Question

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

• A. 13
• B. 8
• C. 7
• D. 9

Answer 1

Answer: (A)

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

$0.2 \,M \,NaOH$ in $100\, cm ^{3}=\frac{0.2 \times 100}{1000}=0.02$ 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 $cm ^{3}$ solution.

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

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

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