Question

On adding $0.1M$ solution each of $[A{g}^{+}],[B{a}^{2+}],[C{a}^{2+}]$ in a $N{a}_{2}S{O}_4$ solution, species first precipitated is
$[K_{sp}BaS{O}_4=10^{-11},K_{sp}CaS{O}_4=10^{-6}$ and $K_{sp}AgS{O}_4=10^{-5}]$

• A. $A{g}_{2}S{O}_4$
• B. $BaS{O}_4$
• C. $CaS{O}_4$
• D. All of these

Answer 1

Answer: (B)

The compound which is having least solubility will be precipitated first
(a) $BaS{O}_4$ Given $K_{sp}=10^{-11}$
Let the solubility $=xmol/L$
$BaS{O}_4⟶\underset{x}{B{a}^{2-}}+\underset{x}{S{O}_4^2}$
$\therefore K_{sp}=\left[B{a}^{2+}\right]\left[S{O}_4^{2-}\right]$
or $K_{sp}=x\times x$
$\therefore K_{sp}=x^2$
or $x=\sqrt{K_{sp}}$
$=\sqrt{10^{-11}}$
$=3.16\times 10^{-6}mol/L$
(b) $CaS{O}_3$ Given, $K_{Sp}=10^{-6}$
Let the solubility $=xmol/L$
$CaS{O}_4⟶C{a}^{2+}+S{O}_4^{2-}$
$\therefore K_{sp}=\left[C{a}^{2+}\right]\left[S{O}_4^{2-}\right]$
or $K_{sp}=x\times x$
$\therefore K_{sp}=x^2$
or $x=\sqrt{K_{sp}}$
$=\sqrt{10^{-6}}$
$=1\times 10^{-3}mol/L$
(c) $A{g}_{2}S{O}_4$ Given, $K_{sp}=10^{-5}$
Let the solubility $=xmol/L$
$A{g}_{2}S{O}_4⟶\underset{2x}{2A{g}^{+}}+\underset{x}{S{O}_4^{2-}}$
$\therefore K_{sp}={\left[A{g}^{2+}\right]}^2\left[S{O}_4^{2-}\right]$
or $=(2x{)}^2(x)$
or $K_{sp}=4x^3$
or $x=\sqrt[3]{\frac{K_{sp}}{4}}$
or $=\sqrt[3]{\frac{10^{-5}}{4}}$
$=10^{-2}mol/L$
$\because BaS{O}_4$ has least solubility.
$\therefore$ It will be precipitated first.