Question

The pH of the solution obtained by mixing $50ml$ of $0.4N$ $HCI$ and $50ml$ of $0.2N$ $NaOH$ is:

• A. $-\log 2$
• B. $-\log 0.2$
• C. $1.0$
• D. $2.0$

Answer 1

The correct option is C $1.0$

$NaOH+HCl\to NaCl+H_{2}O$
1 mole 1 mole

(For $HCl$ and $NaOH$ normality = molarity)
$M_f=\frac{\frac{N_1{V}_1}{n_1}-\frac{N_2{V}_2}{n_2}}{V_1+V_2}$
$\frac{\frac{50\times 0.4}{1}-\frac{50\times 0.2}{1}}{50+50}$
$=\frac{20-10}{100}=\frac{1}{10}or{10}^{-1}M$
$\left[H^{+}\right]={10}^{-1}M$;

$pH=-log\left({10}^{-1}\right)$
$pH=1$