Goes with chapter 18 – Silberberg: Principles of General Chemistry

Mrs. Laura Peck, 2013

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Understand the acid-base theories of Arrhenius, Bronsted-Lowry, and Lewis

Identify strong acids and bases and calculate their pH’s

Calculate the pH of a weak acid or base Calculate the [] of a strong or weak acid or base

from its pH Predict the pH of a salt from its formula and then

calculate the pH of the salt Identify the components of a buffer and perform

calculations involving the preparation of a buffer and the addition of strong acid or strong base to a buffer

Perform calculations involving strong acid-strong base titrations as well as weak acid-strong base and weak base-strong acid calculations.

Be familiar with titatration curves and selection of an acid-base indicator

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Arrhenius Theory States that, in aqueous solution (water), acids produce hydrogen ions

and bases produce hydroxide ions. Bronsted-Lowry theory

Says that an acid is a proton (H+) donor and a base is a proton acceptor In the reaction below, HNO3 transfers a proton to H2O forming H3O+, the

hydronium ion. H3O+ is the conjugate acid of H2O and NO3

- is the conjugate base of HNO3.

The formulas in a conjugate acid-base pair differ by one H+

HNO3 + H2O H3O+ + NO3-

Acid base c.acid c.base Lewis Acid-base model

A lewis acid is an electron pair acceptor and a Lewis base is an electron pair donor.

Remember that this is a much broader definition than the other two, in that the acid-base interaction does not need to involve a transfer of a proton H+.

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Give the formulas for the conjugate base of H2SO4 and the conjugate acid of CH3NH2

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HSO4- is the conjugate base of H2SO4

CH3NH3+ is the conjugate acid of CH3NH2

**Note that each conjugate acid-base pair differs by 1H+

HSO4- + 1H+ = H2SO4; CH3NH3

+ = 1H+ + CH3NH2

The names and formulas of the six strong acids must be memorized.

The six strong acids are: HCl, HI, HNO3, H2SO4, and HClO4

If an acid is not one of the six in the list, then for purposes of the AP exam, you can assume it is a weak acid.

The chart below compares the dissociation of strong acids to weak acids.

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Type of acid, HA

Reversibility of reaction

Ka value Ions existing when acid, HA, dissociates in water

Strong Not reversible Ka value very large

H+ and A- only. No HA present

Weak Reversible Ka is small H+, A-, and HA

Weak acids exist in equilibrium with their ions in aqueous solution. The acid dissociation constant, Ka, measures the

extent to which the acid dissociates in water: HA(aq) + H2O(l) H3O+

(aq) + A-(aq)

The equilibrium expression for the reaction is Ka = [H3O+][A-] [HA]

A table of Ka values for monoprotic acids, containing one acidic hydrogen, appears in the appendix of your textbook and is worth studying. The larger the Ka value, the stronger the acid.

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Example #2: List the acids in order of increasing strength: HCN, HCl, HClO2, HNO2

Example #3: Arrange the following species in order of increasing base strength: NO2

-, ClO2-, CN-, Cl-

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HCN, HNO2, HClO2, and HCl. The first three weak acids are listedIn order of increasing Ka values. HCl is stronger than all of the Weak acids given. HCl is a strong acid

Cl-, ClO2-, NO2

-, CN-; the bases are listed in reverse order of Their conjugate acids in the previous example because the Stronger and acid, the weaker its conjugate base.

Strong bases include groups I and II hydroxides such as NaOH.

Weak bases such as NH3, are not group I or II hydroxides.

The base dissociation constant, Kb, measures the extent to which a base reacts with water.

The reaction of a weak base, B, with water and its corresponding equilibrium expression is:

B(aq) + HOH(l) BH+(aq) + OH-

(aq) Kb = [BH+][OH-] [B]

**AP tip: Writing the reaction for base dissociation can be tricky. Always remember to react the base with water. Use the Arrhenius and Bronsted-Lowry theories to help write the reaction. The base must react with water to produce hydroxide ions according to Arrhenius, and accepts an H+ according to Bronsted-Lowry. Check the charges of the reactant and products and be sure that the sums of the charges on both sides of the reaction are equal.

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For binary acids, HX, the strength of the H-X bond and the polarity of the bond will determine the behavior of the acid. The polarity of the bonds in hydrogen halides become less

polar going down a group. The strength of the H-F bond is what makes it a weak acid,

whereas the rest of the hydrogen halides are strong acids: HI>HBr>HCl

For a given series of Oxoacids such as HClO4, HClO3, HClO2, and HClO, the acid strength increases with increasing number of oxygens attached to the central atom. HClO4 is a strong acid. The remainder of the oxoacids are listed in order of

decreasing strength (decreasing number of oxygen atoms) The O-H bond becomes more polarized and weakened due to

the electron density drawn toward the highly electronegative oxygen atoms.

