Question

On adding $0.1 M$ solution each of $[Ag^{+}], [Ba^{2+}], [Ca^{2+}]$ in a $Na_{2}SO_{4}$ solution, species first precipitated is
$[K_{sp} BaSO_{4}=10^{-11}, K_{sp} CaSO_{4}=10^{-6}$ and $K_{sp} AgSO_{4}=10^{-5}]$

• A. $Ag_{2}SO_{4}$
• B. $BaSO_{4}$
• C. $CaSO_{4}$
• D. All of these

Answer 1

Answer: (B)

The compound which is having least solubility will be precipitated first
(a) $BaSO _{4}$ Given $K_{sp}=10^{-11}$
Let the solubility $=x\,mol/L$
$BaSO _{4} \longrightarrow \underset{x}{ Ba ^{2-}}+\underset{x}{SO_{4}^{2}}$
$\therefore K_{sp}=\left[ Ba ^{2+}\right]\left[ SO _{4}^{2-}\right]$
or $K_{sp}=x \times x$
$\therefore K_{s p}=x^{2}$
or $x =\sqrt{K_{s p}}$
$=\sqrt{10^{-11}}$
$=3.16 \times 10^{-6} mol/L$
(b) $CaSO _{3}$ Given, $K _{ Sp }=10^{-6}$
Let the solubility $=x mol /L$
$CaSO_4 \longrightarrow Ca^{2+} +SO^{2-}_{4}$
$\therefore K_{sp} =\left[ Ca ^{2+}\right]\left[ SO _{4}^{2-}\right]$
or $K_{s p} =x \times x$
$\therefore K_{s p} =x^{2}$
or $x =\sqrt{K_{s p}}$
$=\sqrt{10^{-6}}$
$=1 \times 10^{-3} mol/L$
(c) $Ag _{2} SO _{4}$ Given, $K_{ sp }=10^{-5}$
Let the solubility $=x mol / L$
$Ag_2SO_4 \longrightarrow \underset{2x}{2Ag^+} +\underset{x}{SO^{2-}_{4}}$
$\therefore K_{ sp } =\left[ Ag ^{2+}\right]^{2}\left[ SO _{4}^{2-}\right]$
or $=(2 x)^{2}(x)$
or $K_{ sp } =4 x^{3}$
or $x =\sqrt[3]{\frac{K_{ sp }}{4}}$
or $=\sqrt[3]{\frac{10^{-5}}{4}}$
$=10^{-2} mol / L$
$\because BaSO _{4}$ has least solubility.
$\therefore $ It will be precipitated first.