H2O + NH3 NH4+(aq) + OH–(aq) -5

bK = 1.8 10×

[NH3] [NH4+] [OH–]

I x – 1.0 × 10–7 C –1.2 × 10–6 + 1.2 × 10–6 + 1.2 × 10–6 E ? 1.2 × 10–6 1.3 × 10–6

4b

3

NH OHK

NH

+ −⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦=

⎡ ⎤⎣ ⎦

-6 -6-5

3

1.2 10 1.3 101.8 10

NH

⎡ ⎤ ⎡ ⎤× ×⎣ ⎦ ⎣ ⎦× =⎡ ⎤⎣ ⎦

[NH3] = 8.7 × 10–8 Therefore the initial [NH3] should be 8.7 × 10–8 + 1.2 × 10–6 = 1.3 × 10–6. So we want to reduce [NH3] by 0.24 – 1.3 × 10–6 = 0.23999 M, essentially by 0.24 M. This would require adding equimolar amounts of HCl, or: 0.500 L(0.24 mol NH3/L)(1 mol HCl/1 mol NH3)(36.5 g HCl/1 mol HCl) = 4.4 g HCl. (The difficulty here arises from the fact that such a small amount of OH– is required to precipitate the Mg2+ from solution that even a minimal amount of NH3 produces enough hydroxide ion to do so.)

Chapter 17

435

17.95 The reaction that will dissolve the Mg(OH)2 is: Mg(OH)2 + 2H+ Mg2+ + 2H2O

The concentration of H+ in the solution, before any reaction has occurred between the acid and the Mg(OH)2, is (0.025 L)(0.10 M HCl) = 2.5 × 10–3 mol HCl 2.5 × 10–3 mol/1.000 L = 2.5 × 10–3 M

Since all of the H+ will react with the solid Mg(OH)2, the amount of Mg2+ in solution will be:

2.5 × 10–3 mol HCl2

+1 mol Mg

2 mol H

+⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

= 1.3 × 10–3 mol Mg2+

Find the equilibrium concentration of OH– with 1.3 × 10–3 M Mg2+ Ksp = [Mg2+][OH–]2 = 7.1 × 10–12 7.1 × 10–12 = (1.3 × 10–3)(x)2 x = 7.4 × 10–5 = [OH–] pOH = –log[OH–] = –log(7.4 × 10–5) = 4.13 pH = 14 – pOH = 14 – 4.13 = 9.87

17.96 There are two reactions that have to be considered here: the dissociation of CaCO3 in water,

CaCO3(s) Ca2+(aq) + CO32–(aq) Ksp = [Ca2+][CO3

2–] = 4.5 × 10–9 and the ionization of carbonate ion in water,

CO32–(aq) + H2O HCO3

–(aq) + OH–(aq) 43b 2

3

[OH ][HCO ]K 1.8 10

[CO ]

− −−

−= = ×

Assuming that all of the CO32– reacts with the water, the net reaction is:

CaCO3(s) + H2O(l) Ca2+(aq) + HCO3–(aq) + OH–(aq)

Kc = Ksp × Kb = [Ca2+][HCO3–][OH–] = 8.1 × 10–13

We can obtain [OH–] from the pH: pOH = 14 – pH = 14 – 8.50 = 5.50 [OH–] = 10–pOH = 10–5.50 = 3.2 × 10–6 M Assume that [Ca2+] = [HCO3

–] = x Kc = 8.1 × 10–13 = (x)(3.2 × 10–6 M)(x) = (x)2(3.2 × 10–6 M) Solve for x: x = 5.0 × 10–4 M = [Ca2+] = [HCO3

–] The [Ca2+] is equal to the molar solubility. Thus, the molar solubility of CaCO3 is 5.0 × 10–4 M. 17.97 First, let’s examine the question to make clear what is happening. A solution contains 0.20 M Ag+ ions and

0.10 M acetate ions. The ion product of these two (0.020) is less than Ksp for silver acetate (2.3 × 10–3), so the silver acetate remains in solution. There are also H+ ions (H3O+) in the solution as a result of the following equilibrium (OAc– will symbolize acetate):

H2O + HOAc H3O+(aq) + OAc–(aq) The amount of H3O+ may be found by using the Ka for acetic acid:

[HOAc] [H3O+] [OAc–] I 0.10 – – C – x + x + x E 0.10 – x X x

[ ]3

a

H O OAcK

HOAc

+ −⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦=

[ ][ ][ ]

-5 x x1.8 10

0.10 x× =

−

x ≈ 1.3 × 10–3 So [H3O+] = 1.3 × 10–3 M.

