Solutions for the problems about
„Calculation of pH in the case of polyprotic acids and bases”
1. 20 cm3 of phosphoric acid solution is titrated with 0.075 M sodium hydroxide using
phenolphthalein indicator ( pK aphenolphtalein = 9.8 ). The average volume of sodium
hydroxide solution needed to reach the equivalence point of the titration is 13.25 cm3.
What is the molarity of the phosphoric acid solution?
14
100%
10
70%
pKa2 = 7,21
8
60%
50%
pH
Molar fraction
80%
pKa3 = 11,75
12
A
(purple)
HA
(colorless)
90%
pKaphenolphtalein = 9.8
40%
6
30%
4
20%
pKa1 = 2,12
10%
2
0%
0
2
4
6
pH
8
10
12
14
0
0
The distribution curve of the indicator,
together with its colors as a function of pH
1
2
equivalents
3
4
The titration curve of phosphoric acid,
showing also the colors of the indicator
We can see from the titration curve of phosphoric acid that phenolphthalein indicator
changes its color after adding two equivalents of strong base to the titrated solution (that is
pK a3 + pK a2 11.75 + 7.21
pK aphenolphtalein is around the pH for Na2HPO4 which is
=
= 9.48).
2
2
So, the reaction we can monitor by this titration is:
H3PO4 + 2 NaOH → Na2HPO4 + 2 H2O
The molar amount of NaOH is: nNaOH = cNaOH × VNaOH = 0.075 M × 13.25 cm3 = 0.99375
mmol = 0.00099375 mol. From the previous chemical equation, nH3PO4 = 0.5 × nNaOH =
0.496875 mmol. This amount is in 20 mL of phosphoric acid solution, so the concentration
n H PO
0.496875 mmol
of this solution is: c H 3 PO 4 = 3 4 =
= 0.02484 mol/L
VH 3 PO 4
20 mL
2. Calculate the pH when mixing...
a) 10 cm3 of 0.04 M sulphuric acid and 10 cm3 of 0.115 M sodium hydroxide solutions
b) 10 cm3 of 0.12 M sodium carbonate and 20 cm3 of 0.06 M sodium hydroxide
solutions
Does the pH of the solution mixtures change by 10-fold dilution?
a) The molar amounts are: nH2SO4 = cH2SO4 × VH2SO4 = 0.04 M × 10 cm3 = 0.4 mmol and
nNaOH = cNaOH × VNaOH = 0.115 M × 10 cm3 = 1.15 mmol
We have a more then 2-fold excess of NaOH, so the acid-base reaction is:
H2SO4 + 2 NaOH → Na2SO4 + 2 H2O
The amount of NaOH excess (which determines the pH of the solution) is: n excess
NaOH = nNaOH
− (2 × nH2SO4) = 0.35 mmol. The total volume of the resulting solution is 20 mL
VH2SO4 + VNaOH), so the concentration of the NaOH excess is:
c
excess
NaOH
n excess
0.35 mmol
= NaOH =
= 0.0175 M.
Vtotal
20 mL
(=
In the case of a strong base (like NaOH), [OH–] = cstrong base = 0.0175 M
pOH = -log[OH-] = -log(0.0175) = 1.757
pH = 14.00 – pOH = 14.00 – 1.757 = 12.243
This is a strong base solution, so the pH does change by 10-fold dilution. The pH of the
solution after the dilution is:
[OH–]10-fold dilution = 0.00175 M
pOH = -log[OH-] = -log(0.00175) = 2.757
pH = 14.00 – pOH = 14.00 – 2.757 = 11.243.
b) The molar amounts are: nNa2CO3 = cNa2CO3 × VNa2CO3 = 0.12 M × 10 cm3 = 1.2 mmol and
nNaOH = cNaOH × VNaOH = 0.06 M × 20 cm3 = 1.2 mmol
Na2CO3 is a weak base and NaOH is a strong base, so to calculate the pH we need to take
into account only the strong base. The total volume of the resulting solution is 30 mL (=
VNa2CO3 + VNaOH), so the concentration of the NaOH is: c NaOH =
n NaOH 1.2 mmol
=
=0.04 M.
