# CHM 235 Quantitative Analysis ```CHM 235 Quantitative Analysis
Spring 2007
Dr. S.A. Skrabal
SOLUTIONS TO PROBLEM SET 11
Polyprotic acid-base equilbria
24 April 2007
Please refer to Appendix G in your textbook for required values of Ka.
Note: I have expressed values of pH and pOH to only two decimal places for convenience.
1. Indicate whether an aqueous solution of the following compounds is acidic, neutral, or basic. Explain
your answer. (A) CH3CO2NH4 (ammonium acetate) (B) Na2C2O4 (disodium oxalate) (C) NaNO2
(sodium nitrite) (D) Na2HPO4 (disodium hydrogen phosphate) (E) Na3PO4 (trisodium phosphate
(F) NaHC2O4 (sodium hydrogen oxalate)
(A) CH3CO2NH4 --&gt; CH3CO2- + NH4+. CH3CO2- is a weak acid with Kb = Kw/Ka = 1.00 x 10-14/1.75 x 10-5
= 5.71 x 10-10 and NH4+ is a weak acid with Ka = 5.70 x 10-10. Values of K are essentially the same, so
substance should be essentially neutral (same tendency to act as acid as base). (B) Na2C2O4 dissolves
to give 2 Na+ + C2O42-; C2O42- is fully basic form of diprotic base, so will form basic solution. (C) NaNO2
dissolves to form Na+ + NO2-; NO2- is conjugate base to HNO2 (nitrous acid; Ka = 7.1 x 10-4), so solution
should be basic. (D) Na2HPO4 dissolves to form 2 Na+ + HPO42-, the second intermediate form of
H3PO4. Solutions of this intermediate form have pH  (pKa2 + pKa3)/2  (7.199 + 12.15)/2  9.67, so
basic. (E) Na3PO4 dissolves to give 3 Na+ + PO43-, the fully basic form of H3PO4. This is weak base, so
basic. (F) NaHC2O4. This dissolves to give HC2O4-, the intermediate form of the diprotic acid, oxalic
acid. Solutions of intermediate forms have pH  (pKa1 + pKa2)/2  (1.252 + 4.266)/2  2.759, so acidic.
2. Calculate the pH of a solution that is 0.0500 M in: (A) H3PO4 (B) H3PO3 (C) H2S (D) H2C2O4
(E) H2SO3
(A) 0.0500 M H3PO4 (phosphoric acid). Fully acidic form of triprotic weak acid. Ka1 = x2/(F – x) = 7.11 x
10-3 = x2/(0.0500 – x); x2 + 7.11 x 10-3x – 3.56 x 10-4 = 0; x = [H3O+] = 1.56 x 10-2 M; pH = 1.81.
(B) 0.0500 M H3PO3 (phosphorous acid); Ka1 = 3.0 x 10-2 = x2/(0.0500 – x); x2 + 3.0 x 10-2x – 1.50 x 10-3
= 0; x = [H3O+] = 2.65 x 10-2 M; pH = 1.58 (C) 0.0500 M H2S (hydrogen sulfide);
Ka1 = 9.5 x 10-8 = x2/(0.0500 – x); x2 + 9.5 x 10-8x –4.75 x 10-9 = 0; x = [H3O+} = 6.89 x 10-5 M; pH = 4.16.
(D) 0.0500 M H2C2O4 (oxalic acid); Ka1 = 5.60 x 10-2 = x2/(0.0500 – x); x2 + 5.60 x 10-2 – 2.80 x 10-3 = 0;
x = [H3O+] = 3.19 x 10-2 M; pH = 1.50 (E) 0.0500 M H2SO3 (sulfurous acid); Ka1 = 1.23 x 10-2 =
x2/(0.0500 – x); x2 + 1.23 x 10-2x – 6.15 x 10-4 = 0; x = [H3O+] = 1.94 x 10-2 M; pH = 1.71.
3. Calculate the pH of a solution that is 0.0500 M in (A) NaH2PO4 (B) NaHC2O4 (C) NaH2PO3
(D) NaHSO3 (E) NaHS.
