The pH of a solution obtaine by mixing `50 mL` of `0.4 N HCl` and `50 mL` of `0.2 N NaOH` is

Options

A. `-log 2`
B. `-log 0.2`
C. `1.0`
D. `2.0`

Answer

C

Explanation

M. eq. of 50 ml 0.4 N HCl `= 0.4 xx 50 = 20`
M. eq. of 50 ml, 0.2 N NaOH `= 0.2 xx 50 = 10`
Remaining M.eq. of HCl = 20 - 10 = 10
`[H^(+)]` in `HCl = (10)/(50+50) = 0.1 n = 10^(-1)`
`pH = - log 10^(-1) = 1`.