Question

What will be the resultant $pH$ when $200mL$ of, an aqueous solution of $HCl(pH=2.0)$ is mixed with $300mL$ 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 $=200mL$
$pH$ 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 $=300mL$
$pH$ of $NaOH=12$
$pOH$ of $NaOH=14-12=2$
$\left[O{H}^{-}\right]$ in $NaOH=14-12=2$
$\left[O{H}^{-}\right]$ 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 $=\left[O{H}^{-}\right]$
$=3-2=1$
Remaining concentration of $\left[O{H}^{-}\right]$
$=\frac{1}{500}=2\times 10^{-3}$
$\therefore \left[H^{+}\right]=\frac{10^{-14}}{2\times 10^{-3}}=5\times 10^{-12}$
$pH=-\log \left[H^{+}\right]$
$=-\log \left(5\times 10^{-12}\right)$
$=11.3010$