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Final Exams for Chem 122 - Mathews

2:30 Class: Thursday, Mar. 17 at 1:30 in IH 0100

6:30 Class: Tuesday, Mar. 15 at 5:30pm in MP 1000

Chapter 17 Additional Aspects of Aqueous Equilibria

17.1 The Common-Ion Effect17.2 Buffered Solutions

Composition and Action of Buffered SolutionsBuffer Capacity and pHAddition of Strong Acids or Bases to Buffers

17.3 Acid-Base TitrationsStrong Acid-Strong Base TitrationsWeak Acid-Strong Base TitrationsTitrations of Polyprotic Acids

Common Ion Effect:

This is the same as other equilibrium calculations,

except there is an initial concentration of

two species prior to establishing equilibrium.

Consider a solution which is prepared by mixing 0.10 mole of acetic acid and 0.010 mole of HCl in enough water to make 1.000 L of solution.

What are the concentrations of each species?

HOAc H2O H3O+ OAc-

I 0.10 0.01

d -x +x +x

eq 0.10-x 0.01+x x

0.10 0.01 1.7x10-4

Note that for a solution of 0.10 M acetic acid alone, [OAc- ] = [H3O+] = 1.3 x 10-3 M

HOAc + H2O = H3O+ + OAc-

i 0.30 M ~ 0 0.30 Mδ -x x x

eq (0.30-x) x (0.30+x)

Ka = 1.8 x 10-5 = [(x)(0.30+x)] / [(0.30-x)]

leads to x = [H3O+ ] = 1.8 x 10-5 M compared to

2.3 x 10-3 M for the acid alone!!!

Consider a solution made from dissolving 0.30 moles of acetic acid and 0.30 moles of sodium acetate in enough water to make 1.00 L of solution.

What is the concentration of Hydronium ions and the pH?

Consider 1.00L of solution containing 1.00 mol of formic acid (HCOOH) and 0.500 mol of sodiumformate (NaHCOO). What is the pH? Ka = 1.77 x 10-4

HA H2O H3O+ A-

I 1.00 0.500

d -x +x +x

eq 1.00-x x 0.500+x

1.77 x 10-4 = (x)(0.500+x) / (1.00-x) ≈ (x)(0.500) / (1.00)

so x = 3.54 x 10-4 = [H3O+] and pH = 3.45

Note, however, that we’re doing the same things again!

The expression Ka = [H3O+][A-] / [HA]

can be written [H3O+] = Ka [HA] / [A-] ≈ Ka [HA]i / [A-]i

or take the negative log of both sides:

-log [H3O+] = -logKa - log [HA]i / [A-]i

or pH = pKa - log [HA]i / [A-]i

or pH = pKa + log[A-] / [HA]

Henderson-Hasselbalch Equation(s)Valid ONLY when [H3O+] and [OH-] are very small in comparison to [A-]i and [HA]i

Now let’s add 0.10 mol of strong acid (eg HCl) to the solution (without changing its volume).

HA H2O H3O+ A-

I 1.00 0.500

d -x +x +x

eq 1.00-x x 0.500+x

eq1 1.00 3.54x10-4 0.500

add +0.10

rxn 1.10 ~0 0.400

d -x +x +x

eq2 1.10 x 0.400

And pH =pKa+log(0.400)/(1.10) = 3.75+(-0.44) = 3.31=pH

or [H3O+] = 4.89x10-4

A similar calculation results when we add enough strongbase to the original buffer solution to make the solution0.10 M in OH- .

HA H2O H3O+ A-

I 1.00 0.500

d -x +x +x

eq 1.00-x x 0.500+x

eq1 1.00 3.54x10-4 0.500

add -0.10 +0.10

rxn 0.90 ~0 0..600

d -x +x +x

eq2 0.90 x 0.600

In this case, [H3O+] = 2.66x10-4 and pH = 3.58

Chapter 17 Additional Aspects of Aqueous Equilibria

17.1 The Common-Ion Effect17.2 Buffered Solutions

Composition and Action of Buffered SolutionsBuffer Capacity and pHAddition of Strong Acids or Bases to Buffers

17.3 Acid-Base TitrationsStrong Acid-Strong Base TitrationsWeak Acid-Strong Base TitrationsTitrations of Polyprotic Acids

Final Exams for Chem 122 - Mathews

2:30 Class: Thursday, Mar. 17 at 1:30 in IH 0100

6:30 Class: Tuesday, Mar. 15 at 5:30pm in MP 1000

Titration curves: Strong Acid – Strong Base

titrate 100.0 mL of 0.1000 M HCl with 0.1000 M NaOH

Recognize that if c0 and Vo = concentration and volume of initial acid(basd)

and ct and Vt = concentration and volumeof base (acid) added

Then the ‘equivalence point’ corresponds to the condition

coVo = ctVe, where Ve = Vt when

nacid = nbase

1. V = 0 mL of base addedn(H30+) = (0.1000M)(0.1000 L) = 1.000 x 10-2 mol H30+

and pH = 1.00

2. V = 30 mL of NaOH added (i.e. 0 < V < Ve )n(OH-) = (0.1000 M)(0.0300 L) = 3.000 x 10-3 mol OH-

which reacts with (neutralizes) the same amount of H30+

i.e. n (H+) = (1.000x10-2 – 3.000x10-3) mol H30+

= 7.000 x 10-3 mol H30+ remaining(+ 3.000 mol Na+ and Cl-)

and [H30+] = (7.000x10-3)/(Va + Vb) = (7.000 x 10-3) / (0.1300 L) = 0.05385 M

and pH = 1.27

3. V = 100.0 mL NaOH added (i.e. Vb = Ve )

n(OH- added) = (0.1000 M)(0.1000 L) = 1.000 x 10-2 mol which neutralizes the same amount (all) of H30+ .

