First slide

Solubility product(KSP)

Question

The molar solubility of M(OH)₃ in 0.4 M MX₃ solution in terms of solubility product of M(OH)₃ is

Moderate

A

${(K_{sp} / 10.8)}^{1 / 3}$
B

${(K_{sp} / 10.8)}^{1 / 4}$
C

${(K_{sp} / 0.4)}^{1 / 3}$
D

${(K_{sp} / 3.6)}^{1 / 3}$

Solution

Following equilibria gets simultaneous established
$M (OH)_{3} (s) ⇌ M (\underset{S}{OH})_{3} (aq) \to \underset{S + 0.4}{M^{+ 3}} (aq) + \underset{3 s}{3 {OH}^{-}} (aq)$
And the solubility of M(OH)₃ will get suppressed in presence of 0.4 M MX₃ solution.
${MX}_{3} \to M_{0.4 + S^{'}}^{+ 3} (aq) + 3 X_{3 \times 0.4}^{-} (aq)$
the solubility of M(OH)₃ is suppressed, S' will be negligible. i.e. 0.4 + S' ~ 0.4
K_sp = (S' + 0.4) x (3S')³
⇒ K_sp = 0.4 x S'³ x 27
$\Rightarrow \frac{K_{sp}}{10.8} = {(S^{'})}^{3}$
$\Rightarrow S^{'} = {(\frac{K_{s p}}{10.8})}^{\frac{1}{3}}$

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