Question

The pH of the solution containing $10mL$ of $0.1NNaOH$ and $10mL$ of $0.05N$ $H_{2}S{O}_4$ would be:

• A. \N
• B. 1
• C. >7
• D. 7

Answer 1

The correct option is C >7

$\text{milliequivalent of}NaOH=10\times 0.1=1$
$\text{milliequivalent of}{H}_{2}S{O}_4=10\times 0.05=0.5$
$\text{Left milliequivalent of NaOH}=1-0.5=0.5M$
$\therefore \left[O{H}^{-}\right]=\frac{0.5}{10+10}=2.5\times {10}^{-2}$
we know the relation
$pOH=-log\left[O{H}^{-}\right]=-log(2.5\times {10}^{-2})=1.6020$
Also,
$pH=14-pOH$
$pH=14-1.6020=12.398$