The significant reaction that has taken place isH+ + OH- H2O

therefore [OH-] = [H30+] = 1.000 x 10-7 M

and pH = 7.00

Note also that [Na+] = [Cl-] = (1.000 x 10-2 mol) / (0.2000 L)= 0.5000 M

SA – SB, continued:4. V = 100.05 mL (i.e. Vb > Ve)

also, this is about one drop excess

easiest to treat by realizing that 100.00 mL gavethe result of previous calculation, i.e.neutralized the acid.

then n (OH- excess) = [(100.05 - 100.00)mL][0.1000 M]= 5.0 x 10-6 mol OH- ‘extra’

and [OH-] = (5.0 x 10-6 mol) / (100.00 + 100.05) = 2.5 x 10-5 M OH-

so that pOH = 4.60 and pH = 9.40

similar calculations can proceed for all excess base

Titration of Strong Acid with Strong Base

Weak Acid + Strong Base100.0 mL 0.1000 M HOAc titrated with 0.1000 M NaOH

Note the definition of the equivalence point remains the same.

1. V = 0.00 mL NaOHusual problem of weak acid, with Ka = 1.8 x 10-5

n(HOAc) = (0.1000 M)(0.1000 L) = 1.000 x 10-2 mol HOAc

so [H30+] = 1.3 x 10-3 M and pH = 2.88

2. 0 < V < Veacid will be partially neutralized. Since OH- is a strongerbase than water,

HOAc + OH- = H2O + OAc-

K = 1/Kb = Ka / Kw = 2 x 109 >>1

example, 30 mL of NaOHn(OH-) = (0.1000 M)(0.0300 L) = 3.000 x 10-3 mol OH-n(HOAc) = (0.1000 M)(0.1000 L) = 1.000 x 10-2 mol HOAc

so after all the OH- has reacted with HOAc, n(OAc-) = n(OH- added) = 3.00 x 10-3 moln(HOAc remaining) = (1.000x10-2 – 3.000x10-3) mol HOAc

= 7.000 x 10-3 mol HOAc

2. (continued)

[OAc-] = (3.000 x 10-3 mol OAC-) / (Va + Vb)= (3.000 x 10-3 mol OAC-) / (0.1300 L)= 2.31 x 10-2 M OAc-

and [HOAc] = (7.000 x 10-3) / (0.1300 L) = 5.38 x 10-2 M HOAc

Now we’re able to calculate the pH of this buffer soln

pH = pKa + log{[OAc-] /[HOAc]} = 4.75 + log{(2.31 x 10-2) / (5.38x10-2)} pH = 4.38

Notice also that when Vb = ½ Va, [HA]=[A-] and pH=pKa

WA-SB titration (cont)3. V = 100mL NaOH, V = Ve

we now have n(NaOAc) = n(OAc-] = (0.1000M)(0.1000L) = 1.000 x 10-2 mol OAc-

which is the same as having 1.000 10-2 mol of NaOAc in soln

[OAc-] = (1.000x10-2) / (0.2000 L) = 0.05000 M

As before, we can calculate the pH based on hydrolysis

OAc- + H2O = HOAc + OH- Kb = Kw/Ka = 5.7 x 10-10

and pH = 8.73

WA-SB titration (cont)4. V >100mL NaOH, i.e, V = Ve

just adding more OH- to the solution of NaOAc,

results primarily in the [OH-] being dominated by the NaOH being added

i.e. this portion of the titration curve looksvery similar to the SA-SB titration curve

Titration of Weak Acid with Strong Base, Ve = 50 mL

Titration of Weak Acid with Strong Base, Ve = 50 mL

pH=pKapH=pKa

Anything wrong with this picture???

Discuss weak polyprotic acids, eg H2CO3

Weak polyprotic acids: typical H2CO3 and H3PO4

H2CO3 + H2O = H3O+ + HCO3- Ka1 = 4.3 x 10-7

HCO3- + H2O = H3O+ + CO3

2- Ka2 = 4.8 x 10-11

Example: Aqueous sol’n sat’d in CO2; [H2CO3] = 0.034 M

(1) [H3O+] = [HCO3-] = 1.2 x 10-4; [H2CO3] = 0.034 M

(2) [H3O+] = [HCO3-] = same ; [CO3

2-] = 4.8 x 10-11

Note also we can rewrite the eq expressions:

][][

][

3

1

32

3

OH

K

COH

HCO a

][][

][

3

2

3

23

OH

K

HCO

CO a

37

3

1

32

3 103.410100.1

103.4

][][

][

xx

x

OH

K

COH

HCO a

Example: at pH = 10.00, [H3O+] = 1.00 x 10-10

48.0100.1

108.4

][][

][10

4

3

2

3

23

x

x

OH

K

HCO

CO a

We can calculate the fraction of any species, eg H2CO3

][][][

][][ 2

3332

3232

COHCOCOH

COHCOHfraction

][

][

][

][

][

][

][

][

][

3

23

3

3

3

32

3

32

32

HCO

CO

HCO

HCO

HCO

COH

HCO

COH

COHfraction

44

4

32 106.148.01103.2

103.2][

xx

xCOHfraction

similarly for HCO3- = 0.68 and CO3

2- = 0.32

And, this could be repeated for all pH values

100 mL of 0.1000 M H3PO4 titrated with NaOHpKa1 = 2.27, pKa2 = 7.21, pKa3 = 12.25