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Question: In \(1L\) saturated solution of \[AgCl\] \(\left[ {{K_{sp}}(AgCl) = 1.6 \times {{10}^{ - 10}}} \righ...

In 1L1L saturated solution of AgClAgCl [Ksp(AgCl)=1.6×10−10]\left[ {{K_{sp}}(AgCl) = 1.6 \times {{10}^{ - 10}}} \right] , 0.1mol0.1mol of CuCl[Ksp(CuCl)=1×10−6]CuCl\left[ {{K_{sp}}(CuCl) = 1 \times {{10}^{ - 6}}} \right]is added. The resultant concentration of Ag+A{g^ + } in the solution is 1.6×10−x1.6 \times {10^{ - x}} . The value of xx is:

Explanation

Solution

Ksp{K_{sp}} , also known as solubility product constant, is the equilibrium constant for a solid substance dissolving in an aqueous solution. It represents the level at which a solute dissolves in solution. More soluble substances have higher Ksp{K_{sp}} value.

Complete step by step answer:
We have given the solubility product constants for silver chloride and copper chloride which are as following:
Ksp(AgCl)=1.6×10−10{K_{sp}}(AgCl) = 1.6 \times {10^{ - 10}} and Ksp(CuCl)=1×10−6{K_{sp}}(CuCl) = 1 \times {10^{ - 6}}.
And the final or resultant concentration of Ag+A{g^ + } is 1.6×10−x1.6 \times {10^{ - x}}. -------------(1)
So, we have to calculate the value of xx.
Now, for copper chloride
Initially we have given 0.1M0.1Mof copper chloride which dissociates and becomes 0.1−yM0.1 - yMat equilibrium.
CuCl→Cu++Cl−CuCl \to C{u^ + } + C{l^ - }

initially0.1M0.1M0000
At. eqm.0.1−y0.1 - yyyyy

So, we can write Ksp{K_{sp}} for this,
Ksp=[Cu+][Cl−]{K_{sp}} = \left[ {C{u^ + }} \right]\left[ {C{l^ - }} \right]
At equilibrium the concentrations of both copper ion and chloride ion becomes yy.
⇒Ksp=y×y\Rightarrow {K_{sp}} = y \times y
⇒y=Ksp=10−6=10−3M\Rightarrow y = \sqrt {{K_{sp}}} = \sqrt {{{10}^{ - 6}}} = {10^{ - 3}}M
Similarly the Ksp{K_{sp}} for AgClAgCl will be
Ksp=[Ag+][Cl−]{K_{sp}} = \left[ {A{g^ + }} \right]\left[ {C{l^ - }} \right]
⇒[Ag+]=Ksp[Cl−]=1.6×10−1010−3=1.6×10−7\Rightarrow \left[ {A{g^ + }} \right] = \dfrac{{{K_{sp}}}}{{\left[ {C{l^ - }} \right]}} = \dfrac{{1.6 \times {{10}^{ - 10}}}}{{{{10}^{ - 3}}}} = 1.6 \times {10^{ - 7}} --------------(2)
Now comparing (1) with (2) we get,
1.6×10−x=1.6×10−71.6 \times {10^{ - x}} = 1.6 \times {10^{ - 7}}
⇒x=7\Rightarrow x = 7
Hence the concentration of silver ion is 1.6×10−71.6 \times {10^{ - 7}} and the value of xx is 77 .

Note:
Ksp{K_{sp}} is used to describe the saturated solution of ionic compounds. In order to calculate the solubility product constant we need to convert all the concentrations of each and every ion in terms of molarity or moles per liter.