Chapter 17

436

However, when F– is added to the solution (in the form of KF), the following equilibrium takes place:

F–(aq) + H3O+(aq) H2O + HF

This depletes H3O+ ions from the solution, which causes the first equilibrium above to move to the right, producing more acetate ions. When the acetate ion concentration hits some minimum value (determined by Ksp) silver acetate will precipitate. That value may be found as follows:

+spK = Ag OAc−⎡ ⎤ ⎡ ⎤

⎣ ⎦ ⎣ ⎦

[ ][ ]32.3 10 = 0.20 x−× x = 0.012 mol/L

So the problem becomes…How many grams of KF must be added such that the acetate concentration increases to 0.012 M? This is now a simultaneous equilibrium problem:

H2O + HOAc H3O+(aq) + OAc–(aq) 5

aK = 1.8 10−×

F–(aq) + H3O+(aq) H2O + HF 3c 4

a

1 1K ’ = = = 1.5 10K 6.8 10−

××

(Note the above equation is simply the reverse of that for the Ka of HF, so Kc’ = 1/Ka.) Combined, this becomes: F–(aq) + HOAc HF + OAc–(aq)

( )( )5 3 2c a bK = K K 1.8 10 1.5 10 2.7 10− −⋅ = × × = ×

Recall that we have already found the initial concentrations of OAc– and HOAc above. Using this information, and the fact that we want the final [OAc–] to be 1.2 × 10–2, we can begin to fill out the table below.

[F–] [HOAc] [HF] [OAc–] I x 0.099 – 1.3 × 10–3 C – 0.011 – 0.011 + 0.011 + 0.011 E ? 0.088 0.011 0.012

[ ][ ]

c

HF OAcK

F HOAc

−

−

⎡ ⎤⎣ ⎦=

⎡ ⎤⎣ ⎦

[ ][ ][ ]

2 0.011 0.0122.7 10

F 0.088−

−× =

⎡ ⎤⎣ ⎦

[F–] = 0.056 M

Placing this value into the table as the equilibrium concentration of [F–], we find the initial [F–] must be 0.056 + 0.011 = 0.067 M. Therefore the amount of KF needed in the 200 mL solution is: 0.200 L(0.067 mol KF/L)(58.01 g KF/1 mol KF) = 0.78 g KF

17.98 (1) In order, Figure C Figure B Figure E

(2) Figures A and D are excluded because PbBr2 will precipitate before PbCl2.

Bringing It Together: Chapters 13 –17

437

1. The five factors that affect the rates of reactions are: (a) Chemical nature of the reactants (b) Ability of the reactants to come in contact with each other (c) Concentrations of the reactants (d) Temperature (e) Availability catalysts

2. rate = (0.60 L mol–1 s–1)[I–][OCl–]

(a) rate = (0.60 L mol–1 s–1)[0.0100 mol L–1][0.0200 mol L–1] = 1.2 × 10–4 mol L–1 s–1

(b) rate = (0.60 L mol–1 s–1)[0.100 mol L–1][0.0400 mol L–1] = 2.4 × 10–3 mol L–1 s–1

3. (a) rate = k[A]n[B]m

First, compare the first and third experiments, in which there has been an increase in concentration by a factor of 2 for A while the concentration of B is held constant. This has caused an increase in rate of:

n m nn

n m nrate of reaction 3 k[0.020] [0.010] [0.020] 2rate of reaction 1 k[0.010] [0.010] [0.010]

= = =

=4 1 1

24 1 1

8.0 10 mol L s = 4.00 = 22.0 10 mol L s

− − −

− − −

××

n = 2 rate = k[A]2[B]m Second, compare the second and third experiment, in which there has been an increase in concentration by a factor of 2 for B while the concentration of A is held constant. This has caused an increase in rate of:

n m mm

n m mrate of reaction 2 k[0.020] [0.020] [0.020] 2rate of reaction 3 k[0.020] [0.010] [0.010]

= = =

=4 1 1

04 1 1

8.0 10 mol L s = 1.00 = 28.0 10 mol L s

− − −

− − −

××

m = 0 rate = k[A]2[B]0

(b) The rate constant:

8.0 × 10–4 mol L–1 s–1 = k[0.020 mol L–1]2[0.010]0 k = 2 L mol–1 s–1

(c) rate = 2 L mol–1 s–1[0.017 mol L–1]2[0.033 mol L–1]0

rate = 5.8 × 10–4 mol L–1 s–1 4. The order of the reaction with respect to that reactant is –1

