Question

The solubility of $PbS{O_4}$ water is 0.0608 g/l. calculate the solubility product constant of $PbS{O_4}$ . molar mass of $PbS{O_4}$ =304 g/mole.

Answer 1

Hint: The maximum number of moles of a salt that can be dissolved per litre of solution at a given temperature is called the solubility of the given salt at the given temperature. This can also be expressed as g/L.

Step by step answer:we are given with a salt of $PbS{O_4}$that is dissolved in water. The salt starts ionizing to pass $P{b^{2 + }}$ and $SO_4^{2 - }$ into solution which become surrounded by solvent molecules, we indicate it by:
$PbS{O_4}(s) \to P{b^{2 + }}(aq) + SO_4^{2 - }(aq)$
The $P{b^{2 + }}$ and $SO_4^{2 - }$ions present in solution combine again
 $P{b^{2 + }}(aq) + SO_4^{2 - }(aq) \to PbS{O_4}(s)$
The reaction therefore proceeds in both the directions and can be written as
At equilibrium, ${{\rm K}_{sp}} = \dfrac{{[P{b^{2 + }}][SO_4^{2 - }]}}{{[PbSO]}}$
Where ${{\rm K}_{sp}}$is called a solubility product. Solubility products of an electrolyte at a specified temperature may be defined as the product of the molar concentration of ions in a saturated solution, each concentration raised to the power equal to the number of ions produced on dissociation of one molecule of the electrolyte.
If activity coefficients for the pure and actively solid $PbS{O_4}$is taken as unity. We are given with solubility of lead sulphate in water and its molar mass as well, therefore we can calculate the number of moles of dissociation by
${{\rm K}_{sp}} = \dfrac{{0.0608}}{{304}} = 0.0002moles$
Hence substituting the above value in the equation of solubility product we get, the solubility product as,
${{\rm K}_{sp}} = {(0.0002)^2} = 4.0 \times {10^{ - 8}}$

Note: If ionic product is less than the solubility product then more of the salt dissolves. If both are equal then the solution is saturated and if the ionic product has a higher value then the salt would precipitate out.