Solubility product constants ($\mathrm{K}_{\text{sp}}$) of salts of types $\mathrm{MX}$, $\mathrm{MX}_{2}$ and $\mathrm{M}_{3}\mathrm{X}$ at temperature 'T' are $4.0 \times 10^{-8}$, $3.2 \times 10^{-14}$ and $2.7 \times 10^{-15}$, respectively. Solubilities (mol dm⁻³) of the salts at temperature 'T' are in the order

(D) $\mathrm{MX} > \mathrm{M}_{3}\mathrm{X} > \mathrm{MX}_{2}$

(A) $\mathrm{MX} > \mathrm{MX}_{2} > \mathrm{M}_{3}\mathrm{X}$

(B) $\mathrm{M}_{3}\mathrm{X} > \mathrm{MX}_{2} > \mathrm{MX}$

(C) $\mathrm{MX}_{2} > \mathrm{M}_{3}\mathrm{X} > \mathrm{MX}$

JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Equilibrium

Solubility product constants (

K_{sp}

) of salts of types

MX

MX_{2}

and

M_{3} X

at temperature 'T' are

4.0 \times 1 0^{- 8}

3.2 \times 1 0^{- 14}

and

2.7 \times 1 0^{- 15}

, respectively. Solubilities (mol dm⁻³) of the salts at temperature 'T' are in the order

Answer: (D)

For

MX

K_{s p} = s^{2}

, so

s = 4.0 \times 1 0^{- 8} = 2 \times 1 0^{- 4}

mol dm⁻³

For

MX_{2}

K_{s p} = 4 s^{3}

, so

s = 3 \frac{3.2 \times 1 0 ^{- 14}}{4} = 2 \times 1 0^{- 5}

mol dm⁻³

For

M_{3} X

K_{s p} = 27 s^{4}

, so

s = 4 \frac{2.7 \times 1 0 ^{- 15}}{27} = 1 \times 1 0^{- 4}

mol dm⁻³

Therefore, the order is

MX > M_{3} X > MX_{2}

JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Single correct option