Question

The pH of a solution obtained by mixing $50$ mL of $0.4NN{H}_{4}Cl$ and $50$ mL of $0.2NNaOH$ is

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

Answer 1

Answer: (C)

$1.0$

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

Here, $\left[H^{+}\right]=N_1\times V_1-N_2\times V_2=50\times 0.4-50\times 0.2=10$

Volume of solution $=100$ mL

So, $\left[H^{+}\right]=\frac{10}{100}=0.1M$

$pH=1$