A weak acid of dissociation constant `10^(-5)` is being titrated with aqueous NaOH solution . The pH at the point of one third of neutralization of the acid will be

Options

A. 5 log 2 - log 3
B. 5 - log 2
C. 5 - log 3
D. 5 - log 6

Answer

B

Explanation

On partial neutralization of the weak acid, salt is formed. Hence, it becomes a buffer
`pH = pK_(a) + log. (["Salt"])/(["Acid"])`
1/3rd neutralisation of the acid means out of 1 mole of the acid , salt formed = 1/3 mole and acid left = 2/3 mole
Hence, `pH = - log(10^(-5))+log.(1//3)/(2//3)`
`=5+ log .(1)/(2) = 5 - log2`.