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

The solubility of AgCl(s)AgCl(s) with solubility product 1.6×10−101.6 \times {10^{ - 10}} in 0.1M0.1M NaClNaCl solution would be
a) 1.6×10−111.6 \times {10^{ - 11}}
b) Zero
c) 1.26×10−51.26 \times {10^{ - 5}}
d) 1.6×10−91.6 \times {10^{ - 9}}

Explanation

Solution

The representation of concentration of dissolved ions raise to the power their stoichiometric coefficients is known as the solubility product. When two salts with the same ion will be given, we have to consider the total value of that ion in the solution media.

Complete step by step solution:
Solubility products can be defined as the product of the concentration of ions when dissolved in a solution raised to the power of their stoichiometric coefficients. It is denoted by Ksp{K_{sp}} and the relation can be written as;
AaBb(s)→a[A]+b[B]{A_a}{B_b}(s) \to a[A] + b[B] . For this reaction, Ksp=[A]a[B]b{K_{sp}} = {[A]^a}{[B]^b} where aa and bb are the stoichiometric coefficients of the reactants.
According to the question, the concentration of NaClNaCl is 0.1M0.1M . Thus after dissociation in solution the two ions will have 0.1M0.1M concentration.
NaCl(s)→Na+(0.1M)+Cl−(0.1M)NaCl(s) \to N{a^ + }(0.1M) + C{l^ - }(0.1M)
The next reaction of dissociation of silver chloride will be as AgCl(s)→Ag++Cl−AgCl(s) \to A{g^ + } + C{l^ - }
Initially at some concentration: ss −- −-
After dissociation: −- ss (s+0.1)(s + 0.1)
Let’s say ss is the concentration of ion from AgClAgCl , the solubility product will be written as
KspAgCl=[Ag+][Cl−]{K_{s{p_{AgCl}}}} = [A{g^ + }][C{l^ - }]
On substituting the values of concentration we get
KspAgCl=[s][s+0.1]{K_{s{p_{AgCl}}}} = [s][s + 0.1]
When two salts with the same ion are given, we have to consider the total value of concentration of that ion in the solution media. This is because in the solution as the same ion will be coming from both the salts the concentration of that ion will change drastically. Thus the total added quantity is considered to calculate the concentration of corresponding ions.
It is given that KspAgCl=1.6×10−10{K_{s{p_{AgCl}}}} = 1.6 \times {10^{ - 10}} , solving the equation
1.6×10−10=s(s+0.1) ⇒1.6×10−10=s2+0.1s  1.6 \times {10^{ - 10}} = s(s + 0.1) \\\ \Rightarrow 1.6 \times {10^{ - 10}} = {s^2} + 0.1s \\\
As the value of Ksp{K_{sp}} is very small we can neglect the value of ss , thus the equation became

⇒1.6×10−10=s×0.1 ⇒s=1.6×10−100.1 ⇒s=1.6×10−9  \Rightarrow 1.6 \times {10^{ - 10}} = s \times 0.1 \\\ \Rightarrow s = \dfrac{{1.6 \times {{10}^{ - 10}}}}{{0.1}} \\\ \Rightarrow s = 1.6 \times {10^{ - 9}} \\\

Hence the solubility of AgClAgCl is 1.6×10−91.6 \times {10^{ - 9}} .
The correct option is (a).

Note:
The precipitate silver chloride is white in colour, formed when salt containing chloride ion is mixed with silver nitrate solution. This test is used in the qualitative analysis of anions whereas bromide and iodide form pale yellow and orangish-yellow precipitate of silver bromide and silver iodide.