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Question: The \(K_{sp}\) of \[A{g_2}Cr{O_4}\] is \[1.1 \times {10^{ - 12}}\]at\[298{\text{ }}K\]. The solubili...

The KspK_{sp} of Ag2CrO4A{g_2}Cr{O_4} is 1.1×10−121.1 \times {10^{ - 12}}at298 K298{\text{ }}K. The solubility (in mol/L) of Ag2CrO4A{g_2}Cr{O_4} in a 0.1{\text{ }}M$$$$AgN{O_3} solution is:

A.          1.1×10−11 B.          1.1×10−10 C.          1.1×10−12 D.          1.1××10−9  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}} \\\
Explanation

Solution

In the above question, common ion effect occurs, as both Ag2CrO4A{g_2}Cr{O_4} and AgNO3AgN{O_3} have Ag+  A{g^ + }\; in common. Use the solubility equation to solve the above problem. Here Ksp means solubility product constant.

Complete answer: Concentration of AgNO3=0.1 MAgN{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: Ag2CrO4⇌2Ag++Cr2O42−A{g_2}Cr{O_4} \rightleftharpoons 2A{g^ + } + C{r_2}{O_4}^{2 - }
Another reaction of AgNO3AgN{O_3} is AgNO3→Ag++NO3−AgN{O_3} \to A{g^ + } + N{O_3}^ -
So we can see that Ag+  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 Ag2CrO4A{g_2}Cr{O_4}
Now let us solve the problem,
The equation given is Ag2CrO4⇌2Ag++Cr2O42−A{g_2}Cr{O_4} \rightleftharpoons 2A{g^ + } + C{r_2}{O_4}^{2 - } . Here 11 mole of Ag2CrO4A{g_2}Cr{O_4} gives 22 moles of Ag+  A{g^ + }\; and 11 moles of Cr2O42-
Thus solubility of Ag2CrO4A{g_2}Cr{O_4} will be‘s’ moles/liter, solubility of Ag+  A{g^ + }\; will be ‘2s’ moles/liter and solubility of Cr2O42−C{r_2}{O_4}^{2 - } will be‘s’ moles/liter.
Hence, Ksp=[Ag+]2[CO32−]{K_{sp}} = {[A{g^ + }]^2}[C{O_3}^{2 - }]
⇒Ksp=[2s]2[s]\Rightarrow {K_{sp}} = {[2s]^2}[s]
Next we have another reaction, AgNO3→Ag++NO3−AgN{O_3} \to A{g^ + } + N{O_3}^ -
In this reaction, concentration of AgNO3AgN{O_3} is given as 0.1 M0.1{\text{ }}M in the question. Here 11 mole of AgNO3AgN{O_3} gives11mole of Ag+  A{g^ + }\; and 11 mole of NO32−N{O_3}^{2 - } and so 0.10.1 mole of AgNO3AgN{O_3} will give 0.10.1 mole of Ag+ and 0.10.1 mole of NO32−N{O_3}^{2 - } .
Therefore Ksp=[Ag+][NO32−]{K_{sp}} = [A{g^ + }][N{O_3}^{2 - }]
Since both the reactions have common ion Ag+  A{g^ + }\; so now [Ag+]=[2s+0.1][A{g^ + }] = [2s + 0.1] and [NO32−]=s[N{O_3}^{2 - }] = s
Thus, Ksp=[Ag+][NO32−]{K_{sp}} = [A{g^ + }][N{O_3}^{2 - }]
⇒Ksp=[2s+0.1]2[s]\Rightarrow {K_{{\text{sp}}}} = {[2s + 0.1]^2}[s]
As 2s is very smaller than0.10.1so it can be neglected.
So Ksp=[0.1]2[s]{K_{sp}} = {[0.1]^2}[s]
⇒1.1×10−12=[0.1]2[s]\Rightarrow 1.1 \times {10^{ - 12}} = {[0.1]^2}[s]
⇒1.1×10−12[0.01]=[s]\Rightarrow \dfrac{{1.1 \times {{10}^{ - 12}}}}{{[0.01]}} = [s]
⇒[s]=1.1×10−10M\Rightarrow [s] = 1.1 \times {10^{ - 10}}M
Therefore the solubility of Ag2CrO4A{g_2}Cr{O_4} is 1.1×10−10M1.1 \times {10^{ - 10}}M
So the correct option is B.    1.1×10−10B.\;\;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.