Question

$1$mL of $0.01N$ $HCl$ is added to $999$ mL solution of $0.1N$ $N{a}_{2}S{O}_4$. The $pH$ of the resulting solution will be:

• A. $2$
• B. $7$
• C. $5$
• D. $1$

Answer 1

The correct option is C $5$

Given that

$HCl=0.01N\left(1mL\right)$; $N{a}_{2}S{O}_4=0.1N\left(999mL\right)$

As we know that the $pH$ of any solution only depend on $H^{+}$ ion concentration. The concentration of $H^{+}$ in above solution,

$H^{+}=0.01mol$ in $1mL$

Total volume of solution $=1+999=1000mL$

So, $H^{+}ions=\frac{0.01}{1000}={10}^{-5}N$

$pH=-\log \left[H^{+}\right]$

$pH=-\log \left({10}^{-5}\right)$

$pH=5$

Option C is correct.