The % change of ph

Question

There is 1 L of buffer containing 0.22 mol of HA(weak acid) and 0.020 mol of A^ - The initial pH of this solution was 4.01 and p*K_{a} of this solution is 5.30. Calculate the % change of pH after adding 0.020 mol of OH^ - . New pH should be calcualted by Henderson -Hesselbach equation , The % change of pH is given by: % change of pH = 100[neWpH - initialpH] / (initialpH)

Answer 1

Henderson Hasselbalch equation:

pH = pKa + log [conj.base] / [ weak acid]

Check pH of initial solution:

pH = 5.30 + log (0.020/0.22) = 5.30 - 1.04 = 4.26 This does NOT agree with the stated pH = 4.01.

Not sure there's any sense in proceeding, but if you want to, then pretend you didn't check the initial pH and move on, as follows:

When adding 0.02 mols of OH-, it will react with the weak acid, HA

HA + OH- ==> H2O + A-. So, the [HA] will decrease by 0.02 M and the [A-] will increase by 0.02 M

This leaves us with [HA] = 0.22 - 0.02 = 0.20 M

This leaves us with [A-] = 0.020 + 0.02 = 0.04 M

pH = pKa + log (0.04/0.20)

pH = 5.30 - 0.70

pH = 4.60

% change of pH = new pH - initial pH / initial pH (x100%)

% change of pH = 4.60 - 4.01 / 4.01 (x100%) = 14.7% change

If you use the calculated pH of the initial solution of 4.26 instead of 4.01, we get the following % change:

4.60 - 4.26 / 4.26 (x100%) = 7.98% change