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Question: If the maximum concentration of \[PbC{l_2}\]in water is 0.01 M at 298 K, Its maximum concentration i...

If the maximum concentration of PbCl2PbC{l_2}in water is 0.01 M at 298 K, Its maximum concentration in 0.1 M NaCl will be:
A.4×10−3M4 \times {10^{ - 3}}M
B.0.4×10−4M0.4 \times {10^{ - 4}}M
C.4×10−9M4 \times {10^{ - 9}}M
D.4×10−4M4 \times {10^{ - 4}}M

Explanation

Solution

The mole ratio between the dissolved material and the cations and anions it produces in solution determines the ion concentration in solution. The number of the actual ion concentrations is the overall ion concentration. Molarity, percent by volume, and percent by mass are both ways to convey concentration.

Complete answer: The solubility product is a type of equilibrium constant whose value is temperature dependent. Due to enhanced solubility, Ksp{K_{sp}} typically rises as the temperature rises.
In water, the dissociation reaction forPbCl2PbC{l_2} is written as,
PbCl2→Pb2+(aq)+2Cl−(aq){\text{PbC}}{{\text{l}}_2} \to {\text{P}}{{\text{b}}^{2 + }}({\text{aq}}) + 2{\text{C}}{{\text{l}}^ - }({\text{aq}})
PbCl2PbC{l_2} has a maximum concentration of 0.01 M in water.
Using the stoichiometry from the above reaction, we can calculate the ion concentration.
Pb2+P{b^{2 + }} ion concentration is 0.01 M, and Cl−C{l^ - } ion concentration is 2 × 0.01 = 0.02M.
Solubility product:
The solubility product of PbCl2PbC{l_2} can be written as,
KSP=[Pb2+][Cl−]2{K_{SP}} = [P{b^{2 + }}]{[C{l^ - }]^2}
Let's put in the Pb and Cl concentrations.
KSP{K_{SP}}= (0.01)(0.02)2\left( {0.01} \right){\left( {0.02} \right)^2}
KSP{K_{SP}}= 4 × 10−64{\text{ }} \times {\text{ }}{10^{ - 6}}
Concentration of Pb2+P{b^{2 + }} using common ion effect:
There is a typical ion Cl−C{l^ - } when PbCl2PbC{l_2} is dissolved in 1MNaCl. Due to the common ion impact, PbCl2PbC{l_2} solubility is reduced. In 0.1M NaCl, the concentration of Cl−C{l^ - } is 0.1. Let's use this value to calculate the latest Pb2+P{b^{2 + }} ion concentration.
KSP=[Pb2+][Cl−]2{K_{SP}} = [P{b^{2 + }}]{[C{l^ - }]^2}
4×10−6=[Pb2+][Cl−]24 \times {10^{ - 6}} = [P{b^{2 + }}]{[C{l^ - }]^2}
4×10−6=[Pb2+][0.1]24 \times {10^{ - 6}} = [P{b^{2 + }}]{[0.1]^2}
4×10−6[0.1]2=[Pb2+]\frac{{4 \times {{10}^{ - 6}}}}{{{{[0.1]}^2}}} = [P{b^{2 + }}]
[Pb2+]=0.0004M[P{b^{2 + }}] = 0.0004M
⇒[Pb2+]=4×10−4\Rightarrow \left[ {P{b^{2 + }}} \right] = 4 \times {10^{ - 4}}
In 0.1M NaCl, the maximum concentration of PbCl2PbC{l_2} is 0.0004M.
Hence option D is correct.

Note:
The ability of a substance called a solute to dissolve in a solvent and form a solution is known as solubility. Ionic compounds (which dissociate to form cations and anions) have a wide range of solubility in water. Some compounds are very soluble and can also absorb moisture from the air, while others are extremely insoluble.