The pH of a solution obtained by mixing equal volumes of `(N)/(10)NaOH` and `(N)/(20) HCl`

Answer

12.398

Explanation

Suppose 1 L of N/10 NaOH is mixed with 1 L of N/20 HCl.
1 L of N/10 NaOH contains NaOH `=(1)/(10)" g eq." = 0.1 g eq`.
1 L of N/20 HCl contains HCl `=(1)/(20) g eq. = 0.05 g eq`.
0.05 g eq. of HCl will neutralise 0.05 g eq. of NaOH.
`:. ` NaOH left unneutralised = 0.05 g eq.
Volume of solution = 2 L
`:. `Concentration of NaOH in the final solution `= (0.05)/(2)" g eq" L^(-1) = 0.025 N`
`[NaOH]=[OH^(-)] = 0.025 M = 2.5 xx 10^(-2) :. [H_(3)O^(+)]=(10^(-14))/(2.5xx10^(-2))=4xx10^(-13) M`