At 298, K , solubility product of PbC{l₂} is 1.0 × {10^{ - 6}} . What will be solubility of PbC{l

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6-2.Equilibrium-II (Ionic Equilibrium)

medium

At $298\, K$ , the solubility product of $PbC{l_2}$ is $1.0 \times {10^{ - 6}}$. What will be the solubility of $PbC{l_2}$ in moles/litre

A

$6.3 \times {10^{ - 3}}$

B

$1.0 \times {10^{ - 3}}$

C

$3.0 \times {10^{ - 3}}$

D

$4.6 \times {10^{ - 14}}$

Solution

(a) ${K_{sp}} = 4{s^3}$

$S = \sqrt[3]{{\frac{{{K_{sp}}}}{4}}} = \sqrt[3]{{\frac{{1.0 \times {{10}^{ – 6}}}}{4}}}$$ = 6.3 \times {10^{ – 3}}$.

Standard 11

Chemistry

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