Solubility of PbC{l₂} at {25,^o}C is 6.3 × {10^{ - 3}} mole/litre . Its solubility product at tha

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6-2.Equilibrium-II (Ionic Equilibrium)

medium

The solubility of $PbC{l_2}$ at ${25\,^o}C$ is $6.3 \times {10^{ - 3}}$ $mole/litre$. Its solubility product at that temperature is

A

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

B

$(6.3 \times {10^{ - 3}}) \times (12.6 \times {10^{ - 3}})$

C

$(6.3 \times {10^{ - 3}}) \times {(12.6 \times {10^{ - 3}})^2}$

D

$(12.6 \times {10^{ - 3}}) \times (12.6 \times {10^{ - 3}})$

Solution

(c) $PbC{l_2} \to \mathop {P{b^{ + + }}}\limits_S + \mathop {2C{l^ – }}\limits_{2S} $

${K_{sp}} = S \times {(2S)^2}$$ = [6.3 \times {10^{ – 3}}] \times {[12.6 \times {10^{ – 3}}]^2}$.

Standard 11

Chemistry

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