Question

$100ml$ of $0.1N$ $NaOH$ is mixed with $100ml$ of $0.1N{H}_{2}S{O}_4$. The $pH$ of the resultant solution is :

• A. $<7$
• B. $>7$
• C. $=7$
• D. Cannot be predicated

Answer 1

The correct option is A $<7$

equivalent of $NaOH=0.1\times 0.1=0.01$
equivalents of $H_{2}S{O}_4=0.1\times 0.1=0.01$
moles of $H_{2}S{O}_4=\frac{0.01}{2}=0.005$
So $0.05$ moles $H_{2}S{O}_4$ will react with $0.05$ mole $NaOH$ and $\frac{0.005}{0.2L}=\mathrm{0.025.}$
$P^{o}H=-log\left(O{H}^{-}\right)=1.60$
$pH=13.39$
So$pH>7$