Question

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

Answer 1

Hint: 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 \[PbC{l_2}\]. \[PbC{l_2}\] dissociates in water as \[PbC{l_2} \rightleftharpoons P{b^{2 + }} + 2C{l^ - }\]. This will give us the relation \[{K_{sp}} = [S]{[2S]^2}\]. Find the value of \[S\], which is the solubility of the compound.

Complete step by step answer:
Solubility product of \[PbC{l_2}\]is given to be \[1.7 \times {10^{ - 5}}\]. We have to find its solubility in water.
Solubility product of \[PbC{l_2} = 1.7 \times {10^{ - 5}}\].
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 \[{K_{sp}}\].
Now, \[PbC{l_2}\] dissociates in water as
\[PbC{l_2} \rightleftharpoons P{b^{2 + }} + 2C{l^ - }\]
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 \[{K_{sp}}\] and the solubility of the compound \[S\], is
\[
  \therefore {K_{sp}} = [S]{[2S]^2} \\
  1.7 \times {10^{ - 5}} = 4{S^3} \\
  S = \sqrt[3]{{\frac{{1.7 \times {{10}^{ - 5}}}}{4}}} \\
  S = 0.0162moles/l \\
  S = 0.0162 \times 277 = 4.4875g/l \\
\]
Note:
The solubility product is a kind of equilibrium constant and its value depends on temperature. \[{K_{sp}}\] 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.