Question

What is the concentration of $A{g^ + }$ in a solution made by dissolving both $A{g_2}Cr{O_4}$ and $A{g_2}{C_2}{O_4}$ until saturation is reached with respect to both the salts?
[${K_{sp}}(A{g_2}{C_2}{O_4}) = 2 \times {10^{ - 11}}$, ${K_{sp}}(A{g_2}Cr{O_4}) = 2 \times {10^{ - 12}}$]
(A) $2.80 \times {10^{ - 4}}$
(B) $7.6 \times {10^{ - 5}}$
(C) $6.63 \times {10^{ - 6}}$
(D) $3.52 \times {10^{ - 4}}$

Answer 1

In order to find the concentration of $A{g^ + }$ in a solution, we must know about the solubility product equation. Solubility product constant is similar to equilibrium constant but this involves the dissolution of the solid substance in aqueous solution.

Complete Solution :
Before moving onto the problem, we must know what a solubility product constant is. Solubility product constant is a type of equilibrium constant for dissolving the solid substance into aqueous solution. It is represented by the symbol ${K_{sp}}$.
Let us now consider $A{g_2}Cr{O_4}$,
$A{g_2}Cr{O_4}$ will get dissociated into $2A{g^ + }$and $CrO_4^{2 - }$ions. The reaction is given below:
$AgCr{O_4}(s) \rightleftharpoons 2A{g^ + }(aq) + CrO_4^{2 - }(aq)$
The solubility product for $A{g_2}Cr{O_4}$ is written as
${K_{sp}} = {[A{g^ + }]^2}[CrO_4^{2 - }]$
Where ${K_{sp}}(A{g_2}Cr{O_4}) = 2 \times {10^{ - 12}}$
$2 \times {10^{ - 12}} = {[A{g^ + }]^2}[CrO_4^{2 - }]$……… (1)

- Let us now consider $A{g_2}{C_2}{O_4}$,
$A{g_2}{C_2}{O_4}$ will be dissociated into $2A{g^ + }$and ${C_2}O_4^{2 - }$ ions. The reaction is given below:
$Ag{C_2}{O_4}(s) \rightleftharpoons 2A{g^ + }(aq) + {C_2}O_4^{2 - }(aq)$
The solubility product of $A{g_2}{C_2}{O_4}$ is written as
${K_{sp}} = {[A{g^ + }]^2}[{C_2}O_4^{2 - }]$
Where ${K_{sp}}(A{g_2}{C_2}{O_4}) = 2 \times {10^{ - 11}}$
$2 \times {10^{ - 11}} = {[A{g^ + }]^2}[{C_2}O_4^{2 - }]$………. (2)

Now consider,
$[CrO_4^{2 - }] = x$
$[{C_2}O_4^{2 - }] = y$
$[A{g^ + }] = 2x + 2y$
Substituting these values in equation (1) and (2)
From equation (1), we get
$2 \times {10^{ - 12}} = {(2x + 2y)^2}x$………. (3)
$2 \times {10^{ - 11}} = {(2x + 2y)^2}y$……… (4)
Divide equation (3) and (4), we get
$\dfrac{{2 \times {{10}^{ - 12}}}}{{2 \times {{10}^{ - 11}}}} = \dfrac{{{{(2x + 2y)}^2}x}}{{{{(2x + 2y)}^2}y}}$
$0.1 = \dfrac{x}{y}$
$x = 0.1y$

Substituting the above value in equation (4)
$2 \times {10^{ - 11}} = {(2 \times 0.1y + 2y)^2}y$
$2 \times {10^{ - 11}} = 4.84{y^3}$
Therefore, we get
$y = 1.6 \times {10^{ - 4}}$
$x = 0.16 \times {10^{ - 4}}$
Total $[A{g^ + }] = 2x + 2y$
$[A{g^ + }] = 2 \times 0.16 \times {10^{ - 4}} + 2 \times 1.6 \times {10^{ - 4}}$
$[A{g^ + }] = 3.52 \times {10^{ - 4}}$
Hence the concentration of $A{g^ + }$ in a solution is $3.52 \times {10^{ - 4}}$
So, the correct answer is “Option D”.

Additional information - Solubility of a substance depends on the following parameters
- Solubility of a substance will depend on the temperature. The temperature will be varying for all salts.
- In order to dissolve a solute in a solvent, the solvation enthalpy should be greater than the lattice enthalpy.
- Non-polar solvents will have only low solvation enthalpy.
- solvation enthalpy of ions is always negative i.e.; it means energy will be released during the process.

Note: Solubility of a substance is determined based on the table given below:

Soluble	Solubility > 0.1M
Sparingly Soluble	0.001M < Solubility <0.1M
Insoluble	Solubility < 0.1M