The solubility of $P b S O_{4}$ water is 0.0608 g/L. Calculate the solubility product constant of $P b S O_{4}$ . Molar mass $P b S O_{4}$ = 304 g/mole

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Solution

When the solubility of $P b S O_{4}$ water is 0.038g/l. To Calculate the solubility product constant of lead (II) sulphate, we need to find the number of moles of of lead (II) sulphate then write the equation for the dissociation of it as well.
Number of moles= mass/ relative formula mass
Relative formula mass of $P b S O_{4} = 208 + 32 + 64 = 304$
Therefore moles = 0.038/304 = 0.000125 moles
Dissociation of $P b S O_{4} \to P b^{+ 2} + S O_{4}^{- 2}$
$k_{s p} = (0.000125)^{2}$
$= 1.5 \times 10^{- 8}$

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