Question

At ${90}^o$C, pure water has $\left[H^{+}\right]={10}^{-6}$M, if $100$ mL of $0.2$ M $HCl$ is added to $200$ mL of $0.1$ M $KOH$ at ${90}^o$C then pH of the resulting solution will be:

• A. $7$
• B. $8$
• C. $4$
• D. $6$

Answer 1

The correct option is D $6$

100 mL of 0.2 M $HCl$ will completely neutralise 200 mL of 0.1 M $KOH$. The resulting solution will be neutral with $\left[H^{+}\right]={10}^{-6}M$ and
$pH=-log\left[H^{+}\right]=-log{10}^{-6}=6$