Question

To a $10mL$ of ${10}^{-3}N{H}_{2}S{O}_4$ solution, water has been added to make the total volume of one litre. Its $pOH$ would be

• A. $3$
• B. $12$
• C. $9$
• D. $5$

Answer 1

Answer: (C)

$9$

$pH=-log\left[H_3{O}^{+}\right]=-log\left[H^{+}\right]$

$10$ mL of ${10}^{-3}$ N $H_{2}S{O}_4$ has final

$\left[H^{+}\right]=\frac{{10}^{-3}\times 10}{1000}={10}^{-5}M$

$\left[H^{+}\right]={10}^{-5}M$ and $pH=-log\left[H^{+}\right]=5$

$pOH=14-5=9$