Vtotal
30 mL
In the case of a strong base (like NaOH), [OH–] = cstrong base = 0.04 M
pOH = -log[OH-] = -log(0.04) = 1.398
pH = 14.00 – pOH = 14.00 – 1.398 = 12.602
This is a strong base solution, so the pH does change by 10-fold dilution. The pH of the
solution after the dilution is:
[OH–]10-fold dilution = 0.004 M
pOH = -log[OH-] = -log(0.00175) = 2.398
pH = 14.00 – pOH = 14.00 – 2.398 = 11.602.
3. We have 21.50 g of sodium phosphate and an optional amount of concentrated
hydrochloric acid (36 %(m/m), ρ = 1.18 g/cm3) to prepare 100.00 cm3 pH = 7.000
buffer. What volume of concentrated hydrochloric acid should be used if we use up the
total amount of sodium phosphate?
100%
H3PO4
90%
H2PO4-
HPO42-
PO43-
Molar fraction
80%
70%
60%
pH = 7.000
50%
40%
30%
20%
10%
0%
0
2
4
6
pH
8
10
12
14
Distribution curve of phosphoric acid
At pH = 7.000, [H+] = 10–pH = 1.00×10–10 M.
From this chart, we can see that at pH 7.00 there is a H 2 PO −4 / HPO 24 − (or
NaH2PO4/Na2HPO4) buffer system. To calculate the pH of this buffer, we use the
c
n
n
−
−
−
[H + ]
following formula: [H+] = Ka2 H 2 PO 4 = Ka2 H 2 PO 4 . So, H 2 PO 4 =
= 1.621 and nNaH2PO4
c HPO 2−
n HPO 2−
n HPO 2−
K a2
4
= 1.621 × nNa2HPO4.
4
4
At the same time, we know that the molar amount of the original Na3PO4 (nNa3PO4
=
m Na
3 PO 4
M .W .Na
=
3 PO 4
21.50 g
= 0.13114 mol) equals the sum of the molar amount of
163.95 g/mol
NaH2PO4 and Na2HPO4 in the resulting buffer solution. That is, nNa3PO4 = 0.13114 mol =
nNaH2PO4 + nNa2HPO4 = (1.621 × nNa2HPO4) + nNa2HPO4 = 2.621 × nNa2HPO4. From this, nNa2HPO4 =
0.13114 mol / 2.621 = 0.05004 mol and nNaH2PO4 = 1.621 × nNa2HPO4 = 1.621 × 0.0500 mol =
0.08110 mol. The reactions between HCl and Na3PO4 are the following:
Na3PO4 + HCl → Na2HPO4 + H2O
Na3PO4 + 2 HCl → NaH2PO4 + 2 H2O
to make Na2HPO4 and
to make NaH2PO4.
The total molar amount of HCl needed is: nHCl = nNa2HPO4 + (2 × nNaH2PO4 ) = 0.21224 mol.
The mass of HCl is: mHCl = nHCl × M.W.HCl = 0.21224 mol × 36.458 g/mol = 7.7385 g
HCl. The mass of the solution containing this mass of HCl is: msolution =
7.7385 g × 100%
= 21.494 g solution. The volume of this solution is: Vsolution =
36%
msolution
21.494 g
=
= 18.21 cm3.
ρsolution 1.18 g/cm3
4. A 0.1 M glycine (amino acetic acid) solution is prepared by weighing the calculated
amount of glycine and dissolving it in water. What is the pH of the solution?
