Question

What will be the resultant $pH$ when $200\, mL$ of, an aqueous solution of $HCl(p H=2.0)$ is mixed with $300\, mL$ of an aqueous solution of $NaOH (pH=12)$ ?

• A. 2.699
• B. 13.30
• C. 11.3010
• D. 1.330

Answer 1

Answer: (C)

Given, volume of $HCl$ solution $=200\, mL$
$p H$ of $HCl$ solution $=2.0$
$\therefore \left[H^{+}\right]$ in $HCl$
solution $=1 \times 10^{-2}$
$\therefore $ Milliequivalents of $HCl =200 \times 1 \times 10^{-2} =2$
Similarly, volume of $NaOH$ solution $=300\, mL$
$pH$ of $NaOH =12$
$pOH$ of $NaOH =14-12=2$
$\left[O H^{-}\right]$ in $NaOH =14-12=2$
$\left[ OH ^{-}\right]$ in $NaOH$ solution $=1 \times 10^{-2}$
$\therefore $ Milliequivalents of $NaOH =300 \times 1 \times 10^{-2}=3$
Since, $NaOH$ is in excess.
$\therefore $ Remaining milliequivalents $=\left[ OH ^{-}\right]$
$=3-2=1$
Remaining concentration of $\left[ OH ^{-}\right]$
$=\frac{1}{500}=2 \times 10^{-3}$
$\therefore \left[H^{+}\right]=\frac{10^{-14}}{2 \times 10^{-3}}=5 \times 10^{-12}$
$p H=-\log \left[H^{+}\right]$
$=-\log \left(5 \times 10^{-12}\right)$
$=11.3010$