Question

$0.2\, M$ solution of a weak acid $HA$ is $1 \%$ ionized $25^{\circ} C . K_{a}$ for the acid is equal to

• A. $\frac{0.002 \times 0.002}{0.198}$
• B. $\frac{0.02 \times 0.02}{0.18}$
• C. $\frac{0.01 \times 0.01}{0.19}$
• D. $\frac{0.19}{0.01 \times 0.01}$

Answer 1

Answer: (A)

$HA + H _{2} O \rightleftharpoons H _{3} O ^{+}+ A ^{-}$

$[ HA ] =0.2-\frac{0.2 \times 1}{100}=0.198 \,M$

$[ A ^{-1}] =0.002 \,M$

$\left[ H _{3} O ^{+}\right] =0.002 \,M $

$K_{ a }=\frac{\left[ H _{3} O ^{+}\right][ A -]}{[ HA ]}=\frac{0.002 \times 0.002}{0.198}$