(A) [H3O+] 
K a1 K a 2 F  K a1 K w
=
K a1  F
(7.11x10 3 )(6.32 x10 8 )(0.0500)  (7.11x10 3 )(1.00 x10 14 )
 1.983x10 5 M ; pH  4.70
7.11x10 3  0.0500
or pH = (pKa1 + pKa2) / 2
(B) 0.0500 M NaHC2O4; HC2O4- is intermediate species (H2C2O4 is oxalic acid); same approach as
above:
[H3O+] 
(5.60 x10 2 )(5.42 x10 5 )(0.0500)  (5.60 x10 2 )(1.00 x10 14 )
 1.197 x10 3 M ; pH  2.92
2
5.60 x10  0.0500
(C) 0.0500 M NaH2PO3; H2PO3- is amphiprotic species (H3PO3 is phosphorous acid); same approach as
above:
(3.0 x10 2 )(1.62 x10 7 )(0.0500)  (3.0 x10 2 )(1.00 x10 14 )
 5.511x10 5 M ; pH  4.26
2
3.0 x10  0.0500
(D) 0.0500 M NaHSO3; HSO3- is intermediate species (H2SO3 is sulfurous acid); same approach as
above:
(1.23x10 2 )(6.6 x10 8 )(0.0500)  (1.23x10 2 )(1.00 x10 14 )
 2.553x10 5 M ; pH  4.59
2
1.23x10  0.0500
(E) 0.0500 M NaHS; HS- is intermediate species (H2S is hydrogen sulfide); same approach as above:
(9.5 x10 8 )(1.3x10 14 )(0.0500)  (9.5 x10 8 )(1.00 x10 14 )
 1.42 x10 10 M ; pH  9.85
8
9.5 x10  0.0500
4. Calculate the pH of a solution that is 0.0500 M in (A) Na3PO4 (B) Na2C2O4 (C) Na2HPO3
(D) Na2SO3 (E) Na2S
(A) 0.0500 M Na3PO4; PO43- is base; Kb1 = Kw/Ka3 = (1.00 x 10-14/7.1 x 10-13) = 1.41 x 10-2 = x2/(0.0500
– x); x2 + 1.41 x 10-2x – 7.05 x 10-4 = 0; x = [OH-] = 2.04 x 10-2 M; pOH = 1.69; pH = 12.31 (B) 0.0500
M Na2C2O4; C2O42- is base; Kb1 = Kw/Ka2 = (1.00 x 10-14/5.42 x 10-5) = 1.85 x 10-10 = x2/(0.0500 – x); x2 +
1.85 x 10-10x – 9.25 x 10-12 = 0; x = [OH-] = 3.04 x 10-6 M; pOH = 5.52; pH = 8.48 (C) 0.0500 M
Na2HPO3; HPO32- is fully basic form of the diprotic acid, H3PO3, phosphorous acid. Kb1 = Kw/Ka2 = (1.00
x 10-14/1.62 x 10-7) = 6.17 x 10-8 = x2/(0.0500 – x); x2 + 6.17 x 10-8x – 3.08 x 10-9 = 0; x = [OH-] = 5.55 x
10-5 M; pOH = 4.26; pH = 9.74 (D) 0.0500 M Na2SO3; SO32- is fully basic form of H2SO3, sulfurous acid;
Kb1 = Kw/Ka2 = (1.00 x 10-14/6.6 x 10-8) = 1.52 x 10-7 = x2/(0.0500 – x); x2 + 1.52 x 10-7x – 7.60 x 10-9 = 0;
x = [OH-] = 8.71 x 10-5 M; pOH = 4.06; pH = 9.94
(E) 0.0500 M Na2S; S2- is fully basic form of H2S; Kb1 = Kw/Ka2 = (1.00 x 10-14/1.3 x 10-14) = 7.69 x 10-1 =
x2/(0.0500 – x); x2 + 7.69 x 10-1x – 3.84 x 10-2 = 0; x = [OH-] = 4.71 x 10-2 M; pOH = 1.33; pH = 12.67.
