`20 mL` of `0.2M` sodium hydroxide is added to `50 mL` of `0.2 M `acetic acid to give `70 mL` of the solution. What is the `pH` of this solution. Calculate the additional volume of `0.2M NaOh` required to make the `pH` of the solution `4.74`. (Ionisation constant of `CH_(3)COOh` is `1.8 xx 10^(-5))`

Explanation

20 ml of 0.2 M NaOH would react with 20 ml of 0.2 M acetic acid.
[Acid] = 30 ml of 0.2 M present in 70 ml
`=(30xx0.2)/70=6/70` mole
[Salt] = 20 ml of 0.2 M present in 70 ml
`=(20xx0.2)/70=6/70` mole
`pK_a=-log1.8xx10^(-5)` = 4.74
`pH=pK_a+"log""[salt]"/"[Acid]"=4.74+"log" 4/70xx70/6`
pH = 4.74 + log 0.66 = 4.74 – 0.18 = 4.56
To make a solution of pH = 4.74, [Acid] = [Salt]
So 25 ml of 0.2 M NaOH must be added to 50 ml of 0.2 M acetic acid.
Addition volume of NaOH to be added = 25 – 20 = 5 ml