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Solubility

The solubilit...

Question

The solubility of $P b I_{2}$ at $25^{o}$ C is $0.7$ g $L^{- 1}$ . The solubility product of $P b I_{2}$ at this temperature is:
(Molar mass of $P b I_{2} = 461.2$ g $m o l^{- 1}$ )

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Solution

$P b I_{2} ⇌ P b^{2 +} + 2 I^{-}$
$x$ $2 x$
given solubility of $P b I_{2}$ is $0.7 \frac{g}{L}$
also, molar mass of $P b I_{2} = 461.2 g / m o l$
$\Rightarrow [P b I_{2}] = \frac{0.7}{461.2} m o l / L i t r e = 1.5 \times 10^{- 3}$
solubility product, $K_{s p} = x (2 x)^{2}$
$K_{s p} = 4 x^{3}$
$K_{s p} = 4 (1.5 \times 10^{- 3})^{3}$
$∴$ Solubility product of $P b I_{2} = 135 \times 10^{- 9} m o l^{2} / l i t^{2}$

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