`pH` of a mixture of `1 M` benzoic acid (`pK_(a)=4.20`) and `1 M C_(6)H_(5)COONa` is `4.5`. In `300ml` buffer, benzoic acid is `[log2=0.3]`

Options

A. 200 mL
B. 150 mL
C. 100 mL
D. 50 mL

Answer

C

Explanation

`pH=pK_(a)+log ((300-V_(1))xx1)/(1xxV_(1))` ,
Let `V_(1)` be volume of benzoic acid and thus `(300-V_(1))` is volume of benzoate.
`:. 4.5=4.2+log (300-V_(1))/(V_(1))`
`:. (300-V_(1))/(V_(1))=2`
`:. V_(1) = 100 mL`