5. Calculate the pH of a solution that is made up to contain the following formal concentrations:
(A) 0.0400 M H3AsO4 and 0.0250 M NaH2AsO4 (B) 0.0500 M NaH2AsO4 and 0.0300 M Na2HAsO4
(C) 0.0300 M Na2CO3 and 0.0600 M NaHCO3 (D) 0.0500 M H3PO4 and 0.0200 M NaH2PO4
(E) 0.0600 M NaHSO4 and 0.0300 M Na2SO4
(A) pH = pKa1 + log ([H2AsO4-]/[H3AsO4] = 2.24 + log (0.0250/0.0400) = 2.04
(B) pH = pKa2 + log([HAsO42-]/[H2AsO4-] = 6.96 + log(0.0300/0.0500) = 6.74
(C) pH = pKa2 + log ([CO32-]/[HCO3-] = 10.33 + log(0.0300/0.0600) = 10.03
(D) pH = pKa1 + log([H2PO4-]/[H3PO4] = 2.15 + log(0.0200/0.0500) = 1.75
(E) pH = pKa2 + log([SO42-]/[HSO4-] = 1.99 + log(0.0300/0.0600) = 1.69
6. Identify the dominant conjugate acid/base pair and calculate the ratio between them (not the actual
concentrations of each!) in a solution that is buffered to pH 6.00 and contains (A) H2SO3 (B) citric acid
(C) malonic acid (D) succinic acid.
Remember, best buffering when pH is near pKa. So acid-base pair that constitutes buffer of pH 6.00
will be determined by pKa of appropriate acid form. (A) H2SO3, pKa1 = 1.91, pKa2 = 7.18; pH = 6.00 =
pKa2 + log([SO32-]/[HSO3-]); 6.00 = 7.18 + log([SO32-]/[HSO3-]); -1.18 = log([SO32-]/[HSO3-]; 10-1.18 = [SO32]/[HSO3-] = 0.066 (B) Citric acid, pKa1 = 3.128, pKa2 = 4.761, pKa3 = 6.396;
6.00 = pKa3 + log([citr3-]/[Hcitr2-]); 6.00 + 6.396 + log([citr3-]/[Hcitr2-]); -0.396 = log([citr3-]/[Hcitr2-]); 10-0.396
= [citr3-]/[Hcitr2-] = 0.402 (C) Malonic acid, pKa1 = 2.847, pKa2 = 5.696; pH = pKa2 + log([malonate2]/[Hmalonate-]) = 6.00 = 5.696 + log([malonate2-]/[Hmalonate-]); 0.304 = log([malonate2-]/[Hmalonate-]);
100.304 = [malonate2-]/[Hmalonate-] = 2.01 (D) Succinic acid, pKa1 = 4.207, pKa2 = 5.636; pH = pKa2 +
log([succinate2-]/[Hsuccinate-]) = 6.00 = 5.636 + log([succinate2-]/[Hsuccinate-]); 0.364 = log([succinate2]/[Hsuccinate-]); 100.364= [succinate2-]/[Hsuccinate-] = 2.31.
7. How many grams of potassium hydrogen phthalate must be added to 500.0 mL of 0.0650 M phthalic
acid to prepare a buffer of pH 3.90?
Initial mols of H2Phth = (500.0 x 10-3 L)(0.0650 mol/L) = 3.25 x 10-2 mol H2Phth
pH = pKa1 + log (mol HPhth-/mol H2Phth); pKa1 = 2.95;
3.90 = 2.95 + log (mol HPhth-/3.25 x 10-2 mol H2Phth); 0.95 = log(mol HPhth-/3.25 x 10-2);
100.95 = mol HPhth-/3.25 x 10-2 mol HPhth- = (100.95)(3.25 x 10-2) = 0.290 mol HPhth-. Since 1 mol
HPhth- per mol KHP and FW of KHP = 204.23, g KPH = (0.290 mol KHP)(204.23 g KHP/mol KHP) =
59.2 g KHP
8. How would you prepare 1.00 L of a buffer with a pH of 9.60 from 0.300 M Na2CO3 and 0.200 M HCl?
(What volumes of each solution would you need?)
