pH Calculations
1)Calculate PH of a solution of strong acid
pH of HCl solution with concentration 0.1 M
[H_{3}O^{+}]= 0.1 M (total dissociation)
pH = log[H_{3}O^{+}] = log[0.1] = 1
the calculation for the strong base is similar
2)Calculate PH of a solution of weak acid
Calculate the pH of a 0.01 M solution of butanoic acid
a) using an approximation
butanoic acid has Ka = 1.5 * 10 ^{5} and concentration = 10 ^{2} M = 0.01
Ka = [H_{3}O^{+}] [C_{3}H_{7}CO_{2}^{}] / [C_{3}H_{7}CO_{2}H]
[H_{3}O^{+}] and [C_{3}H_{7}CO_{2}^{}] are equal, because every acid molecule that dissocates provides 1 one of each. .Hence, [H_{3}O^{+}] [C_{3}H_{7}CO_{2}^{}]
= [H_{3}O^{+}] * [H_{3}O^{+}]
[C_{3}H_{7}CO_{2}H] is only a little bit lower than the initial concentration, because the acid is weak. Hence, we consider its concentration to be the same as the initial one (this is an approximation).
So, [H_{3}O^{+}] * [H_{3}O^{+}] = [C_{3}H_{7}CO_{2}H] * Ka
Substituing we get:
[H_{3}O^{+}] = 3.87 * 10 ^{4 } M so that pH= 3.41
b) doing it exactly
Now we no longer make the approximation that the final concentration of acid equals the original: it diminishes by x, so that the concentration of H_{3}O^{+} is also x:
x^{2} = (0.01  x) Ka
x^{2} = 1.5 *10 ^{7}  1.5 * 10 ^{5} x
x^{2 }+ 1.5 * 10 ^{5} x  1.5 *10 ^{7} = 0
Solving this quadratic equation gives:
x = [H_{3}O^{+}] = 3.79 * 10 ^{4} so that pH = 3.42
In this case, t he result is almost identical to the one obtained using the approximation.
2)Calculate PH of a solution of a diprotic acid: 0.1 M solution of sulphuric acid, considering double ionization
Consider a 0.1 M solution of sulphuric
acid. The equilibrium in this solution are:
H_{2}SO_{4} + H_{2}O >H_{3}O^{+} +
HSO_{4}^{} K_{1} = very large
H_{2}O + HSO_{4}^{} > H_{3}O^{+}
+ SO_{4}^{2} K_{2 }=
0.01
If the concentration of the hydrogen ions from the second ionisation only is x, then the concentration of HSO_{4}^{} ions will be 0.1  x and the total concentration of hydrogen ions from both ionisations will be 0.1 + x. Since it is the total hydrogen ion concentration that appears
in the evaluation of K_{2}, we have:
K_{2} =

[H_{3}O^{+}] [SO_{4}^{2}] 
= (0.1 + x) x 
= 0.01 mol dm^{3}

Thus after a little algebra we obtain
x^{2} + 0.11 x – 0.001 = 0
x = 8.44 x 10^{3}M.
The total hydrogen ion concentration is 0.01 + x = 0.10845 M, which, using a sensible number of significant figures (0.109 M)
gives a pH of 0.96.