Question

The $pH$ of a solution obtained by mixing $50ml$ of $0.4N$ $N{H}_{4}Cl$ and $50ml$ of $0.2N$ $NaOH$ is:

• A. $-log2$
• B. $-log\left(0.2\right)$
• C. $1.0$
• D. $2.0$

Answer 1

The correct option is C $1.0$

The expression for the hydrogen ion concentration is as given below.

$\left[H^{+}\right]=\frac{V_b{N}_b-V_a{N}_a}{V_a+V_b}$
Substitute values in the above expression,

$\left[H^{+}\right]=\frac{50\times 0.4-50\times 0.2}{50+50}=0.1$
$pH=-log\left[H^{+}\right]=-log\left(0.1\right)=1$