Question

The $pH$ of a solution formed by mixing $40$ ml of $0.10M$ $HCl$ with $10$ ml of $0.45M$ of $NaOH$ is:

• A. $12$
• B. $10$
• C. $8$
• D. $6$

Answer 1

The correct option is A $12$

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

The number of mmol of $HCl$ $=40mL\times 0.10M=4mmol$

The number of mmol of $NaOH$ $=10mL\times 0.45M=4.5mmol$

4 mmol of $HCl$ will neutralise 4 mmol of $NaOH$.

$4.5-4=0.5$ mmol of $NaOH$ will remain.

Total volume $=40mL+10mL=50mL$

$\left[O{H}^{-}\right]=\left[NaOH\right]=\frac{0.5mmol}{50mL}=0.01M$

$pOH=-lo{g}_{10}\left[O{H}^{-}\right]=-lo{g}_{10}\left(0.01\right)=2$

$pH=14-pOH=14-2=12$

Hence, option $A$ is correct.