Glycine is an ampholite, so its pH is: pH =
pK a1 + pK a2 2.350 + 9.550
=
= 5.950
2
2
5. What is the pH of a sodium hydrogen carbonate solution?
NaHCO3 is an ampholite, so its pH is: pH =
pK a1 + pK a2 6.367 + 10.252
=
= 8.309
2
2
6. What is the pH of a 0.05 M sulphuric acid solution?
Sulphuric acid has a strong first and a weak second deprotonation step. So, its reactions in
water are:
H2SO4 → HSO −4 + H+
HSO −4
SO 24 − + H+
Ka2 =
[H + ][SO 24 − ]
[HSO −4 ]
The concentration of H+ ion formed in the first deprotonation reaction is 0.05 M, and the
concentration of H+ ion formed in the second deprotonation step is X M. So:
[H+]total = (0.05 + X) M
[ HSO −4 ]total = (0.05 – X) M
[ SO 24 − ]total = X M
Ka2
[H + ][SO 24 − ]
(0.05 + X) ⋅ X
=
, so 1.2·10–2 =
. Solving this equation will give X =
−
[HSO 4 ]
(0.05 − X)
0.0085095 M, so [H+]total = (0.05 + X) M = 0.0585095 M. From this, pH = 1.233.
7. Specify the following acid/base systems and calculate their pH!
a) 15 cm3 of 0.05 M phosphoric acid solution + 37.5 cm3 of 0.04 M sodium hydroxide
solution
b) 10 cm3 of 0.05 M sodium carbonate solution + 5 cm3 of 0.05 M of sodium chloride
solution
a) The molar amounts are: nH3PO4 = cH3PO4 × VH3PO4 = 0.05 M × 15 cm3 = 0.75 mmol and
nNaOH = cNaOH × VNaOH = 0.04 M × 37.5 cm3 = 1.5 mmol
We have a 2-fold excess of NaOH, so the acid-base reaction is:
H3PO4 + 2 NaOH → Na2HPO4 + 2 H2O
pK a2 + pK a3 7.210 + 11.750
Na2HPO4 is an ampholite, so its pH is: pH =
=
= 9.480
2
2
b) The molar amounts are: nNa2CO3 = cNa2CO3 × VNa2CO3 = 0.05 M × 10 cm3 = 0.5 mmol.
NaCl has no effect on the pH. So, the resulting solution is a weak base (Na2CO3), and its
n Na CO
0.5 mmol
2
3
=
=
concentration after mixing the two solutions is: cNa2CO3 =
Vtotal
10 cm3 + 5 cm3
10−14
K
0.03& M. For Na2CO3, Kb1 = W =
= 1.786·10–4. In this case, there is no 3 log
K a2 5.60 ⋅ 10−11
unit difference between cweak
1.786 ⋅ 10 − 4
acid
c weak base − [OH − ]
and Kb1, so [OH ] = K b1 ⋅
=
[OH − ]
−
0.03& − [OH − ]
. From this, [OH–] = 0.002352 M.
−
[OH ]
pOH = -log[OH-] = -log(0.002352) = 2.629
pH = 14.00 – pOH = 14.00 – 2.629 = 11.371
8. Nicotine (C10H14N2) is a diprotic weak base.
a) What is the pH of a 0.05 M nicotine solution?
b) What is the pH after adding one equivalent of hydrochloric acid?
c) Is it possible to determine the concentration of a nicotine solution by acid-base
titration? If yes, how can it be done?
a) Nicotine is a weak base, and cweak
base
>>> Kb1, so [OH–] = K b1 ⋅ c weak base =
(7.0 ⋅ 10−7 ) ⋅ 0.05 = 1.87·10–4 M.
pOH = -log[OH-] = -log(1.87·10–4) = 3.728
pH = 14.00 – pOH = 14.00 – 3.728 = 10.272
b) After adding one equivalent of strong acid to a diprotic weak base, we have an
pK b1 + pK b2 6.155 + 10.854
ampholite and its pOH is: pOH =
=
= 8.504
2
2
pH = 14.00 – pOH = 14.00 – 8.505 = 5.496
c) It is possible to determine the concentration of a nicotine by acid-base titration, using a
strong acid (e.g. HCl solution) to titrate the nicotine solution with an indicator changing its
color around pH 5.496. Using this indicator, the acid-base reaction is:
nicotine + HCl → nicotine·HCl
So, one molecule of nicotine reacts with one equivalent of strong acid.
