Chemistry - Ionic Equilibrium Question with Solution | TestHub

Let $s_1$ be the solubility of Ag₂CrO₄ and $s_2$ be the solubility of Ag₂C₂O₄.

The solubility equilibrium for Ag₂CrO₄ is:

Ag₂CrO₄(s) <=> 2Ag⁺(aq) + CrO₄²⁻(aq) $K_{sp1}=2\times 10^{-12}$

The solubility equilibrium for Ag₂C₂O₄ is:

Ag₂C₂O₄(s) <=> 2Ag⁺(aq) + C₂O₄²⁻(aq) $K_{sp2}=2\times 10^{-11}$

Let [Ag⁺] = $x$. Then,

$K_{sp1}=[A{g}^{+}{]}^2[Cr{O}_4^{2-}]=x^2{s}_1=2\times 10^{-12}$

$K_{sp2}=[A{g}^{+}{]}^2[C_2{O}_4^{2-}]=x^2{s}_2=2\times 10^{-11}$

Also, $s_1=\frac{2\times 10^{-12}}{x^2}$ and $s_2=\frac{2\times 10^{-11}}{x^2}$

Since both salts are present, the total silver ion concentration is $x$, and the individual solubilities are related to $x$ by:

$[Cr{O}_4^{2-}]=\frac{2\times 10^{-12}}{x^2}$ and $[C_2{O}_4^{2-}]=\frac{2\times 10^{-11}}{x^2}$

The total silver concentration is given by:

$x/2=s_1+s_2$ (INCORRECT - the silver ion concentration is $x$, not $x/2$)

The correct equations are:

$K_{sp1}=x^2[Cr{O}_4^{2-}]=2\times 10^{-12}$

$K_{sp2}=x^2[C_2{O}_4^{2-}]=2\times 10^{-11}$

$[Cr{O}_4^{2-}]=\frac{2\times 10^{-12}}{x^2}$

$[C_2{O}_4^{2-}]=\frac{2\times 10^{-11}}{x^2}$

Since the solution is saturated with respect to both salts, we can write:

$2s_1+2s_2=x$ (INCORRECT - this assumes the solubility is directly related to the silver ion concentration which is not true in this case)

Instead, we use the $K_{sp}$ expressions directly.

Let $[A{g}^{+}]=x$. Then $[Cr{O}_4^{2-}]=\frac{2\times 10^{-12}}{x^2}$ and $[C_2{O}_4^{2-}]=\frac{2\times 10^{-11}}{x^2}$.

Since the solution is saturated with both salts, we have:

$K_{sp1}=(x{)}^2[Cr{O}_4^{2-}]=2\times 10^{-12}$

$K_{sp2}=(x{)}^2[C_2{O}_4^{2-}]=2\times 10^{-11}$

We need to find $x$.

Let $s_1$ be the molar solubility of $A{g}_{2}Cr{O}_4$ and $s_2$ be the molar solubility of $A{g}_2{C}_2{O}_4$.

Then $[A{g}^{+}]=2s_1+2s_2$.

$K_{sp1}=[A{g}^{+}{]}^2[Cr{O}_4^{2-}]=(2s_1+2s_2{)}^2{s}_1=2\times 10^{-12}$

$K_{sp2}=[A{g}^{+}{]}^2[C_2{O}_4^{2-}]=(2s_1+2s_2{)}^2{s}_2=2\times 10^{-11}$

Dividing the two equations:

$\frac{s_1}{s_2}=\frac{2\times 10^{-12}}{2\times 10^{-11}}=\frac{1}{10}$

$s_2=10s_1$

Substituting into the first equation:

$(2s_1+20s_1{)}^2{s}_1=2\times 10^{-12}$

$(22s_1{)}^2{s}_1=484s_1^3=2\times 10^{-12}$

$s_1^3=\frac{2\times 10^{-12}}{484}=4.132\times 10^{-15}$

$s_1=\sqrt[3]{4.132\times 10^{-15}}=1.605\times 10^{-5}$

$s_2=10s_1=1.605\times 10^{-4}$

$$\begin{gathered} \text{displaylines}[A{g}^{+}]=2s_1+2s_2=2(1.605\times 10^{-5})+2(1.605\times 10^{-4})=3.21\times 10^{-5}+3.21\times 10^{-4} \\ \left[A{g}^{+}\right]=3.531\times 10^{-4} \end{gathered}$$

Final Answer: The final answer is $\boxed{3.52\times 10^{-4}}$