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JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Equilibrium

A_{3} B_{2}

is a sparingly soluble salt of molar mass

M (g mol^{- 1})

and solubility

x g L^{- 1}

. The solubility product satisfies

K_{s p} = a (\frac{x}{M})^{5}

. The value of

a

is ______. (Integer answer)

Answer: 108

Solution

Dissolution gives

A_{3} B_{2} ⇌ 3 A^{2 +} + 2 B^{3 -}

, so a molar solubility

s

produces

[A] = 3 s

and

[B] = 2 s

. Then

K_{s p} = (3 s)^{3} (2 s)^{2} = 27 \times 4 \times s^{5} = 108 s^{5} .

With

s = x / M

in

mol L^{- 1}

,

K_{s p} = 108 (\frac{x}{M})^{5}

, so the constant equals

108

.

JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Integer answer

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