Question

The pH of a solution obtaine by mixing `50 mL` of `0.4 N HCl` and `50 mL` of `0.2 N NaOH` is
A. `log 2`
B. `-log0.2`
C. `1.0`
D. `2.0`

Answer 1

Correct Answer - C
`[H^(+)] = (V_(a)N_(a) - V_(b) N_(b))/(V_(a) + V_(b))`
`[H^(+)] = (50 xx 0.4 - 50 xx 0.2)/(100)`
`[H^(+)] = (20 - 10)/(100) = 10^(-1)`
`pH = -log 10^(-1) , pH = 1`