`20ml` of `0.2 M NaOH` is added to `50 ml`, of `0.2 M CH_(3)COOH` to give `70 ml`, of the solution. What is the `pH` of the solution? The ionization constant of acetic acid is `2xx10^(-5)`

Options

A. `4.522`
B. `5.568`
C. `6.522`
D. `7.568`

Answer

A

Explanation

The addition of `NaOH` converts equivalent amount of acetic acid into sodium acetate. Hence,
Concentration of acetic acid after the addition of
`NaOH=(30)/(70)xx0.2 M`
Concentration of `CH_(3)COONa` after the addition of
`NaOH=(20)/(70)xx0.2 M`
Hence, using the expression
`pH=pK_(a)+log``([Salt])/([Acid])`
`= -log(2xx10^(-5))+log``((20)/(30))`
`=4.699-0.177=4.522`