Question

$0.365g$ of $HCl$ gas was passed through $100c{m}^3$ of $0.2M$ $NaOH$ solution. The $pH$ of the resulting solution would be

• A. 13
• B. 8
• C. 7
• D. 9

Answer 1

Answer: (A)

$0.365gHCl=\frac{0.365}{36.5}=0.01$ moles of HCl

$0.2MNaOH$ in $100c{m}^3=\frac{0.2\times 100}{1000}=0.02$ moles of $NaOH$

0.01 moles of $HCl$ get neutralized with 0.01 moles of $NaOH$, thus resulting solution contains 0.01 moles of $NaOH$ in 100 $c{m}^3$ solution.

Concentration of $NaOH=\frac{0.01}{100}\times 1000=0.1M$

$\left[O{H}^{-}\right]=0.1M=10^{-1}M,\left[H^{+}\right]\left[O{H}^{-}\right]=10^{-14}$

$\left[H^{+}\right]=\frac{10^{-14}}{10^{-1}}=10^{-13}M,pH=-\log 10^{-13}=13$