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The pH of a strong acid can be calculated directly from the hydrogen ion concentration pH = -log[H+] [H+]. The molar concentration of the hydrogen ion,

is obtained from the molarity of the acid. The pH of a strong base can be calculated

from its hydroxide ion concentration. pOH = -log[OH-]

pOH + pH = 14.00 The concentration of a strong acid or strong

base can be determined from the solution’s pH.

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Example #4: Calculate the pH of 0.010M HCl

Example #5: The pH of a Sr(OH)2 solution is 13.50. Calculate the concentration of Sr(OH)2

**AP tip: the number of sig figs in a pH measurement is equal to the number of decimal places in the pH. For example, 1.70 has 2 sig figs.

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The pH equals 2.00; -log(0.010) = 2.00

pOH = 14 – pH 14 – 13.5 = 0.50

[OH] = inv log(-pOH) inv log (-0.50) = 0.32M OH-

0.32 mol OH x 1 mol Sr(OH)2 = 0.16M Sr(OH)2

1 L 2 mol OH-

The pH of a weak acid cannot be calculated directly from the concentration of the acid since all of the acid does not dissociate to form H+.

The equilibrium reaction of the acid must be considered.

Example #6: Calculate the pH of 0.25M HCN

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First, write the reaction of the acid with water. Use the Arrhenius and Bronsted-Lowry theories to help you write the products. Check that youHave the correct charges on the products.

HCN + H2O H3O+ + CN- now, set up an ‘ICE’ chart for equilibrium

I. 0.25 0 0C. -x +x +xE. 0.25-x x x

Third, write the equilibrium expression forKa in the same manner as you did in the Last topic. Plug in the values from the Equilibrium line of the ICE chart.

Ka = 6.2x10-10 = [H3O+][CN-] x2 ~ x2 [HCN] (0.25-x) 0.25

Example #6 cont…. Ka = 6.2x10-10 = [H3O+][CN-] x2 ~ x2 [HCN] (0.25-x) 0.25

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Fourth, solve for x which is equal to H3O+

(6.2x10-10)(0.25) = x2 x = 1.2x10-5M

Finally, calculate the pH from the value of x, the [H3O+]

pH = -log(1.2x10-5) = 4.90

*AP tips: Always follow these steps when performing calculations involving theDisassociation of weak acids or weak bases:1.Write the reaction of the acid or base with water. 2.Set up an ICE chart3.Write the equilibrium expression in terms of reactants and products.4.Solve for x, using the method of approximation. Test approximation5.Solve for pH. (be careful for base equilibria, x = OH-. You need to find pOH and then pH6. Always remember that for weak acids and bases at eq: pH =-log[H3O+]

The percent dissociation of an acid (or base) is the amount of the acid, HA, which has dissociated, x, divided by the acid’s initial concentration, HA0, multiplied by 100 x/[HA0] * 100%

When making assumptions in an equilibrium calculation, it is best to test the assumption by making sure that the percent dissociation is less than or equal to 5%. The test for the assumption is the same as the calculation for

the percent dissociation. Example #7: The percent dissociation of an acid, HA, which

is 0.100M is 2.5%. Calculate the Ka of the acid.

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x/0.100M x 100% = 2.5% x = 2.5 x 10-3M

Ka = [H3O+][A-] Ka = (2.5x10-3)2 = 6.4x10-5

[HA] (0.100 – 2.5x10-3)

The calculations involving weak base equilibria are similar to the weak acid equilibria problems except that the equation is written for a base reacting with water and the calculation initially involves finding [OH-]. You will need to find the pOH and then the pH.

Example #8: The pH of a 0.20M solution of H2NNH2 is 11.38. Calculate Kb for H2NNH2

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Write the reaction with water. Bases accept H+. Watch charges! One hint toHelp in writing the reaction is that the pH is 11.38. The basic pH indicatesThat OH- must be one of the products.

H2NNH2 + HOH OH- + H2NNH3+

I. 0.20 0 0C. -x +x +xE. 0.20-x x x

You are given the pH, but x=[OH-]Find pOH; pH+pOH=14.00pOH= 14.00 – 11.38 = 2.62

pOH = -log[OH-] find [OH-] = inv log(-pOH)

Inv log(-2.62) = 2.4x10-3M (you can type this Into your calculator as 10-pOH

Plug this value of x into the Kb expression: Kb = [OH-][H2NNH3

+]/[H2NNH2] = x2/(0.20-x)

Kb = (2.4x10-3)2 / 0.20 = 2.9x10-5

Polyprotic acids can donate more than one proton, H+, and dissociate by losing 1H+ at a time.

