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[0.1] = 1

 

the calculation for the strong base is similar

 
2)Calculate PH of a solution of weak acid

 

butanoic acid , K = 1.51 E-5 , concentration 10 E-2 M = 0.01

 

K = [H3O+] [C3H7CO2-] / [C3H7CO2H]

 

[H3O+] and [C3H7CO2-] are equal, because they come from the same molecule .Hence, [H3O+] [C3H7CO2-]

= [H3O+]* [H3O+]

 

[C3H7CO2H] is 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] * K

 

Substituing we get:

 

[H3O+] = 3.89 E-4 M and pH= 3.42

 

More accurate pH calculations:

 

Consider a 0.1 mol dm-3 solution of sulphuric acid. The equilibrium in this solution are:

H2SO4(aq) + H2O(l) à H3O+(aq) + HSO4- (aq)       K1 = very large

H2O(aq) + HSO4- (aq)  Û   H3O+(aq) + SO42-(aq)      K2 = 0.01 mol dm-3

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 =

- 0.11 ± Ö(0.112 + 0.004)  =
2

- 0.11 ± 0.1269
2

 

Since a concentration must be a positive quantity the negative root makes no physical sense; thus

x = 8.45 x 10-3 mol dm-3.

The total hydrogen ion concentration is therefore 0.10845 mol dm-3 which, using a sensible number of significant figures (0.109 mol dm-3) gives a pH of 0.96.