Question

Consider two sparingly soluble salts $AB$ and $AC$.
If $AB$ has solubility as $xmol{L}^{-1}$ in presence of $AC$ , and $AC$ has solubility as $ymol{L}^{-1}$ in presence of $AB$.
Which of the following relations are correct?

• A. $\frac{K_{s{p}_1}}{K_{s{p}_2}}=\frac{x}{y}$
• B. $x-y=\sqrt{K_{s{p}_1}+K_{s{p}_2}}$
• C. $x+y=\sqrt{K_{s{p}_1}+K_{s{p}_2}}$
• D. $\frac{K_{s{p}_2}}{K_{s{p}_1}}=\frac{x}{y}$

Answer 1

Answer: (A), (C)

$x+y=\sqrt{K_{s{p}_1}+K_{s{p}_2}}$

Simultaneous solubility:

Sparingly soluble salt 1: $AB$ having $xmol{L}^{-1}$ solubility in the presence of $AC$

Sparingly soluble salt 2: $AC$ having $ymol{L}^{-1}$ solubility in the presence of $AB$

$AB\left(s\right)\rightleftharpoons A^{+}\left(aq\right)+B^{-}\left(aq\right)with{K}_{s{p}_1}$

$Equilibrium:C_1-xx+yx$

$AC\left(s\right)\rightleftharpoons A^{+}\left(aq\right)+C^{-}\left(aq\right)with{K}_{s{p}_2}$

$Equilibrium:C_2-yy+xy$

From the above reactions the $K_{sp}$ values are

$K_{s{p}_1}=\left(x\right)(x+y)$ $-\left(1\right)$

$K_{s{p}_2}=\left(y\right)(x+y)$ $-\left(2\right)$

Dividing both the equations

$\frac{K_{s{p}_1}}{K_{s{p}_2}}=\frac{x}{y}$

Adding both the equations

$K_{s{p}_1}+K_{s{p}_2}=(x+y{)}^2$

$x+y=\sqrt{K_{s{p}_1}+K_{s{p}_2}}$

$\therefore$ Option A and option C are correct.