5. (a) t1/2 = ln 2k

t1/2 = 6 1ln 2

2.41 10 s− −× = 2.88 × 105 s

t1/2 = 4 1ln 2

2.22 10 s− −× = 3.12 × 105 s


438

(b) [C2(NO2)6] = 0.100 M

[ ][ ] ( )6 10.100 60 sln 2.41 10 s 500 minx 1 min

− − ⎛ ⎞⎛ ⎞= × ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

x = 9.30 × 10–2 mol L–1

(c) a2–1 –1

1 2 1

Ek 1 1lnk T T8.314 J mol K

⎛ ⎞ ⎛ ⎞−= −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

4 –1a

6 –1 –1 –1E2.22 10 s 1 1ln

373 K 343 K2.41 10 s 8.314 J mol K

−

−

⎛ ⎞ −× ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎜ ⎟× ⎝ ⎠⎝ ⎠

37.6 J mol–1 = –Ea (–2.34 × 10–4) Ea = 1.60 × 105 J mol–1

(d) a2–1 –1

1 2 1

Ek 1 1lnk T T8.314 J mol K

⎛ ⎞ ⎛ ⎞−= −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

4 –1 5 –1

–1 –1 –12.22 10 s 1.60 10 J mol 1 1ln

373 K 393 Kx s 8.314 J mol K

−⎛ ⎞× − × ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

x s–1 = 3.07 × 10–3 s–1

6. (a) 1 1 1 12 2 2 8× × = An eight of the 90Sr will be left.

(b) 12

ln 2 kt

=

k = ln 228 y

= 0.025 y–1

[ ][ ]

0

t

Aln kt

A=

[ ][ ] ( )10

t

Aln 0.025 y (168 y)

A−=

[ ][ ]

0

t

A

A = 1

0.015

(c) [ ][ ]

0

t

Aln kt

A=

( )11

16

1ln 0.025 y t−=

t = 111 y

(d) [ ][ ]

0

t

Aln kt

A=

( )116

1ln 0.025 y t−=

t = 71.7 y

[ ][ ]

0

t

Aln kt

A=


439

7. (a) t1/2 = 1k (initial concentration of reactant)×

t1/2 = –4 –1 –11

1.0 10 L mol s (0.250 M)× × = 4 × 104 s

(b) [ ] [ ]t 0

1 1 ktA A

− =

[ ] ( )–4 –1–1

t

1 1 60 s1.0 10 s 30 minA 1 min0.250 mol L

⎛ ⎞− = × ⎜ ⎟⎡ ⎤ ⎝ ⎠⎣ ⎦

[A]30 min = 0.239 mol L–1 [B]30 min = 0.150 M – (0.250 M – 0.239 M) = 0.139 M

8. Reaction mechanism is an entire series of elementary processes to describe the steps involved in a reaction.

Rate-determining step is the slowest step in a reaction mechanism Elementary process is a reaction whose rate law can be written from its own chemical equation, using its coefficients as exponents for the concentration terms.

9. rate = k[O3] 10.

The activation energy for the forward reaction is the difference in energy between the reactants and the transition state. The activation energy for the reverse reaction is the difference in energy between the products and the transition state. The heat of reaction is ∆H.

11. A heterogeneous catalyst increases the rate of a chemical reaction by providing another reaction pathway

with a lower activation energy. 12. (a) The initiation step is C2H6 2CH3·

(b) The propagation steps are CH3· + C2H6 CH4 + C2H5· C2H5· C2H4 + H· H· + C2H6 C2H5· + H2

(c) The termination step is H· + C2H5· C2H6


440

13. (a) [ ]

[ ][ ]

3

2 2

NONO N O

(b) [ ]2SO

(c) 2 23Ni CO+ −⎡ ⎤ ⎡ ⎤

⎣ ⎦ ⎣ ⎦

14. The equilibrium constant is very small, therefore the reaction does not proceed appreciably to products.

[HF] = 0.010 M [HF] [H2] [F2] I 0.010 – – C –2x +x +x E 0.010–2x +x +x

Kc = [ ][ ][ ]

2 22

H F

HF =

[ ][ ][ ]2

x x

0.010 2x− = 1 × 10–13

x << 0.010

1 × 10–13 = [ ][ ][ ]2

x x

0.010

Take the square root of both sides and solve for x x = 3.2 × 10–9 [H2] = [F2] = 3.2 × 10–9

15. gn

p cK K (RT)∆= 2NO2(g) N2O4(g) ∆ng = –1 6.5 × 10–2 = Kc[(0.0821 L atm mol–1 K–1)(393 K)]–1 Solve for Kc Kc = 2.1

16. NO2(g) + SO2(g) NO(g) + SO3(g)

[NO2] [SO2] [NO] [SO3] I 0.0200 0.0200 – – C –x –x +x +x E 0.0200–x 0.0200–x +x +x

