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Question: Calculate the solubility of lead chloride in water, if its solubility product is \[1.7 \times {10^{ ...

Calculate the solubility of lead chloride in water, if its solubility product is 1.7×1051.7 \times {10^{ - 5}}.
(Pb=206,Cl=35.5)(Pb = 206,Cl = 35.5)

Explanation

Solution

In questions where there is no chemical formula of the compound given, first formulate the formula of the compound. It will give us immense insight into what has to be found. The formula of lead chloride is PbCl2PbC{l_2}. PbCl2PbC{l_2} dissociates in water as PbCl2Pb2++2ClPbC{l_2} \rightleftharpoons P{b^{2 + }} + 2C{l^ - }. This will give us the relation Ksp=[S][2S]2{K_{sp}} = [S]{[2S]^2}. Find the value of SS, which is the solubility of the compound.

Complete step by step answer:
Solubility product of PbCl2PbC{l_2}is given to be 1.7×1051.7 \times {10^{ - 5}}. We have to find its solubility in water.
Solubility product of PbCl2=1.7×105PbC{l_2} = 1.7 \times {10^{ - 5}}.
The solubility product constant is defined as the equilibrium constant for the dissolution of a solid substance into its aqueous solution. It is denoted by the symbol Ksp{K_{sp}}.
Now, PbCl2PbC{l_2} dissociates in water as
PbCl2Pb2++2ClPbC{l_2} \rightleftharpoons P{b^{2 + }} + 2C{l^ - }
Solubility of a solute is it’s property to dissolve in a solvent and form a solution. Different ionic compounds (which dissociate to form cations and anions) in water have varied solubilities. Some compounds are highly soluble and can turn aqueous in the presence of air, and some are not at all soluble.
What is the significance of Solubility products? Solubility of a solid substance depends on a number of parameters which includes the lattice enthalpy of the salt and solvation enthalpy of ions in the solution.
The relation between the solubility constant Ksp{K_{sp}} and the solubility of the compound SS, is

Ksp=[S][2S]2 1.7×105=4S3 S=1.7×10543 S=0.0162moles/l S=0.0162×277=4.4875g/l  \therefore {K_{sp}} = [S]{[2S]^2} \\\ 1.7 \times {10^{ - 5}} = 4{S^3} \\\ S = \sqrt[3]{{\frac{{1.7 \times {{10}^{ - 5}}}}{4}}} \\\ S = 0.0162moles/l \\\ S = 0.0162 \times 277 = 4.4875g/l \\\

Note:
The solubility product is a kind of equilibrium constant and its value depends on temperature. Ksp{K_{sp}} usually increases with an increase in temperature due to increased solubility. So, make sure the calculation is done at room temperature or make further alterations to the formula as required.