JEE MainChemistryToughPhysical Chemistry

Calculate in a 0.2 M solution of dichloroacetic acid
(K_a = 5 × 10^-2) that also contains 0.05 M H_2SO_4. Report your
answer after multiplying it by 400.

Show answer and solution

\begin{matrix} \ HA\ \ \ \ \ \ \ \ \ \ \ \ \rightleftharpoons \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ A^{-}\ \ \ \ + \ \ \ \ H^{+} \end{matrix}

t = 0 0.2 0.1
t = eq 0.2(1 - α) 0.2 α 0.1 + 0.2 α

K_{a} = \frac{\left\lbrack H^{+} \right\rbrack\left\lbrack A^{-} \right\rbrack}{\lbrack HA\rbrack} = \frac{0.2\alpha(0.1 + 0.2\alpha)}{0.2(1 - \alpha)}

....(1)
Expecting α < < 1, 5×10^-2 =

\frac{0.2\alpha \times 0.1}{0.2}

\therefore

a = 0.5 (not negligible)
So, solve quadratic (1) & get α = 0.25

\therefore

= 0.1 + 0.2 × 0.25
= 0.15 M
0.15 × 400 = 60

Source: Grass-G / CH-24 Chemistry Paper 31 Dec..docx

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