Question

What mass of \[P{b^{2 + }}\] is left in solution when \[50.0ml\] of \[0.20M\] \[Pb{(N{O_3})_2}\] is added to \[50.0ml\] of \[1.50M\] \[NaCl\] ? [Given \[{K_{sp}}\] for \[PbC{l_2} = 1.7 \times {10^{ - 4}}\] ]

Answer 1

Hint: The solubility product contains the product of concentrations of ions raised to the power of their stoichiometric coefficients. Beyond the solubility product i.e. with a higher concentration the salt begins to precipitate.

Complete answer:
An equal volume of lead nitrate and sodium chloride solutions are mixed which would result in a double displacement reaction. The lead ions formed from the dissociation of lead nitrate would combine with the chloride ions released by the dissociation of sodium chloride and lead chloride is formed as the product.
Lead nitrate forms precipitate to some extent and only dissociates partially in water. Due to the partial dissociation of lead nitrate the unionized salt and the dissociated ions present in aqueous medium establish a dynamic ionic equilibrium.
The equilibrium reaction can be represented as follows:
\[PbC{l_2}(s) \to P{b^{2 + }}(aq) + 2C{l^ - }(aq)\]
In order to determine the amounts of lead ions and chloride ions contributed by their respective solutions, we must find a product of their molarity and volume in liters.
\[n(P{b^{2 + }}) = 0.20M \times 0.05l = 0.01mol\]
\[n(C{l^ - }) = 1.50M \times 0.05l = 0.075mol\]
According to the stoichiometric coefficients, chloride ions will react with twice the amount of lead ions. Therefore the amount of chloride ions utilized in reaction with lead ions is :
\[n(C{l^ - })reacted = 0.01 \times 2 = 0.02mol\]
Thus, the concentration of chloride ions that is left unreacted is :
\[n(C{l^ - })unreacted = 0.075mol - 0.02mol = 0.055mol\]
\[Concentration(C{l^ - }) = \dfrac{{0.055mol}}{{(0.05 + 0.05)l}} = 0.55M\]
The expression for solubility product can be written as follows:
\[{K_{sp}}(PbC{l_2}) = [P{b^{2 + }}]{[C{l^ - }]^2}\]
This can be rearranged to find out the concentration of lead ions left in solution:
\[[P{b^{2 + }}] = \dfrac{{{K_{sp}}(PbC{l_2})}}{{{{[C{l^ - }]}^2}}} = \dfrac{{1.7 \times {{10}^{ - 4}}}}{{{{(0.55)}^2}}} = 5.62 \times {10^{ - 4}}M\]
The amount of lead ions can be calculated by multiplying the molarity with volume.
\[n(P{b^{2 + }})left = 5.62 \times {10^{ - 4}}M \times (0.05 + 0.05)l = 5.62 \times {10^{ - 5}}mol\]
The mass can be determined by finding the product of moles and molar mass of lead ions.
\[mass(P{b^{2 + }})left = 5.62 \times {10^{ - 5}}mol \times 207gmo{l^{ - 1}} = 1.163 \times {10^{ - 2}}g\]
Hence, the mass of lead ions left is \[1.163 \times {10^{ - 2}}g\] .

Note:
The solubility product is a modified or derived equilibrium constant but not an equilibrium constant in itself as its expression does not contain the concentration of the undissociated salt present in the solid form.