[ ][ ][ ]

3c

2 2

NO SOK 3.60

NO SO⎡ ⎤⎣ ⎦= =

Kc = 3.60 = [ ][ ]

[ ][ ]x x

0.0200 x 0.0200 x− −

Take the square root of both sides

1.90 = x0.0200 x−

Solve for x. x = 0.0131 [NO2] = [SO2] = 6.90 × 10–3 M [NO] = [SO3] = 0.0131 M


441

[NO2] [SO2] [NO] [SO3] I 6.90 × 10–3 M 6.90 × 10–3 M 0.0151 0.0151 C +x +x –x –x E 6.90 × 10–3 +x 6.90 × 10–3 +x 0.0151–x 0.0151–x

[ ][ ][ ]

3c

2 2

NO SOK 3.60

NO SO⎡ ⎤⎣ ⎦= =

Kc = 3.60 = [ ][ ]

[ ][ ]0.0151 x 0.0151 x

0.00690 x 0.00690 x− −

+ +

Again take the square root of both sides and solve for x.

1.90 = 0.0151 x0.00690 x

−+

x = 6.86 × 10–4 [NO2] = [SO2] = 7.59 × 10–3 M [NO] = [SO3] = 1.44 × 10–2 M

17. (a) Increase (b) Increase (c) Decrease (d) Decrease (e) No change

18. Kw = [H+][OH–] = 9.5 × 10–14

[H+] = [OH–] = 3.1 × 10–7 pH = 6.51 The solution is neither acidic nor basic because the concentration of the hydronium ion equals the concentration of the hydroxide ion.

19. The concentration of OH– is 4.7 × 10–7 g L–1

The molar concentration of OH– is:

[OH–] = 7 — —

—4.7 10 g OH 1 mol OH

1 L solution 17.01 g OH

−⎛ ⎞⎛ ⎞×⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

= 2.76 × 10–8 M

pOH = –log[OH–] = –log(2.76 × 10–8 M) = 7.56 pH = 14 – pOH = 14 – 7.56 = 6.44

The water is acidic. 20. H3AsO4 is a stronger acid since there are more oxygens on the As. The electrons are more electronegative

and pull electron density away from the OH bonds. 21. H2Te 22. (a) H2SO3

(b) N2H5+

23. (a) SO3

2– (b) N2H3

– (c) C5H5N

24. CH3NH2 and CH3NH3

+ NH4

+ and NH3


442

25. An amphoteric substance can behave as either an acid or a base. A Lewis acid is an electron pair acceptor. A Lewis base is an electron pair donor.

26.

S

O

O O

O HS

O

O O

O H

27. The stronger binary acid than Y would be Z. Binary acid strength increase left to right across a period. 28. (a) C6H5OH(aq) + H2O(l) → C6H5O–(aq) + H3O+(aq)

(b) Ka = 6 5 3

6 5

C H O H O

C H OH

− +⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

⎡ ⎤⎣ ⎦

(c) [C6H5OH] = 0.550 M since the amount of dissociation is negligible [H3O+] = –log(pH) = –log(5.07) = 8.51 × 10–6 M [C6H5O–] = 8.51 × 10–6 M since all of the H3O+ comes from the dissociation of the phenol

Ka = 6 5 3

6 5

C H O H O

C H OH

− +⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

⎡ ⎤⎣ ⎦ =

[ ]

6 68.51 10 8.51 10

0.550

− −⎡ ⎤ ⎡ ⎤× ×⎣ ⎦ ⎣ ⎦ = 1.32 × 10–10

pKa = –log Ka = –log(1.32 × 10–10) = 9.88

(d) Kb = 14 14

10a

1 10 1 10K 1.32 10

− −

−× ×

=×

= 7.58 × 10–5

pKb = –log Kb = –log(7.58 × 10–5) = 4.12 29. (a) pKb = 14 – pKa = 14 – 11.68 = 2.32

(b) C7H3SO3– + H2O C7H3SO3H + OH–

The solution will be basic

Kb = 7 3 3

7 3 3

C H SO H OH

C H SO

−

−

⎡ ⎤⎡ ⎤⎣ ⎦⎣ ⎦⎡ ⎤⎣ ⎦

= 4.79 × 10–3

[C7H3SO3

–] [C7H3SO3H] [OH–] I 0.010 – – C –x +x +x E 0.010–x +x +x

4.79 × 10–3 = [ ][ ]

[ ]x x

0.010 x−

x will be large compared to 0.010 M. Therefore we need to solve it either by successive approximations or the quadratic equation. x = 4.92 × 10–3 = [OH–] pOH = 2.31 pH = 11.69


