0.2 M solution of a weak acid HA is 1 % ionized 25° C . Ka for the acid is equal to

Question

0.2 M solution of a weak acid HA is 1 % ionized 25° C . Ka for the acid is equal to (A) (0.002 × 0.002/0.198) (B) (0.02 × 0.02/0.18) (C) (0.01 × 0.01/0.19) (D) (0.19/0.01 × 0.01)

Tardigrade Admin · Accepted Answer

HA + H 2 O leftharpoons H 3 O ++ A - [ HA ] =0.2-(0.2 × 1/100)=0.198 M [ A -1] =0.002 M [ H 3 O +] =0.002 M K a =([ H 3 O +][ A -]/[ HA ])=(0.002 × 0.002/0.198)

Q. $0.2 M$ solution of a weak acid $H A$ is $1%$ ionized $2 5^{\circ} C . K_{a}$ for the acid is equal to

$\frac{0.002 \times 0.002}{0.198}$

$\frac{0.02 \times 0.02}{0.18}$

$\frac{0.01 \times 0.01}{0.19}$

$\frac{0.19}{0.01 \times 0.01}$

$H A + H_{2} O ⇌ H_{3} O^{+} + A^{-}$

$[H A] = 0.2 - \frac{0.2 \times 1}{100} = 0.198 M$

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

$[H_{3} O^{+}] = 0.002 M$

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

Q. 0.2M solution of a weak acid HA is 1% ionized 25∘C.Ka​ for the acid is equal to

0.1980.002×0.002​

0.180.02×0.02​

0.190.01×0.01​

0.01×0.010.19​