MCQ
At $298\, K$, the solubility product of $PbCl_2$ is $1.0 \times 10^{-6}$. What will be the solubility of $PbCl_2$ in $moles/litre$
- ✓$6.3 \times {10^{ - 3}}$
- B$1.0 \times {10^{ - 3}}$
- C$3.0 \times {10^{ - 3}}$
- D$4.6 \times {10^{ - 14}}$
✓
Answer
Correct option: A.
$6.3 \times {10^{ - 3}}$
a
$PbC{l_2}(s) \rightleftharpoons \mathop {P{b^{2 + }}(aq)}\limits_S + \mathop {2C{l^ - }(aq)}\limits_{2S} $
$PbC{l_2}(s) \rightleftharpoons \mathop {P{b^{2 + }}(aq)}\limits_S + \mathop {2C{l^ - }(aq)}\limits_{2S} $
$\mathrm{K}_{\mathrm{sp}}=4 \mathrm{s}^{3}$
$1 \times 10^{-6}=4 \mathrm{s}^{3}$
$\mathrm{S}^{3}=\frac{1}{4} \times 10^{-6}$
$=\left(\frac{1}{4}\right)^{\frac{1}{3}} \times 10^{-2}$
$=0.63 \times 10^{-2}=6.3 \times 10^{-3}$
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