443

30. Let Cod stand for codeine Cod + H2O CodH+ + OH–

Kb = [ ]

CodH OH

Cod

+ −⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ = 1.6 × 10–6

[Cod] [CodH+] [OH–] I 0.0115 – – C –x +x +x E 0.0115–x x x

1.6 × 10–6 = [ ][ ]

[ ]x x

0.0115 x−

x is large compared ot 0.0115, therefore solve for x using either the quadratic equation or successive approximations x = 1.3 × 10–4 = [OH–] pOH = 5.79 pH = 14 – pH = 8.21

31. CH3NH2(aq) + H2O CH3NH3

+(aq) + OH–(aq)

Kb = 3 3

3 2

CH NH OH

CH NH

+ −⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

⎡ ⎤⎣ ⎦

32. pKb = 3.43

pKa of its conjugate acid = 14 – pKb = 14 – 3.43 = 10.57

33. (a) M ascorbic acid solution = 2 6 6 6 2 6 6 6

2 6 6 6

3.12 g H C H O 1 mol H C H O0.125 L 176.1 g H C H O

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠ = 0.142 M H2C6H6O6

(b) H2C6H6O6 HC6H6O6– + H3O+ Ka1 =

6 6 6 3

2 6 6 6

HC H O H O

H C H O

− +⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

⎡ ⎤⎣ ⎦ = 7.94 × 10–5

HC6H6O6– C6H6O6

2– + H3O+ Ka2 = 2

6 6 6 3

6 6 6

C H O H O

HC H O

− +

−

⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

⎡ ⎤⎣ ⎦

= 1.62 × 10–12

Since all of the H3O+ will come from the first equilibrium reaction, it can be calculated as follows:

[H2C6H6O6] [HC6H6O6–] [H3O+]

I 0.142 – – C –x +x +x E 0.142–x x x

7.94 × 10–5 =

6 6 6 3

2 6 6 6

HC H O H O

H C H O

− +⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

⎡ ⎤⎣ ⎦

7.94 × 10–5 =

[ ][ ][ ]

x x0.142 x−

Solve for x x = 3.32 × 10–3 = [H3O+] pH = –log[H3O+] = –log(3.32 × 10–3) = 2.48

[C6H6O62–] = 1.62 × 10–12 [HC6H6O6

–] = [H3O+] in the Ka2 equation, these cancel


444

34. pH = pKa + log[ ]

initial

initial

A

HA

−⎡ ⎤⎣ ⎦

4.50 = 4.74 + log[ ]

initial

initial

A

HA

−⎡ ⎤⎣ ⎦

–0.24 = log[ ]

initial

initial

A

HA

−⎡ ⎤⎣ ⎦

10–0.24 = [ ]

initial

initial

A

HA

−⎡ ⎤⎣ ⎦

[ ]initial

initial

A

HA

−⎡ ⎤⎣ ⎦ = 0.575

35. (a) C2H3O2

– + H+ HC2H3O2 (b) HC2H3O2 + OH– C2H3O2

– + H2O

36. Ka = 1.8 × 10–5 = 2 3 2 3

2 3 2

C H O H O

HC H O

− +⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

⎡ ⎤⎣ ⎦

[ ][ ]

30.10 H O

0.15

+⎡ ⎤⎣ ⎦ = 1.8 × 10–5

[H3O+] = 2.7 × 10–5 pH = 4.57 [HC2H3O2]final = (0.15 – 0.04)M = 0.11 M [C2H3O2

–]final = (0.10 + 0.04)M = 0.14 M [ ]

[ ]30.14 H O

0.11

+⎡ ⎤⎣ ⎦ = 1.8 × 10–5

[H3O+] = 1.4 × 10–5 pH = 4.85 The change in pH is 0.28 pH units.

37. (a) Propanoic acid, pKa = 4.87 and its salt, sodium propionate

(b) This problem is best solved using two equations and two unknowns:

pH = pKa + log mol Amol HA

−

5.10 = 4.87 + log[ ]A 0.005

HA 0.005

−⎡ ⎤ +⎣ ⎦−

5.00 = 4.87 + log[ ]A

HA

−⎡ ⎤⎣ ⎦


445

0.23 = log[ ]A 0.005

HA 0.005

−⎡ ⎤ +⎣ ⎦−

0.13 = log[ ]A

HA

−⎡ ⎤⎣ ⎦

100.23 = [ ]A 0.005

HA 0.005

−⎡ ⎤ +⎣ ⎦−

100.13 = [ ]A

HA

−⎡ ⎤⎣ ⎦

[A–] = 100.13[HA]

