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Question: In quantitative analysis, the metals of Group 1 can be separated to form other ions by precipitating...

In quantitative analysis, the metals of Group 1 can be separated to form other ions by precipitating them as chloride salts. A solution initially contains Ag+A{{g}^{+}} and Pb2+P{{b}^{2+}} at a concentration of 0.10 M. An aqueous HClHCl is added to this solution until the Cl−C{{l}^{-}} concentration is 0.10 M. What is the concentration of Ag+A{{g}^{+}} and Pb2+P{{b}^{2+}}be at equilibrium?
(Ksp for AgCl=1.8 x 10−10, Ksp for PbCl2=1.7 x 10−5)({{K}_{sp}}\text{ }for\text{ }AgCl=1.8\text{ x 1}{{\text{0}}^{-10}},\text{ }{{K}_{sp}}\text{ }for\text{ }PbC{{l}_{2}}=1.7\text{ x 1}{{\text{0}}^{-5}})
(A) [Ag+]=1.8 x 10−7M;[Pb2+]=1.7 x 10−6M[A{{g}^{+}}]=1.8\text{ x 1}{{\text{0}}^{-7}}M;[P{{b}^{2+}}]=1.7\text{ x 1}{{\text{0}}^{-6}}M
(B) [Ag+]=1.8 x 10−11M;[Pb2+]=8.5 x 10−5M[A{{g}^{+}}]=1.8\text{ x 1}{{\text{0}}^{-11}}M;[P{{b}^{2+}}]=8.5\text{ x 1}{{\text{0}}^{-5}}M
(C) [Ag+]=1.8 x 10−9M;[Pb2+]=1.7 x 10−3M[A{{g}^{+}}]=1.8\text{ x 1}{{\text{0}}^{-9}}M;[P{{b}^{2+}}]=1.7\text{ x 1}{{\text{0}}^{-3}}M
(D) [Ag+]=1.8 x 10−11M;[Pb2+]=8.5 x 10−4M[A{{g}^{+}}]=1.8\text{ x 1}{{\text{0}}^{-11}}M;[P{{b}^{2+}}]=8.5\text{ x 1}{{\text{0}}^{-4}}M

Explanation

Solution

In equilibrium, the given salt will form cations and anions. For salt AgClAgCl, the value of solubility product will be Ksp=[Ag+][Cl−]{{K}_{sp}}=[A{{g}^{+}}][C{{l}^{-}}] and for salt PbCl2PbC{{l}_{2}}, The value of solubility product will be Ksp=[Pb2+][Cl−]2{{K}_{sp}}=[P{{b}^{2+}}]{{[C{{l}^{-}}]}^{2}}.

Complete step by step solution:
First, let us study the solubility product:
If a sparingly soluble salt ABABis stirred with water, only a small amount of it goes into solution while most of the salt remains undissolved. But whenever the little amount of salt dissolves, it gets completely dissociated into ions. In other words, there exists a dynamic equilibrium between the undissolved solid salt and the ions which it furnishes in solution when a sparingly soluble salt is added to water.
Thus, the equilibrium can be represented as:
AB(s)⇌A+(aq)+B−(aq)AB(s)\rightleftharpoons {{A}^{+}}(aq)+{{B}^{-}}(aq)
We can also write according to the law of chemical equilibrium,
K=[A+][B−][AB]K=\dfrac{[{{A}^{+}}][{{B}^{-}}]}{[AB]}
Since the concentration of the undissociated solid remains constant, we may write,
[A+][B−]=K x !![!! AB !!]!! =Ksp[{{A}^{+}}][{{B}^{-}}]=K\text{ x }\\!\\![\\!\\!\text{ AB }\\!\\!]\\!\\!\text{ =}{{\text{K}}_{sp}}
Where Ksp{{K}_{sp}}is the solubility product is equal to the product of[A+][B−][{{A}^{+}}][{{B}^{-}}].
For salt AgClAgCl, the value of solubility product will be:
Ksp=[Ag+][Cl−]{{K}_{sp}}=[A{{g}^{+}}][C{{l}^{-}}]
Given the solubility product of Ksp AgCl=1.8 x 10−10{{K}_{sp}}\text{ }AgCl=1.8\text{ x 1}{{\text{0}}^{-10}}
And the concentration of chloride ion is given 0.10 M
1.8 x 10−10=[Ag+][0.1]1.8\text{ x 1}{{\text{0}}^{-10}}=[A{{g}^{+}}][0.1]
[Ag+]=1.8 x 10−9M[A{{g}^{+}}]=1.8\text{ x 1}{{\text{0}}^{-9}}M
for salt PbCl2PbC{{l}_{2}}, The value of solubility product will be:
Ksp=[Pb2+][Cl−]2{{K}_{sp}}=[P{{b}^{2+}}]{{[C{{l}^{-}}]}^{2}}
Given the solubility product of Ksp PbCl2=1.7 x 10−5{{K}_{sp}}\text{ }PbC{{l}_{2}}=1.7\text{ x 1}{{\text{0}}^{-5}}
And the concentration of chloride ion is given 0.10 M
1.7 x 10−5=[Pb2+][0.1]21.7\text{ x 1}{{\text{0}}^{-5}}=[P{{b}^{2+}}]{{[0.1]}^{2}}
[Pb2+]=1.7 x 10−3M[P{{b}^{2+}}]=1.7\text{ x 1}{{\text{0}}^{-3}}M
So, [Ag+]=1.8 x 10−9M[A{{g}^{+}}]=1.8\text{ x 1}{{\text{0}}^{-9}}Mand [Pb2+]=1.7 x 10−3M[P{{b}^{2+}}]=1.7\text{ x 1}{{\text{0}}^{-3}}M.

Therefore the correct answer is an option (C) [Ag+]=1.8 x 10−9M;[Pb2+]=1.7 x 10−3M[A{{g}^{+}}]=1.8\text{ x 1}{{\text{0}}^{-9}}M;[P{{b}^{2+}}]=1.7\text{ x 1}{{\text{0}}^{-3}}M.

Note: The concentration of the ions must be raised to the power equal to the number of ions produced in the solution. The solubility product is different from the ionic product as the solubility product is only restricted to saturated solutions but the ionic product is for both saturated and unsaturated solutions.