Question

The $K_{sp}$ of \[A{g_2}Cr{O_4}\] is \[1.1 \times {10^{ - 12}}\]at\[298{\text{ }}K\]. The solubility (in mol/L) of \[A{g_2}Cr{O_4}\] in a \[0.1{\text{ }}M\]\[AgN{O_3}\] solution is:
\[
  A.\;\;\;\;\;1.1 \times {10^{ - 11}} \\
  B.\;\;\;\;\;1.1 \times {10^{ - 10}} \\
  C.\;\;\;\;\;1.1 \times {10^{ - 12}} \\
  D.\;\;\;\;\;1.1 \times \times {10^{ - 9}} \\
 \]

Answer 1

Hint: In the above question, common ion effect occurs, as both \[A{g_2}Cr{O_4}\] and \[AgN{O_3}\] have \[A{g^ + }\;\] in common. Use the solubility equation to solve the above problem. Here Ksp means solubility product constant.

Complete answer: Concentration of \[AgN{O_3} = 0.1{\text{ }}M\] and we have to find the solubility of the solution.
Ksp is the solubility product constant of a substance. It is defined as the equilibrium constant for a solid substance dissolving in an aqueous solution.
The equation for the reaction taking place here is: \[A{g_2}Cr{O_4} \rightleftharpoons 2A{g^ + } + C{r_2}{O_4}^{2 - }\]
Another reaction of \[AgN{O_3}\] is \[AgN{O_3} \to A{g^ + } + N{O_3}^ - \]
So we can see that \[A{g^ + }\;\] is common in both of the reactions. So here the common ion effect is taking place. Common ion effect occurs when addition of common ions decreases the solubility of the salt. In our case the salt is \[A{g_2}Cr{O_4}\]
Now let us solve the problem,
The equation given is \[A{g_2}Cr{O_4} \rightleftharpoons 2A{g^ + } + C{r_2}{O_4}^{2 - }\] . Here $1$ mole of \[A{g_2}Cr{O_4}\] gives $2$ moles of \[A{g^ + }\;\] and $1$ moles of Cr2O42-
Thus solubility of \[A{g_2}Cr{O_4}\] will be‘s’ moles/liter, solubility of \[A{g^ + }\;\] will be ‘2s’ moles/liter and solubility of \[C{r_2}{O_4}^{2 - }\] will be‘s’ moles/liter.
Hence, \[{K_{sp}} = {[A{g^ + }]^2}[C{O_3}^{2 - }]\]
 \[ \Rightarrow {K_{sp}} = {[2s]^2}[s]\]
Next we have another reaction, \[AgN{O_3} \to A{g^ + } + N{O_3}^ - \]
In this reaction, concentration of \[AgN{O_3}\] is given as \[0.1{\text{ }}M\] in the question. Here $1$ mole of \[AgN{O_3}\] gives$1$mole of \[A{g^ + }\;\] and $1$ mole of \[N{O_3}^{2 - }\] and so \[0.1\] mole of \[AgN{O_3}\] will give \[0.1\] mole of Ag+ and \[0.1\] mole of \[N{O_3}^{2 - }\] .
Therefore \[{K_{sp}} = [A{g^ + }][N{O_3}^{2 - }]\]
Since both the reactions have common ion \[A{g^ + }\;\] so now \[[A{g^ + }] = [2s + 0.1]\] and \[[N{O_3}^{2 - }] = s\]
Thus, \[{K_{sp}} = [A{g^ + }][N{O_3}^{2 - }]\]
 \[ \Rightarrow {K_{{\text{sp}}}} = {[2s + 0.1]^2}[s]\]
As 2s is very smaller than$0.1$so it can be neglected.
So \[{K_{sp}} = {[0.1]^2}[s]\]
 \[ \Rightarrow 1.1 \times {10^{ - 12}} = {[0.1]^2}[s]\]
 \[ \Rightarrow \dfrac{{1.1 \times {{10}^{ - 12}}}}{{[0.01]}} = [s]\]
 \[ \Rightarrow [s] = 1.1 \times {10^{ - 10}}M\]
Therefore the solubility of \[A{g_2}Cr{O_4}\] is \[1.1 \times {10^{ - 10}}M\]
So the correct option is \[B.\;\;1.1 \times {10^{ - 10}}\]

Note:
Solubility product is a kind of equilibrium constant. Its value depends upon temperature. When there is increase in the temperature the value of Ksp increases and this leads to increase in solubility. The solubility product constant is an important concept in chemistry in studying the solubility of different solutes.