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Questions  >>  JEEMAIN and NEET  >>  Chemistry  >>  Equilibrium
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Q)

The dissociation constant of certain weak acid HA is $1 \times 10^{-4}$. The equilibrium constant for its reaction with strong base BOH is

(a) $1 \times 10^{-4}$

(b) $1 \times 10^{-10}$

(c) $\infty$

(d) $1 \times 10^{10}$

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Answer: $1 \times 10^{10}$
$HA \rightleftharpoons H^+ + A^-$
Ionic product, $K_a = \frac{[H^+][A^-]}{[HA]} = 10^{-4}$.............(i)
The reaction with strong base BOH can be expressed as
$HA + OH^- \rightleftharpoons H_2O + A^-$
$K = \frac{[A^-][H_2O]}{[OH^-][HA]}$.............(ii)
Also $K_w = [H^+][OH^-] = 10^{-14}$...........(iii)
From equations (i), (ii) and (iii),
$K =\large \frac{K_a}{K_w} $$= \large \frac{10^{-4}}{10^{-14}} $$= 1\times 10^{10}$
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