pH Calculations

1)Calculate PH of a solution of strong acid

 

pH of HCl solution with concentration 0.1 M

 

[H3O+]= 0.1 M (total dissociation)

 

pH = -log[H3O+] = -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 = [H3O+] [C3H7CO2-] / [C3H7CO2H]

 

[H3O+] and [C3H7CO2-] are equal, because every acid molecule that dissocates provides 1 one of each. .Hence, [H3O+] [C3H7CO2-]

= [H3O+] * [H3O+]

 

[C3H7CO2H] 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, [H3O+] * [H3O+] = [C3H7CO2H] * Ka

 

Substituing we get:

 

[H3O+] = 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 H3O+ is also x:

x2 = (0.01 - x) Ka

x2 = 1.5 *10 -7 - 1.5 * 10 -5 x

x2 + 1.5 * 10 -5 x - 1.5 *10 -7 = 0

Solving this quadratic equation gives:

x = [H3O+] = 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:

H2SO4 + H2O -->H3O+ + HSO4-      K1 = very large

H2O + HSO4-  --> H3O+ + SO42-     K2 = 0.01

If the concentration of the hydrogen ions from the second ionisation only is x, then the concentration of  HSO4- 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  K2, we have:

K2 = 

[H3O+] [SO42-]
[HSO4-]

= (0.1 + x) x
(0.1 – x)

= 0.01 mol dm-3

Thus after a little algebra we obtain

x2 + 0.11 x – 0.001 = 0

x = 8.44 x 10-3M.

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.

 

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