An explanation that makes reference to the following points:
- (when half the acid has been neutralised)
(1)
[CH3CH2COOH]=[CH3CH2COO−]
(1)
- evaluation of pH
Allow in words Allow pH at half neutralisation =pKa propanoic acid Allow [H+]at half neutralisation =K a propanoic acid
pH=−log101.30×10−5
=4.8861 = 4.89/4.9
Ignore SF except 1SF
Correct answer scores 2