Question

What shall be the $pH$ of a solution formed by mixing $10mL$ of $0.1M{H}_{2}S{O}_4$ and $10mL$ of $\frac{N}{10}$ $KOH$ ?

• A. $11.40$
• B. $8.64$
• C. $3.00$
• D. None of these

Answer 1

Answer: (D)

Milliequivalent of $H^{+}=10\times 0.1\times 2=0$
Milliequivalent of $O{H}^{-}=10\times \frac{1}{10}=1$
Normality of solution $=\frac{2-1}{20}=\frac{1}{20}=0.05$
Molarity of the resulting solution
$=2\times 0.05=0.1$
$pH$ of the solution $=-log[H^{+}]=-log(0.1)$
$=-log(10^{-1})=1$