Remember, the best buffer is formed when pH is near pKa of the conjugate acid in the buffer. Two
conjugate acids are possible here: H2CO3 (carbonic acid) with pKa1 = 6.35 and HCO3- (bicarbonate ion)
with pKa2 = 10.33. The desired pH of 9.60 is closer to pKa2, so the buffer must consist of HCO3- and
CO32-. If you start with a solution of CO32- and add some strong acid, you will produce some HCO3-:
H3O+ + CO32-  HCO3- + H2O.
Using the H-H equation, we have
pH = pKa2 + log[(mol CO32-)/(mol HCO3-)]
9.60 = 10.33 + log[(mol CO32-)/(mol HCO3-)]
-0.73 = log[(mol CO32-)/(mol HCO3-)]
10-0.73 = (mol CO32-)/(mol HCO3-) = 0.186
So now we know the ratio between the number of moles of A2- and HA-; how do we get the moles of
each one? Since all the added moles of H3O+ from HCl turn (approximately) into HCO3-, then:
mols HCO3- = (x) (0.200 mol/L), where x = L of HCl.
The added H3O+ from the HCl reacts with the CO32- to decrease the initial concentration of CO32-.
Since we are limited to a total of 1.00 L,
initial mols of CO32- = (1 – x)(0.300 mol CO32-/L)
final mols of CO32- = (1 – x)(0.300 mol CO32-/L) – (x)(0.200 mol/L).
Plug these into the equation above solve for x, which is L of 0.200 M HCl:
0.186 = (mol CO32-)/(mol HCO3-)
0.186 = ((1 – x)(0.300) – 0.200x)/0.200x
0.186 = (0.300 – 0.300x – 0.200x)/0.200x
0.186 = (0.300 – 0.500x)/0.200x
0.0372x = 0.300 – 0.500x
0.5372x = 0.300
x = 0.558 L or 558 mL of 0.200 M HCl and
(1 – x) = 0.442 L or 442 mL of 0.300 M Na2CO3
9. What is the pH of a buffer prepared by mixing 50.0 mL of 0.200 M NaH2PO4 with (A) 50.0 mL of
0.120 M NaOH (B) 50.0 mL of 0.120 M HCl?
(A) H2PO4- reacts with strong base to form HPO42-. Initial mol OH- = (50.0 x 10-3 L)(0.120 mol/L) = 6.00
x 10-3 mol OH-. Initial mol H2PO4- = (50.0 x 10-3 L)(0.200 mol/L) = 1.00 x 10-2 mol H2PO4-. Final mol
HPO42- = 6.00 x 10-3 mol HPO42-. Final mol H2PO4- = 1.00 x 10-2 – 6.00 x 10-3 = 4.00 x 10-3 mol H2PO4-.
Buffer: pH = pKa2 + log(mol HPO42-/mol H2PO4-) = 7.199 + log(6.00 x 10-3/4.00 x 10-3) = 7.38.
(B) NaH2PO4 + strong acid react to give H3PO4 and H2PO4-. Initial mol H3O+ = (50.0 x 10-3 L)(0.120
mol/L) = 6.00 x 10-3 mol H3O+. Initial mol H2PO4- = (50.0 x 10-3 L)(0.200 mol/L) = 1.00 x 10-2 mol H2PO4-.
Final mol H3PO4 = 6;.00 x 10-3 mol H3PO4. Final mol H2PO4- = 1.00 x 10-2 – 6.00 x 10-3 = 4.00 x 10-3 mol
H2PO4-. Buffer: pH = pKa1 + log (mol H2PO4-/mol H3PO4) =
2.148 + log(4.00 x 10-3/6.00 x 10-3) = 1.97
10. A buffer is prepared by dissolving 5.00 g of Na2CO3 and 5.00 g of NaHCO3 in a total volume of
0.100 L of H2O. (A) What is the pH of the buffer? (B) What would be the pH of the solution after adding
1.00 mL of 0.50 M HNO3? (C) What would be the pH of the solution after adding 1.00 mL of 0.50 M
NaOH?
