JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Equilibrium

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Two salts A₂X and MX have the same value of solubility product of 4.0 × 10⁻¹². The ratio of their molar solubilities i.e S(A₂X)/S(MX) = ____.
(Round off to the Nearest Integer).

Answer: 50

Solution

A saturated solution of

A_{2} X

of molar solubility

s_{1}

holds

2 s_{1}

A^{+}

and

s_{1}

X^{2 -}

, so

K_{s p} = (2 s_{1})^{2} s_{1} = 4 s_{1}^{3} = 4.0 \times 1 0^{- 12} \Rightarrow s_{1} = 1.0 \times 1 0^{- 4} M

For MX the two ions appear in equal concentration, so

K_{s p} = s_{2}^{2} = 4.0 \times 1 0^{- 12} \Rightarrow s_{2} = 2.0 \times 1 0^{- 6} M

Dividing one by the other,

\frac{s _{1}}{s _{2}} = \frac{1.0 \times 1 0 ^{- 4}}{2.0 \times 1 0 ^{- 6}} = 50

The molar solubility of

A_{2} X

is 50 times that of MX.

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Integer answer

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