Example #9: Calculate the [H+] of a 0.20M solution. Also determine the concentrations of H3AsO4, H2AsO4

-, HAsO4

2-, and AsO43-.

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For H3AsO4; Ka1 = 5x10-3, Ka2 = 8x10-8, Ka3 = 6x10-

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H3AsO4 + HOH H3O+ + H2AsO4-

I. 0.20 0 0C. -x +x +xE. 0.20-x x x

H3AsO4 + H3O+ H3O+ + H2AsO4-

Ka1 = [H3O+][H2AsO4-] / [H3AsO4]

5x10-3 = x2 / (0.20-x)X = 3x10-2M[H3O+]=[H2AsO4

-] = 3x10-2

[H3AsO4-] = 0.20 – 0.03 = 0.17M

Example #9 cont…. For H3AsO4; Ka1 = 5x10-3, Ka2 = 8x10-8, Ka3 =

6x10-10

Since Ka3<<<Ka2<<<Ka1, very little of H2AsO4- and HAsO4

2- dissociates compared to H3AsO4, so [H3O+] and [H2AsO4

-] will not change very much by the Ka2 dissociation, and we can use their concentrations to find the concentration of HAsO4

2-.

H2AsO4- + HOH H3O+ + HAsO4

2-

I. 3x10-2 3x10-2 0

C. -x +x +x

E. (3x10-2)-x (3x10-2)+x x

Ka2 = 8x10-8 = (3x10-2)[HAsO42-]

(3x10-2)

[HAsO42-] = 8x10-8M; the assumption that Ka2 does not

contribute significantly to [H3O+] and [H2AsO4-] is good.

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Example #9 cont…. For H3AsO4; Ka1 = 5x10-3, Ka2 = 8x10-8, Ka3 =

6x10-10

Repeat the process to find [AsO43-]

HAsO42- + HOH H3O+ + AsO4

3-

I. 8x10-8 3x10-2 0

C. -x +x +x

E. (8x10-8)-x (3x10-2)+x +x

Ka3 = 6x10-10 = (3x10-2)[AsO43-] / (8x10-8)

[AsO43-] = 2x10-15M. Assumption that x is small is good.

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Determining the Approximate pHOf ions in solution.

Determining theApproximate pHOf a salt.

Determine whether an aqueous solution of KC2H3O2 is acidic, basic or neutral

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KC2H3O2 is basic. Use the method of asking questions outlinedIn the previous slide:

Split the salt into its cation, K+, and its anion, C2H3O2-

Is the cation of Group I metal? Yes; the cation does not affect the pH

Is the anion a conjugate base of a strong acid? No; C2H3O2- is The c.base of a weak acid, HC2H3O2, which makes the solutionBasic.

Because we have a salt with a cation which doesn’t affect the pHAnd a basic anion, an aqueous solution of the salt is basic.

To calculate the pH of a salt, first you must decide whether the salt is acidic, basic, or neutral. If the salt is basic, then its anion is the

c.base of a weak acidThe anion will undergo hydrolysis. You will need to write an equation for the

reaction of that ion with water to form the acid and OH- ions: A- + HOH = HA + OH-.

Then write the equilibrium expression Kb = Kw = [HA][OH-] Ka [A-]

If the salt is basic, then the hydrolysis reaction will produce an acid.

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Determine the pH of a 0.100M aqueous solution of NaCN. The Ka for HCN is 5.8x10-10

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In aqueous solutions, NaCN ionizes completely into Na+ and CN-. However, the cyanide ions ionize in water according to the following: CN-(aq) + HOH(l) HCN(aq) + OH-(aq)

Na+ ions do not affect the pH. Ions from group IA and IIA never undergoHydrolysis because they are cations of strong bases. The equilibrium Expression for the solution is: Kb = [OH-][HCN] [CN-]We need a value for Kb. Since we have Ka for HCN, we can calculate the Value of Kb for CN-: Kb = (1.0x10-14)/Ka = (1.0x10-14)/(5.8x10-10)

CN- + HOH HCN + OH-I. 0.100 0 0C. -x +x +xE. 0.100-x x x

Since K is much smaller than 0.100, Assume that 0.100-x = 0.100

Kb = [OH-][HCN]/[CN-] 1.7x10-5 = x2/0.100X = [OH-] = 1.3x10-3

pOH = 2.89; pH = 14.00 – 2.89pH = 11.11

Buffers resist changes in pH when acids or bases are added.

Calculating the pH of a buffer: The calculation to find the pH of a buffer is

similar to all equilibrium calculations EXCEPT there are now two initial concentrations, one for each part of the buffer pair.

