Buffers, Titration Curves, Complexation and Solubility Equilibria Sample Questions
1. What is the pH of a buffer solution made by dissolving 22.5 g NH4Cl and 32.0 g NH3 to form 650 mL of
solution? Look up any constants you might need.
22.5 g NH4Cl = 0.421 mol; [NH4+] = 0.65M NH4+
32.0 g NH3 = 1.87 mol; [NH3] = 2.89 M NH3
pH  pKa  log
[base]
2.89
 9.25  log
 9.90
[acid ]
0.65
2. 100 mL of each of the two noted solutions are mixed together. Which mixtures will result in buffer
solutions?
a) 0.10 M HCl + 0.10 M NaOH
b) 0.10 M NH3 + 0.10 M NaOH
c) 0.10 M CH3CO2H + 0.07 M NaCH3CO2
d) 0.10 M NaCl + 0.10 M CH3CO2H
e) 0.10 M HCl + 0.27 M NaCH3CO2
ANS = c because it gives a mixture of acetic acid and acetate ion
and e, because the HCl converts 0.10 M or CH3CO2- into CH3CO2H; this results in a mixture of
0.10 M CH3CO2H and the remaining 0.17 M CH3CO2-.
3. Consider this titration curve:
a) What is being titrated?
weak acid, weak base, strong acid, or strong base
A weak base is being titrated. You know it is a base because the pH starts above 7 and
decreases as the titration progresses. You know it is not strong because the pH drops a
lot right at the start.
b) if a weak acid, what is its Ka?; if a weak base, what is its Kb?
The midpoint of the titration is the pKa of the conjugate acid of the weak base. It is
about pH = 9.3, which means Ka = 10-9.3 = 5.0 x 10-10. For an acid-base pair, Ka x Kb = 1.0
x 10-14, so
Kb = 1.0 x 10-14/5.0 x 10-10 = 2.0 x 10-5.
c) what is the “titrant,“ a strong acid or a strong base?
Strong acid.
4. Using the Ksp value given, predict the solubility of Zn3(PO4)2 ; Ksp = 9.1 x 10-33
In grams per liter.
Zn3(PO4)2(s) ===== 3 Zn2+(aq) + 2 PO43-(aq)
Equil
3x
2x
Ksp = (3x)3(2x)2 = 108x5
x5
Ksp
 1.5 107 mol / L
108
solubility = 1.5x10-7 mol/L x 386.1 g/mol = 5.9x10-5 g/L
5. What will the solubility of Zn3(PO4)2 be in a solution of 0.22 M Na3PO4?
Zn3(PO4)2(s) ===== 3 Zn2+(aq) + 2 PO43-(aq)
Equil
3x
0.22 +2x = about 0.22
Ksp = (3x)3(0.22)2 = 1.31x3
x 3
Ksp
 1.9 1011 mol / L
1.31
solubility = 1.9 x 10-11 mol/L x 386 g/mol = 7.4 x 10-9 g/L
6. If 1.0 grams of ZnCl2 and 1.0 g of Na2CO3 are added to 100 mL of water, will a precipitate form?
1.0 g ZnCl2 = 0.00734 mol = 0.0743 M Zn2+
1.0 g Na2CO3 = 0.00944 mol = 0.0944 M CO32Q = [Zn2+][CO32-]= 7.0 x 10-3
Ksp = 1.5 x 10-11, so Q is much larger than Ksp, so, yes, a precipitate does form.