(A) Mols Na2CO3 = (5.00 g Na2CO3)(1 mol Na2CO3 /105.99 g Na2CO3) = 4.72 x 10-2 mol Na2CO3
Mols NaHCO3 = (5.00 g NaHCO3)(1 mol NaHCO3/84.01 g NaHCO3) = 5.95 x 10-2 mol NaHCO3
pKa2 of H2CO3 = 10.33; pH = 10.33 + log(mol Na2CO3/mol NaHCO3) =
10.33 + log(4.72 x 10-2/5.95 x 10-2) = 10.23
(B) Strong acid reacts with weak base (CO32-) to form conjugate weak acid (HCO3-):
Initial mols H3O+ = (1.00 x 10-3 L)(0.50 mol/L) = 5.0 x 10-4 mol H3O+
Final mol HCO3- = 5.95 x 10-2 + 5.0 x 10-4 = 6.0 x 10-2 mol HCO3Final mol CO32- = 4.72 x 10-2 – 5.0 x 10-4 = 4.7 x 10-2 mol CO32pH = 10.33 + log(4.7 x 10-2/6.0 x 10-2) = 10.22
(C) Strong base reacts with weak acid (HCO3-) to form conjugate weak base (CO32-):
Initial mols OH- = (1.00 x 10-3 L)(0.50 mol/L) = 5.0 x 10-4 mol OHFinal mol HCO3- = 5.95 x 10-2 – 5.0 x 10-4 = 5.9 x 10-2 mol HCO3Final mol CO32- = 4.72 x 10-2 + 5.0 x 10-4 = 4.8 x 10-2 mol CO32pH = 10.33 + log(4.8 x 10-2/5.9 x 10-2) = 10.24
11. What is the pH of 0.200 M solutions of the following forms of alanine: (A) H2A+ (B) HA (C) A- ?
(A) Fully acidic form of diprotic weak acid. Ka1 = x2/(F – x) = 4.49 x 10-3 = x2/(0.200 – x);
x2 + 4.49 x 10-3x – 8.98 x 10-4 = 0; x = [H3O+] = 2.78 x 10-2 M; pH = 1.56.
(B) Intermediate form:
(4.49 x10 3 )(1.36 x10 10 )(0.200)  (4.49 x10 3 )(1.00 x10 14 )
 7.73x10 7 M ; pH  6.11
4.49 x10 3  0.200
or pH = (pKa1 + pKa2) / 2 = (2.35 + 9.87) / 2 = 6.11
(C) Fully basic form of diprotic weak acid: Kb1 = Kw/Ka2 = 1.00 x 10-14/1.36 x 10-10 = 7.35 x 10-5 =
x2 / (F – x) = x2 / (0.200 – x ); x2 + 7.35 x 10-5x – 1.47 x 10-5 = 0 ; x = [OH-] = 3.80 x 10-3 M;
pOH = -log(3.80 x 10-3) = 2.42; pH = 14.00 – 2.42 = 11.58.
12. The fully protonated form of the amino acid leucine can be denoted H2Leu+. (A) What is the pH of a
0.126 M solution of H2Leu+? (B) What is the pH of a 0.059 M solution of the species HLeu?
(A) Fully acidic form of diprotic WA:
Ka1 = x2 / (F –x) = 4.69 x 10-3 = x2 / (0.126 – x)
In quadratic form: x2 + 4.69 x 10-3x – 5.91 x 10-4 = 0
Use quadratic eqn. to obtain x = [H3O+] = 2.21 x 10-2 M
pH = -log(2.21 x 10-2) = 1.66
Or assume x &lt;&lt; F, Ka1 = x2 / F to obtain x = 2.43 x 10-2 M; pH = 1.61. Reasonably good assumption.
(B) Intermediate form:
[H3O+] 
K a1 K a 2 F  K a1 K w
K a1  F

(4.69 x10 3 )(1.79 x10 10 )(0.059)  (4.69 x10 3 )(1.00 x10 14 )
(4.69 x10 3  0.059)
pH = -log(9.00 x 10-7) = 6.05
Or use pH  (pKa1 + pKa2) / 2
pKa1 = -log(4.69 x 10-3) = 2.33
pKa2 = -log(1.79 x 10-10) = 9.75
pH  (2.33 + 9.75) / 2 =6.04

9.00 x10 7 M
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