100.23 = [ ]

[ ]0.1310 HA 0.005

HA 0.005+

−

100.23([HA] – 0.005) = 100.13[HA] + 0.005 1.698[HA] – 0.008491 = 1.349[HA] + 0.005 0.3493[HA] = 0.01349 [HA] = 0.03865 M [A–] = 100.13(0.03865 M) = 0.05214 M For 625 mL (0.625 L) of solution:

For determining the mass of propionic acid needed:

g C3H6O2 = 0.625 L 3 6 2 3 6 2

3 6 2

0.03865 mol C H O 74.09 g C H O1 L 1 mol C H O

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠ = 1.79 g C3H6O2

For determining the mass of sodium propionate:

g NaC3H5O2 = 0.625 L 3 5 2 3 5 2

3 5 2

0.05214 mol NaC H O 96.07 g NaC H O1 L 1 mol NaC H O

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠ = 3.13 g

NaC3H5O2 (c) From the above calculation:

The concentration of propionic acid is 0.03865 M. The concentration of sodium propionate is 0.05214 M.

38. (a) Neutral

(b) Acidic (c) Acidic (d) Acidic

39. pKb = 14 – pKa = 14 – 4.87 = 9.13

Kb = [ ]HA OH

A

−

−

⎡ ⎤⎣ ⎦

⎡ ⎤⎣ ⎦

= 7.41 × 10–10

All of the acid has been converted into the salt at the equivalence point. (0.115 M)(50.00 mL) = (0.100 M)(x mL) x = 57.50 mL total volume is 107.50 mL [A–] = 0.05349 M [HA] = [OH–] = x


446

7.41 × 10–10 =( )( )

[ ]x x

0.05349

x = 6.296 × 10–6 M = [OH–] pOH = 5.201 pH = 8.799 Thymol blue or phenolphthalein would be good indicators.

40. Kb1 = 2

14w

12a

K 1.0 10K 1.6 10

−

−×

=×

= 6.3 × 10–3

[C6H6O6

2–] [HC6H6O6–] [OH–]

I 0.050 – – C –x +x +x E 0.050–x x x

6 6 6

26 6 6

HC H O OH

C H O

− −

−

⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

⎡ ⎤⎣ ⎦

= [ ][ ]

[ ]x x

0.050 x− = 6.3 × 10–3

x is too large to make a simplifying assumption, therefore solve for x either using the quadratic equation or by successive approximations. x = 0.015 = [OH–] pOH = 1.82 pH = 12.18

41. Since half of the acid is neutralized the concentration of the acid is equal to the concentration of its

conjugate base, the pKa can be determined:

pH = pKa + log[ ]A

HA

−⎡ ⎤⎣ ⎦

3.56 = pKa + log 1 pKa = pH = 3.56 Ka = 10–pKa = 10–3.56 = 2.75 × 10–4

42. pH = pKa + log[ ]A

HA

−⎡ ⎤⎣ ⎦

pKa = 14 – pKb pKb = –log Kb = –log 1.8 × 10–5 pKa = 14 – 4.74 = 9.26

9.26 = 9.26 + log[ ]A

HA

−⎡ ⎤⎣ ⎦

[A–] = [HA] [NH4

+] = 0.100 before the addition of NaOH. In order for the two concentrations to be equal, half as much NaOH must be added: (0.100 L NH4

+)(0.100 M NH4+ solution) = 0.0100 mol NH4

+ (0.0100 mol NH4

+)(0.5) = 0.00500 mol NaOH

0.00500 mol NaOH 40.00 g NaOH1 mol NaOH

⎛ ⎞⎜ ⎟⎝ ⎠

= 0.200 g NaOH


447

43. Ksp = [Ag+]2[CrO42–]

[Ag+] = 2 × (6.7 × 10–5 M) = 1.34 × 10–4 M [CrO4

2–] = 6.7 × 10–5 M Ksp = (1.34 × 10–4)2(6.7 × 10–5) Ksp = 1.2 × 10–12

44. Mg(OH)2(s) Mg2+(aq) + OH–(aq)

Ksp = 7.1 × 10–12 = [Mg2+][OH–]2 7.1 × 10–12 = [x][2x]2 x = 1.2 × 10–4 2x = 2.4 × 10–4 = [OH–] pOH = –log[OH–] = –log(2.4 × 10–4) = 3.62 pH = 14 – 3.62 = 10.38

