First slide

Solubility product(KSP)

Question

The molar solubility of PbI₂in 0.2M Pb(NO₂)₂ solution in terms of solubility product, K_sp

Easy

A

${(K_{sp} / 0.2)}^{1.2}$
B

${(K_{sp} / 0.4)}^{1 / 2}$
C

${(K_{sp} / 0.8)}^{1 / 2}$
D

${(K_{sp} / 0.8)}^{1 / 3}$

Solution

Molar solubility PbI₂ will get suppressed in presence of Pb(NO₂)₂ due to commonion effect, and the following equilibria gets established.
${PbI}_{2} (s) ⇌ {PbI}_{2} (aq) \to {Pb}^{+ 2} (aq) + 2 I^{-} (aq)$
$Pb {({NO}_{2})}_{2} \to {Pb}^{+ 2} (aq) + {\underset{0.2 + S^{'}}{2 {NO}_{2}}}^{-} (aq)$
K_sp = (S' + 0.2) x (2S')²
⇒ K_sp = 0.2 x 4(S')²
$\Rightarrow \frac{K_{sp}}{0.8} = {(S^{'})}^{2}$
$\Rightarrow S^{'} = {(\frac{K_{sp}}{0.8})}^{\frac{1}{2}}$

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