Question

Salt mixture containing ${Cu}_3{\left(As{O}_4\right)}_2$ $(K_{sp}=8\times {10}^{-36})$ and ${Pb}_3{\left(As{O}_4\right)}_2$ $(K_{sp}=4.096\times {10}^{36})$ is shaken with water. Find the concentration of metal cations in the solution at equilibrium. Neglect any hydrolysis of the dissolved ions.

• A. $\left[{Cu}^{+2}\right]=8.825\times {10}^{-8}$, $\left[{Pb}^{+2}\right]=7.119\times {10}^{-8}$
• B. $\left[{Cu}^{+2}\right]=8.825\times {10}^{-6}$, $\left[{Pb}^{+2}\right]=7.119\times {10}^{-6}$
• C. $\left[{Cu}^{+2}\right]=8.825\times {10}^{-9}$, $\left[{Pb}^{+2}\right]=7.119\times {10}^{-9}$
• D. $\left[{Cu}^{+2}\right]=8.825\times {10}^{-5}$, $\left[{Pb}^{+2}\right]=7.119\times {10}^{-5}$

Answer 1

The correct option is A $\left[{Cu}^{+2}\right]=8.825\times {10}^{-8}$, $\left[{Pb}^{+2}\right]=7.119\times {10}^{-8}$

Let solubility of $C{u}_3(As{O}_4{)}_2$ and $P{b}_3(As{O}_4{)}_2$ be $x$ and $y$ respectively.

$C{u}_3(As{O}_4{)}_2\rightleftharpoons 3C{u}^{+2}+2As{O}_4^{-3}{K}_{sp}=8\times {10}^{-36}$
$3x$ $2x+2y$
$P{b}_3(As{O}_4{)}_2\rightleftharpoons 3P{b}^{+2}+2As{O}_4^{-3}{K}_{sp}=4.096\times {10}^{-36}$
$3y$ $2y+2x$
$108x^3{(x+y)}^2=8\times {10}^{-36}........\left(i\right)$
$108y^3{(x+y)}^2=4.096\times {10}^{-36}.....\left(ii\right)$

Dividing (i) by (ii), we get
$\frac{\left(i\right)}{\left(ii\right)}$ $\Rightarrow$ $\frac{x^3}{y^3}=\frac{8}{4.096}$
$\Rightarrow$ $x=1.25y$
Putting this in equation $\left(ii\right)$, we get
$108y^3{\left(2.25y\right)}^2=4.096\times {10}^{-36}$
$y=2.3\times {10}^{-8}$ $x=1.25$ $y=2.875\times {10}^{-8}$
$\left[{Cu}^{+2}\right]=3x=8.825\times {10}^{-8}$
$\left[{Pb}^{+2}\right]=3y=7.1\times {10}^{-8}$