Solveeit Logo
Login

Question

Question: In 1 L saturated solution of AgCl (\({{K}_{sp,AgCl}}\)=\[1.6\times {{10}^{-10}}\]), 0.1 mol of CuCl ...

In 1 L saturated solution of AgCl (Ksp,AgCl{{K}_{sp,AgCl}}=1.6×10−101.6\times {{10}^{-10}}), 0.1 mol of CuCl (Ksp,CuCl{{K}_{sp,CuCl}}=1×10−61\times {{10}^{-6}}) is added. The resultant concentration of Ag+A{{g}^{+}} in the solution is 1.6×10−xM1.6\times {{10}^{-x}}M. Calculate the value of x.
A. 7
B. 8
C. 9
D. 10

Explanation

Solution

To correctly find the solution of this question we should know about the solubility product constant, Ksp{{K}_{sp}}. We will find the value of x by using the relationship and formula between solubility constant and concentration.

Complete step by step answer:
Solubility product constant Ksp{{K}_{sp}}​, is the equilibrium constant for a solid substance that we are dissolving in an aqueous solution. It represents the level at which a solute dissolves in solution. The more soluble a substance is, the higher the Ksp{{K}_{sp}} value it has. The concentration depends on certain conditions such as temperature, pressure, and composition. It is influenced by surroundings.
First of all we will write the data that is given in the question.

So, solubility product constant of AgCl and CuCl is:
Ksp{{K}_{sp}} AgCl=1.6×10−101.6\times {{10}^{-10}}
Ksp{{K}_{sp}} CuCl=1×10−61\times {{10}^{-6}}

So, now we know that 0.1 mol of CuCl is added to 1L AgCl solution. So, we can write this as follows:

& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,CuCl(s)\to C{{u}^{+}}+C{{l}^{-}} \\\ & Initial\,Conc.\,\text{ }\,\,\,\text{ }\,\,=\,\,\,\,\,\,\,\,0.1M\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,0 \\\ & At\,equilibrium\,\,=\,\,\,\,\,\,0.1-x\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,x \\\ \end{aligned}$$ So we can say that, after addition of CuCl the concentration will decrease on the reactant side and will increase on the product side due to the common ion effect. We can write formula of solubility product constant as: $$\begin{aligned} & {{K}_{sp,\,CuCl}}=[A{{g}^{+}}][C{{l}^{-}}] \\\ & {{K}_{sp,\,CuCl}}=S\times S \\\ & {{K}_{sp,\,CuCl}}={{S}^{2}} \\\ \end{aligned}$$ We should know the concentration of chloride ions. We can find the concentration of chloride ion as follows: $$[C{{l}^{-}}]=S=\sqrt{{{K}_{sp,CuCl}}}=\sqrt{1.0\times 10}{{}^{-6}}=1.0\times {{10}^{-3}}M$$ $$\to {{K}_{sp,\,CuCl}}=[A{{g}^{+}}][C{{l}^{-}}]$$ And from this, we will find the concentration of silver ions. So, the concentration of silver ion will be: $$[A{{g}^{+}}]=\dfrac{{{K}_{sp,AgCl}}}{[C{{l}^{-}}]}$$ Now, by putting the values we will get the concentration of silver ion: $$\begin{aligned} & [A{{g}^{+}}]=\dfrac{{{K}_{sp,AgCl}}}{[C{{l}^{-}}]} \\\ & \to \dfrac{1.6\times {{10}^{-10}}}{1.0\times {{10}^{-3}}}=1.6\times {{10}^{-7}}M=1.6\times {{10}^{-x}}M \\\ & x=7 \\\ \end{aligned}$$ From the above calculation we can say that the value of x is 7. **So, the correct answer is “Option A”.** **Note:** Remember solubility and solubility product constant are different terms. Solubility of an ion is the extent of hydration in that ion in the solution. Solubility product constant is the product of all the ion concentrations of that compound. Higher the solubility product constant value, higher will be the solubility.