On decreasing the \(p \mathrm{H}\) from 7 to 2, the solubility of a sparingly soluble salt (MX) of a weak acid (HX) increased from \(10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}\) to \(10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}\). The \(p \mathrm{~K}_{\mathrm{a}}\) of HX is:

JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Equilibrium

On decreasing the

p H

from 7 to 2, the solubility of a sparingly soluble salt (MX) of a weak acid (HX) increased from

1 0^{- 4} mol L^{- 1}

1 0^{- 3} mol L^{- 1}

. The

p K_{a}

of HX is:

Answer: (B)

At pH = 7 (pure water), solubility

s_{1} = K_{s p} = 1 0^{- 4}

mol/L.

At pH = 2, the common ion effect is suppressed and

X X^{-}

reacts with

H X^{+}

to form HX:

MX (s) M X^{+} + X X^{-}

X X^{-} + H X^{+} HX

The enhanced solubility

s

at pH = 2 can be derived. For the equilibrium:

s^{2} = \frac{K _{s p}}{K _{a}} [H X^{+}]

Given

s = 1 0^{- 3}

at pH = 2 (

[H X^{+}] = 1 0^{- 2}

) and

s_{1} = 1 0^{- 4}

:

From the ratio

\frac{s}{s _{1}} = 10

, and using the relationship:

\frac{s}{s _{1}} = \frac{1}{K _{a} [ H X ^{+} ]}

Solving:

10 = \frac{1}{K _{a} \cdot 1 0 ^{- 2}}

Squaring:

100 = \frac{1}{K _{a} \cdot 1 0 ^{- 2}}

Therefore:

K_{a} = 1 0^{- 4}

, so

p K_{a} = 4

.

Answer: (B)

JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Single correct option