100%
90%
80%
0
12
2
10
4
nicotine·HCl
70%
60%nicotine·2HCl
50%
pK b1 = 6.155
8
pH
pK a2 = 7.845
6
40%
30%
4
pK a1 = 3.146
20%
10%
6
pKaindicator = 5.5
8
pOH
Molar fraction
14
nicotine
10
pK b2 = 10.854
2
12
0%
0
2
4
6
pH
8
10
12
14
0
14
0
Distribution curve of nicotine as a function of pH
1
HCl equivalents
2
3
Titration of nicotine with HCl titrant
9. 50 cm3 of 0.200 M maleic acid solution is titrated by 0.500 M sodium hydroxide
solution. Calculate the volume of titrant and the pH...
a) after adding one equivalent of base.
b) after adding two equivalents of base.
a) Maleic acid is a biprotic weak acid (H2mal), so, after adding one equivalent of base, we
have got an ampholite (Hmal– or NaHmal), and its pH is: pH =
pK a1 + pK a2 2.000 + 6.260
= 4.130
=
2
2
The acid-base reaction is:
H2mal + NaOH → NaHmal + H2O
The amount of maleic acid in the 50 cm3 sample is: nH2mal = cH2mal × VH2mal = 0.2 M × 50
cm3 = 10 mmol. From the acid-base equation, nNaOH = nH2mal = 10 mmol. So, the volume of
n
10 mmol
base added is: VNaOH = NaOH =
= 20 cm3.
c NaOH 0.500 M
b) After adding two equivalents of base, the resulting solution is a weak base (Na2mal), so
to calculate the pH, we need its concentration and its pKb. The acid-base reaction is:
H2mal + 2 NaOH → Na2mal + 2 H2O
The amount of maleic acid in the 50 cm3 sample is: nH2mal = cH2mal × VH2mal = 0.2 M × 50
cm3 = 10 mmol. From the acid-base equation, nNaOH = 2 × nH2mal = 20 mmol. So, the
volume of base added is: VNaOH =
n NaOH 20 mmol
=
= 40 cm3.
c NaOH 0.500 M
The concentration of the formed weak base (Na2mal) after mixing the two solutions is:
n Na 2 mal
10 mmol
cNa2mal =
= 0.1& M.
=
Vtotal
50 cm 3 + 40 cm 3
At the same time, Kb1 =
10 −14
KW
=
= 1.8& 1& ·10–8. In this case, there is more than 3 log
K a2 5.50 ⋅ 10 −7
unit difference between cweak acid and Kb1, so [OH − ] = K b1 ⋅ c weak base = 1.8& 1& ⋅ 10 −8 × ⋅0.1& =
4.495·10−5 M.
pOH = -log[OH-] = -log(4.495·10−5) = 4.347
pH = 14.00 – pOH = 14.00 – 2.629 = 9.653
10. Calculate the % distribution of the species formed in a solution of phthalic acid at pH
= 4.50.
pH = 4.50, so [H+] = 10–pH = 3.16227·10–5 mol/dm3. We also know the Ka1 and Ka2
values for phthalic acid, so:
1
= 0.10705, that is 10.705 % of A2– (pht2–).
X A 2− =
+
[H ] [H + ] 2
+
1+
K a2 K a1 K a2
X HA −
[H + ]
K a2
=
= 0.02495, that is 2.495 % of HA– (Hpht–).
+
+ 2
[H ] [H ]
1+
+
K a2 K a1 K a2
X H 2A
[H + ] 2
K a1 K a2
=
= 0.86800, that is 86.800 % of H2A (H2pht).
[H + ] [H + ] 2
1+
+
K a2 K a1 K a2