45. Fe(OH)2(s) Fe2+(aq) + 2OH–(aq)

ksp = 7.9 × 10–16 = [Fe2+][OH–]2 pH = 10.00 pOH = 14 – 10.00 = 4.00 [OH–] = 10–4.00 = 1 × 10–4 7.9 × 10–16 = [Fe2+][1 × 10–4]2 7.9 × 10–8 M = [Fe2+]

g L–1 =

8 2+2 2

2+2

1 mol Fe(OH) 89.86 g Fe(OH)7.9 10 mol Fe1 L solution 1 mol Fe(OH)1 mol Fe

−⎛ ⎞ ⎛ ⎞× ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠

= 7.10 × 10–6 g L–1

46. (a) This is a limiting reagent problem.

mL KI needed = 30.0 mL Pb(NO3)2 3 2

3 2

0.100 mol Pb(NO )1000 mL Pb(NO ) solution⎛ ⎞⎜ ⎟⎝ ⎠

×3 2

2 mol KI 1000 mL KI solution1 mol Pb(NO ) 0.500 mol KI⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠ = 12.0 mL KI solution

20.0 mL KI solution is supplied. KI is in excess.

g PbI2 = 30.0 mL Pb(NO3)2 3 2

3 2

0.100 mol Pb(NO )1000 mL Pb(NO ) solution⎛ ⎞⎜ ⎟⎝ ⎠

× 2 2

3 2 2

1 mol PbI 461.0 g PbI1 mol Pb(NO ) 1 mol PbI⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠ = 1.38 g PbI2

(b) The concentration of the spectator ions: [K+] = (0.500 M K+)(20.0 mL solution) ÷ 50.0 mL = 0.200 M K+ [NO3

–] = (0.100)(2)(30.0 mL solution) ÷ 50.0 mL = 0.12 M NO3–

[I–]: Total moles of I– = (0.500 M I–)(0.0200 L solution) = 0.0100 mol I–

mol I– used = 30.0 mL Pb2+

2+

2+ 2+0.100 mol Pb 2 mol I

1000 mL Pb solution 1 mol Pb

−⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

= 6.00 × 10–3 mol I–

[I–] = (0.0100 mol I– – 6.00 × 10–3 mol I–)/(0.050 L solution) = 0.0800 M I–

[Pb2+]: This will be the amount of Pb2+ that is in solution after the solid PbI2 reaches equilibrium PbI2(s) Pb2+(aq) + 2I–(aq)


448

[Pb2+] [I–] I – 0.0800 M C +x +x E x 0.0800 + x

Ksp = 7.9 × 10–9 = [Pb2+][I–]2 = (x)(0.0800 + x)2 x << 0.0800 7.9 × 10–9 = (x)(0.0800)2 x = 1.23 × 10–6 M = [Pb2+]

47. This is a simultaneous equilibrium problem.

CuCO3(s) Cu2+(aq) + CO32–(aq) Ksp = [Cu2+][CO3

2–] =2.5 × 10–10

Cu2+(aq) + 4NH3(aq) Cu(NH3)42+(aq) Kform =

( ) 23 4

423

Cu NH

Cu NH

+

+

⎡ ⎤⎢ ⎥⎣ ⎦⎡ ⎤⎡ ⎤⎣ ⎦⎣ ⎦

= 1.1 × 1013

CuCO3(s) + 4NH3(aq) Cu(NH3)4

2+(aq) + CO32–(aq)

Koverall = ( ) 2 2

3 344

3

Cu NH CO

NH

+ −⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎣ ⎦⎣ ⎦

⎡ ⎤⎣ ⎦ = 2.75 × 103

All of the Cu(NH3)42+ comes from the CuCO3

[Cu(NH3)42+] = 1.00 g CuCO3 3

3

1 mol CuCO 1123.6 g CuCO 1.00 L solution⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠ = 8.09 × 10–3

As the Cu2+ is used to form Cu(NH3)42+, the [CO3

2–] = [Cu(NH3)42+] = 8.09 × 10–3

2.75 × 103 = ( ) 2 2

3 344

3

Cu NH CO

NH

+ −⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎣ ⎦⎣ ⎦

⎡ ⎤⎣ ⎦ =

[ ]

3 3

4

8.09 10 8.09 10

x

− −⎡ ⎤ ⎡ ⎤× ×⎣ ⎦ ⎣ ⎦

x = 0.0124 M NH3

mol NH3 = 1.00 L solution 30.0124 mol NH1 L solution

⎛ ⎞⎜ ⎟⎝ ⎠

=0.0124 mol NH3

48. Assume the solution is saturated with CO2 and therefore the concentration of H2CO3 is 0.030 M

H2CO3 2H+ + CO32–

K = Ka1 × Ka2

= (4.3 × 10–7)(4.7 × 10–11) = 2.0 × 10–17 =

2 2–3

2 3

H CO

H CO

+⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

⎡ ⎤⎣ ⎦

First, calculate the [H+] at which thePbCO3 will begin to precipitate: PbCO3(s) Pb2+(aq) + CO3

