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Question: The solubility of AgCl (s) with solubility product \(1.6 \times {10^{ - 10}}\)in 0.1 M NaCl solution...

The solubility of AgCl (s) with solubility product 1.6×10−101.6 \times {10^{ - 10}}in 0.1 M NaCl solution would be
(A) 1.6×10−11M1.6 \times {10^{ - 11}}M
(B) Zero
(C) 1.26×10−5M1.26 \times {10^{ - 5}}M
(D) 1.6×10−91.6 \times {10^{ - 9}}M.

Explanation

Solution

The solubility product constant (Ksp{K_{sp}}) of the compound in equilibrium with the saturated solution is the product of the concentration of both the ions formed after the dissociation of the compound. The concentration of the ionic compound remains the same in its constituent ion at equilibrium.

Complete step by step answer:
Given, the solubility product of AgCl is 1.6×10−101.6 \times {10^{ - 10}}.
The concentration of the NaCl solution is 0.1 M.
The dissociation of sodium chloride NaCl is shown below.
NaCl⇌Na++Cl−NaCl \rightleftharpoons N{a^ + } + C{l^ - }
Here, as sodium chloride completely dissociates into its constituent ions, the concentration of the sodium chloride remains the same.
0.1 M sodium chloride NaCl gives 0.1M sodium ion Na+N{a^ + } and 0.1M chloride ion Cl−C{l^ - }.
The dissociation of silver chloride is shown below.
AgCl⇌Ag++Cl−AgCl \rightleftharpoons A{g^ + } + C{l^ - }
The solubility of silver chloride is given by s, so the solubility of silver ion is s and it is given that the solubility of chloride ion is 0.1 M, so here the solubility of chloride ion will be s + 0.1.
The solubility product constant is the product of its constituent ions.
Hence, the solubility product constant for silver chloride is given as shown below.
Ksp=[Ag+][Cl−]{K_{sp}} = [A{g^ + }][C{l^ - }]
Substitute the given values in the above equation.
1.6×10−10=s(s+0.1)1.6 \times {10^{ - 10}} = s(s + 0.1)
⇒1.6×10−10=s(0.1)\Rightarrow 1.6 \times {10^{ - 10}} = s(0.1) (s < < < < < 0.1)
⇒s=1.6×10−100.1\Rightarrow s = \dfrac{{1.6 \times {{10}^{ - 10}}}}{{0.1}}
⇒s=1.6×10−9M\Rightarrow s = 1.6 \times {10^{ - 9}}M
Thus, the solubility of AgCl (s) with solubility product 1.6×10−101.6 \times {10^{ - 10}}in 0.1 M NaCl solution is 1.6×10−91.6 \times {10^{ - 9}}M.

Thus, the correct option is option (D).

Note:
Make sure that the dissolution of silver chloride is done in sodium chloride. As the concentration of chloride ion remains the same during the dissolution, the concentration value of chloride ion is the same for both the ionic compound.