Solutions for the “Calculation of pH in buffer systems”
1. What is the pH of the buffer solution in which the concentration of acetic acid is
0.200 M and that of sodium acetate is 0.500 M?
cacetic acid = 0.200 M
cNa-acetate = 0.500 M
Ka = 1.86 × 10–5
[H+] = Ka
c weak acid
=
c weak base
1.86 × 10 −5 ×
0.200
= 7.44 × 10–6 M
0.500
pH = – log [H+]
pH = – log 7.44 × 10–6
pH = 5.128
2. What is the pH of the buffer solution that was prepared by mixing 100 mL of 0.200 M
NH3 solution and 40.00 mL of 0.100 M HCl solution?
V(NH3 solution) = 100 mL
n(base) = nbase - nHCl
c(NH3 solution) = 0.200 M
n(base) = 0.02 – 0.004
n(NH3) = 0.1 × 0.2 = 0.02 mol
n(base) = 0.016 mol
V(HCl solution) = 40 mL
n(salt) = 0.004 mol
c(HCl solution) = 0.100 M
n(HCl) = 0.1 × 0.04 = 0.004 mol
[OH–] = Kb
n weak base
=
n salt
1.75 × 10 −5 ×
0.016
= 7 × 10–5 M
0.004
pOH = – log [OH–]
pOH = – log 7 × 10–5
pOH = 4.155
pH = 14.000 – 4.155 = 9.845
3. What is the buffer capacity of an acidic buffer in which csalt = cacid = 1.0 M? What is
the change of pH in this buffer upon 10-fold dilution?
n weak acid
1
= Ka = Ka
n weak base
1
The effect of a strong acid (molar amount added to 1 L buffer is Y):
[H+]starting = Ka
1
[H+] = 10×[H+]starting = 10 × Ka = Ka ×
1+ Y
1- Y
1+ Y
1- Y
10 – 10Y = 1 + Y
Y = 0.81818 M
The effect of a strong base (molar amount added to 1 L buffer is Z):
1− Z
[H+] = 0.1×[H+]starting = 0.1 × Ka = Ka ×
1+ Z
1− Z
0.1 =
1+ Z
0.1 + 0.1Z = 1 – Z
Z = 0.81818 M
10 =
Dilution does not change the pH of this buffer, because [H+] = Ka
n weak acid
and the
n weak base
n weak acid
ratio does not cange with dilution.
n weak base
4. What is the buffer capacity of a 1.0 M HCl solution? What is the change of pH in this
buffer upon 10-fold dilution?
[H+]starting = 1 M, so pHstarting = 0.00
The effect of a strong acid:
[H+] = 10 M
The added amount of strong acid is 10M – 1M = 9 M
The effect of a strong base:
[H+] = 0.10 M
The added amount of strong base is 1M – 0.1M = 0.9 M
After a 10-fold dilution, [H+] = 0.10 M, pH = 1.00
5. What volume of 0.1 M NaOH or 0.1 M HCl causes 1 unit change in the pH of 1 dm3 of
acetate buffer, which has a pH of 5.000 and the sum of the acetate and the acetic acid
concentrations is exactly 1.00 M?
pH = 5.00, so [H+] = 10-5 M
n
[H+] = Ka weak acid
n weak base
n weak acid
n
10-5 = 1.86×10-5×
= 1.86×10-5× weak acid
n weak base (or salt)
1 - n weak acid
By solving this equation, nweak acid = 0.34965 M and nweak base = 0.65035 M
The effect of a strong acid (molar amount: X):
pH = 5.00 – 1 = 4.00, so [H+] = 10-4 M
0.34965 + X
10-4 = 1.86×10-5×
0.65035 - X
X = 0.49352 mol strong acid can be added to decrease the pH by 1 unit.
We know the concentration (0.1 M), so the volume can be calculated:
2
n
0.49352 mol
=
= 4.9352 L
0.1 mol/L
c
The effect of a strong base (molar amount: Y):
pH = 5.00 + 1 = 6.00, so [H+] = 10-6 M
0.34965 − Y
10-6 = 1.86×10-5×
0.65035 + Y
Y = 0.2986 mol strong base can be added to increase the pH by 1 unit.
We know the concentration (0.1 M), so the volume can be calculated:
n
0.2986 mol
V= =
= 2.986 L
0.1 mol/L
c
V=
6. A buffer with pH = 9.25 is needed from 100 cm3 of 2.00 M ammonia solution and an
unlimited amount of solid NH4Cl. What is the mass of ammonium chloride needed?
What is the final pH of this buffer solution after the addition of 50 or 500 cm3 of 1.00
M HCl?
pH = 9.25
pOH = 14 – pH = 4.75
[OH-] = 10-pOH = 1.778×10-5 M
c
[OH-] = Kb ⋅ weak base
c salt
2M
1.778 × 10-5 = 1.75 × 10-5 ⋅
c salt
From this, csalt = 1.9682 M. We know the volume (100 mL), so the molar amount of the
salt can be calculated: n = c × V = 1.9682 M × 0.1 L = 0.19682 mol.
We also know the M.W. for NH4Cl (53.5 g/mol), so m = n × M.W. = 10.53 g NH4Cl
After the addition of 50 mL 1 M HCl:
nHCl = 0.050 mL × 1 M = 0.050 mol
n
0.2 - 0.050
[OH-] = Kb ⋅ weak base = 1.75 × 10-5 ⋅
=1.06352 × 10-5 M
0.19682 + 0.050
n salt
-5
pOH = -log(1.06352 × 10 ) = 4.973, pH = 14.00 – pOH = 9.027
After the addition of 500 mL 1 M HCl:
nHCl = 0.50 mL × 1 M = 0.50 mol
The total volume of the resulting solution is 600 mL (500 mL + 100 mL)
In this solution, we have 0.39682 mol NH4Cl (0.2 + 0.19682) and 0.3 mol HCl (0.5 – 0.2).
So, take only the strong acid into consideration.
0.3 mol
cHCl =
= 0.5 M = [H+], and pH = -log(0.5) = 0.301
0.6 L
3