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. `-log2`
B. `-log0.2`
C. `1.0`
D. `2.0`

Answer

C

Explanation

`[H^(+)]=(V_(a)N_(a)-V_(b)N_(b))/(V_(a)+V_(b))`
`[H^(+)]=(50xx0.4-50xx0.2)/(100)`
`[H^(+)]=(20-10)/(100)=10^(-1)`
`pH=-log10^(-1),pH=1`