2–(aq) Ksp = 7.4 × 10–14 = [Pb2+][CO32–]

[Pb2+] = 0.010 M [CO3

2–] = (7.4 × 10–14)/(0.010) = 7.4 × 10–12 M

[H+] = ( ) ( )[ ]

( )

1 1–17 –172 2

2 3

2– –123

2.0 10 H CO 2.0 10 0.030

CO 7.4 10

⎛ ⎞ ⎛ ⎞× ×⎡ ⎤⎣ ⎦⎜ ⎟ ⎜ ⎟=⎜ ⎟ ⎜ ⎟⎡ ⎤⎜ ⎟ ⎜ ⎟×⎣ ⎦⎝ ⎠ ⎝ ⎠

= 2.8 × 10–4

pH = 3.55


449

Now, determine the [H+] at which the BaCO3 will begin to precipitate: BaCO3(s) Ba2+(aq) + CO3

2–(aq) Ksp = 5.0 × 10–9 = [Ba2+][CO32–]

[Ba2+] = 0.010 M [CO3

2–] = (5.0 × 10–9)/(0.010) = 5.0 × 10–7 M

[H+] = ( ) ( )[ ]

( )

1 1–17 –172 2

2 3

2– –73

2.0 10 H CO 2.0 10 0.030

CO 5.0 10

⎛ ⎞ ⎛ ⎞× ×⎡ ⎤⎣ ⎦⎜ ⎟ ⎜ ⎟=⎜ ⎟ ⎜ ⎟⎡ ⎤⎜ ⎟ ⎜ ⎟×⎣ ⎦⎝ ⎠ ⎝ ⎠

= 1.1 × 10–6

pH = 5.96 The pH range is between 3.55 and 5.96, above 5.96, BaCO3 will begin to precipitate.

49. The less soluble substance is PbS. We need to determine the minimum [H+] at which NiS will precipitate.

[ ]2+2

spa 2 + 2+

+

Ni H S (0.100)(0.1)K = = = 40 (from Table 17.2)[H ]H

(0.10)(0.1)[H ] = = 0.01640

⎡ ⎤⎣ ⎦

⎡ ⎤⎣ ⎦

pH = –log[H+] = 1.80. At a pH lower than 1.80, PbS will precipitate and NiS will not. At larger values of

pH, both PbS and NiS will precipitate.

We also need to determine the [H+] at which PbS will start to precipitate

[ ]2+2 7

spa 2 + 2+

+7

Pb H S (0.100)(0.1)K = = = 3 10 (from Table 17.2)[H ]H

(0.10)(0.1)[H ] = = 1823 10

−

−

⎡ ⎤⎣ ⎦ ×

⎡ ⎤⎣ ⎦

×

pH = –log[H+] = –2.26. Any acid in water will precipitate the PbS. The pH range is –2.26 to 1.80 to allow the PbS to precipitate without the NiS.

50. The less soluble substance is SnS. We need to determine the minimum [H2S] at which FeS will precipitate.

[ ] [ ]

[ ]

2+2 2

spa 2 3 2+

3 23

2

Fe H S (0.10) H SK = = = 600 (from Table 17.2)

[1 10 ]H

(600)(1 10 )H S = = 6 10(0.1)

−

−−

⎡ ⎤⎣ ⎦

×⎡ ⎤⎣ ⎦

××

The concentration of Sn2+ can now be determined.

[ ]2+ 2+ 32 5

spa 2 —3 2+

5 3 22+ 9

3

Sn H S Sn 6 10K = = = 1 10 (from Table 17.2)

[1 10 ]H

(1 10 )(1 10 )Sn = = 1.7 10(6 10 )

−−

− −−

−

⎡ ⎤ ⎡ ⎤ ⎡ ⎤×⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ××⎡ ⎤

⎣ ⎦

× ×⎡ ⎤ ×⎣ ⎦ ×


450

51. (a) MS(s) M2+(aq) + S2–(aq) Ksp = [M2+][S2–] = 4.0 × 10–29 MS(s) + H2O M2+(aq) + HS2–(aq) + OH–(aq)

Kspa = [ ]2

2

2

M H S

H

+

+

⎡ ⎤⎣ ⎦

⎡ ⎤⎣ ⎦

= (4.0 × 10–29)(1021) = 4.0 × 10–8

(b) [ ]

2

2x

0.30 = 4.0 × 10–8

x = 6.0 × 10–5 M

Ch.17

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Transcript of Ch.17