Question

To $100 \,mL$ of an aqueous solution of $0.1 M CH _{3} COOH$ $\left( K _{ a }=2 \times 10^{-5}\right), 0.01 mol$ of $HCl ( g )$ is passed. Select correct options regarding the resulting solution.

• A. Degree of dissociation of acetic acid in resulting solution is $10^{-4}$
• B. $pH$ of resulting solution is nearly $1$
• C. Degree of dissociation of water in resulting solution is $1.8 \times 10^{-15}$
• D. Concentration of $OH ^{-}$ ions contributed by water in resulting solution is $10^{-7}\, M$

Answer 1

Answer: (C)

${ }^{n} CH _{3} COOH =0.1 \times 0.1=0.01 \,mol$

${ }^{ n} HCl =0.01 \,mol$

$\left[ H ^{+}\right]_{ f }=\frac{0.01}{0.1}=0.1 \,M$

$pH =1$



$2 \times 10^{-5}=\frac{X \times 0.1}{0.1}$

$x=2 \times 10^{-5}$

$\alpha CH _{3} COOH =\frac{2 \times 10^{-5}}{0.1}=2 \times 10^{-4}$

$H _{2} O \rightleftharpoons H ^{+}+ OH ^{-}, 10^{-14}=0.1 \times\left[ OH ^{-}\right]$ $\left[ OH ^{-}\right]=10^{-13} \,M$

$\alpha\left( H _{2} O \right)=\frac{10^{-13}}{\frac{1000}{18}}$

$=18 \times 10^{-16}=1.8 \times 10^{-15}$