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Components Examples

Weak acid + salt containing the c.base

HCN & NaCN

Weak base + salt containing the c.acid

CH3NH2 & CH3NH3Cl

Weak acid + excess strong base *or* weak base + excess strong acid

2 mol of HCN + 1mol NaOH react to yield 1mol HCN and 1mol NaCN

2mol NH3 + 1mol HCl react to yield 1mol NH3 and 1mol NH4Cl

Calculate the pH of a solution that is 0.60M HF and 1.00M KF. Ka for HF is 7.2x10-4

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First, write the reaction of the acid with water. (You are given Ka, andThe buffer is made of a weak acid and its c.base.)

HF + HOH H3O+ + F- Second, set up an ICE chart

I. 0.60 0 0C. -x +x +xE. 0.60-x x x

Third, write the equilibrium Expression for Ka. Plug in the valuesFrom the equilibrium line (E) of theICE chart. Check to see if x is small.

Ka = 7.2x10-4 = [H3O+][F-] = x(1.00-x) = x(1.00) [HF] (0.60-x) 0.60

Fourth, solve for xWhich equalsH3O+

Ka = 7.2x10-4 = x(100)/0.60 x = [H3O+] = 4.3x10-4

pH = -log[H3O+] pH = -log(4.3x10-4) = 3.37

A buffer can be made from a weak acid and a salt containing its c.base or from a weak base and a salt containing its c.acid

Henderson-Hasselbalch equation: pH = pKa + log([c.base]/[acid])

Example #13: Calculate the mass of NaC2H3O2 required to prepare a buffer of pH 4.55 when added to 0.500L of 0.67M acetic acid. (assume no change in volume) Ka = 1.8x10-5 for HC2H3O2

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Answer is 1.8g NaC2H3O2……. This problem can be solved by either using an ICE chart or by using the Henderson-Hasselbalch equation.

pKa = -logKa pH = pKa + log([C2H3O2-] / [HC2H3O2])

4.55 = 4.74 + log ([C2H3O2-] / 0.67M) -0.19 = log([C2H3O2-]/0.67M)

0.65 = [C2H3O2-]/0.67M 0.44M = [C2H3O2-]

Mass NaC2H3O3 = 0.44mol[C3H3O2-] x 1mol NaC2H3O2 x 0.500L x 82.0g 1 L 1 mol [C2H3O2-] 1mol NaC2H3O2

You will solve these problems in the same manner as the previous problem #12, with the exception of the first three steps which are new.

Example #14: calculate the pH when 100.0 mL of 0.50M HCl are added to the buffer consisting of 20.0g HC2H3O2 and 18.0g of NaC2H3O2 dissolved in 5.00x102 mL of water.

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First, Calculate the initial mols of weak acid and c.acid present in the buffer. Write These amounts under the reaction, above the ICE table.Second, calculate the moles of strong acid, H+, added. The added strong Acid will react with the basic part of the buffer, C2H3O2 to produce the otherPart of the buffer HC2H3O2. Make a new line under the initial mole called‘add mol H+’. Calculate the new initial moles of weak acid and it c.baseAfter the c.base has reacted with the strong acid, by adding or subtracting the number of moles H+ added.

HC2H3O3 + HOH H3O+ + C2H3O2

-

Initial mols 0.33 0 0.21Add mol H+ +0.05 0 -0.05New initial mol 0.38 0 0.16I. 0.63 0.27C. -x +x +x E. 0.63-x x 0.27+x

Initial [] of weak acid and c.base: HC2H3O2 = 0.38mol/0.600L = 0.63M C2H3O2- = 0.16mol/0.600L = 0.27M

Plug values intoHenderson-Hasselbalch

Ka = [H+][C2H3O2] [HC2H3O2]

1.8x10-5=[H+](0.27+x)~[H+]0.27 (0.63+x) 0.634.2x10-5M=[H+] pH=-log[H+] 4.38

Example #15: Calculate the pH when 100.0ml 0.50M NaOH are added to the buffer consisting of 20.0g of HC2H3O2 and 18.0g of NaC2H3O2 dissolved in 5.00x102mL of water.

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1st, Calculate the initial moles of weak acid and c.acid that are present in theBuffer. Write these amounts under the reaction, above the ICE table. 2nd, caclulate the mole of strong base, OH- added. The added strong baseWill react with the acidic part of the buffer, HC2H3O2 to produce the otherPart of the buff, C2H3O2-. Make a new line under the initial mols called ‘addMol OH-’. 3rd, calculate the new initial concentration of the acid and its c.base and fillOut the ICE chart as in example #14.

HC2H3O2 + HOH H3O+ + C2H3O2-

Initial mols 0.33 0 0.21Add mol OH- -0.05 0 +0.05New initial mol 0.28 0 0.26I 0.47 0.43C -x +x +x E 0.47-x x 0.43+x

[H+] = 2.0x10-5M pH = 4.71

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