A Level chem- enthalpy of neutralisation

This is because as the base (say NaOH) is continuously added, the H+ ions are continuously removed, thus in order to maintain equilibrium, the CH3COOH should keep dissociating so as to compensate those H+ ions removed.
Now, let's see a numerical example:
Suppose 1M of ethanoic acid is neutralized by equal volume of 1M NaOH, then after mixing, there will be 0.5M of CH3COONa (since the volume of mixture is doubled) at the neutralization point and hence:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Chem/Buffer14.jpg

From the result above, we can see that only 5.13 % of ethanoic acid molecules remain unionized and hence 94.87% of H+ ions have been evolved.
Therefore the enthalpy of neutralization is -57 × 94.87% = -54.1 kJ/mol
which closely agrees with the value given in the question.