To determine the molar solubility [tex]\( S \)[/tex] of lead(II) carbonate ([tex]\( PbCO_3 \)[/tex]), given the solubility product constant [tex]\( K_{sp} = 7.40 \times 10^{-14} \)[/tex], we can follow these steps:
1. Write the dissociation equation:
[tex]\( PbCO_3(s) \leftrightarrow Pb^{2+}(aq) + CO_3^{2-}(aq) \)[/tex]
2. Express the concentrations at equilibrium:
Let [tex]\( S \)[/tex] be the molar solubility of [tex]\( PbCO_3 \)[/tex].
At equilibrium:
[tex]\[ [Pb^{2+}] = S \][/tex]
[tex]\[ [CO_3^{2-}] = S \][/tex]
3. Write the expression for the solubility product constant ([tex]\( K_{sp} \)[/tex]):
[tex]\[ K_{sp} = [Pb^{2+}][CO_3^{2-}] \][/tex]
Since both ions have the same concentration [tex]\( S \)[/tex] at equilibrium, the expression becomes:
[tex]\[ K_{sp} = S \times S = S^2 \][/tex]
4. Solve for the molar solubility [tex]\( S \)[/tex]:
[tex]\[ K_{sp} = S^2 \][/tex]
[tex]\[ S = \sqrt{K_{sp}} \][/tex]
Substituting the value of [tex]\( K_{sp} \)[/tex]:
[tex]\[ S = \sqrt{7.40 \times 10^{-14}} \][/tex]
[tex]\[ S = 2.7202941017470887 \times 10^{-7} \][/tex]
Thus, the molar solubility [tex]\( S \)[/tex] of [tex]\( PbCO_3 \)[/tex] is approximately:
[tex]\[ S = 2.7202941017470887 \times 10^{-7} \, M \][/tex]
This indicates that the solubility of lead(II) carbonate in water is very low due to its small solubility product constant.
