Question

On adding $0.1$ $M$ solution each of $\left[{Ag}^{+}\right],\left[{Ba}^{2+}\right],\left[{Ca}^{2+}\right]$ in ${Na}_{2}S{O}_4$ solution, species first precipitated is:
$\left[K_{sp}\right(BaS{O}_4)={10}^{-11},K_{sp}(CaS{O}_4)={10}^{-6}and{K}_{sp}({Ag}_{2}S{O}_4)={10}^{-5}]$

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

Answer 1

The correct option is B $BaS{O}_4$

On adding $0.1$ $M$ solution each of $\left[{Ag}^{+}\right],\left[{Ba}^{2+}\right],\left[{Ca}^{2+}\right]$ in ${Na}_{2}S{O}_4$ solution, species first precipitated is $BaS{O}_4$.
1. For $BaS{O}_4$

$K_{sp}BaS{O}_4=\left[B{a}^{2+}\right]\left[S{O}_4^{2-}\right]$

${10}^{-11}=0.1\times \left[S{O}_4^{2-}\right]$

$\left[S{O}_4^{2-}\right]={10}^{-10}$ M
2. For $CaS{O}_4$

$K_{sp}CaS{O}_4=\left[C{a}^{2+}\right]\left[S{O}_4^{2-}\right]$

${10}^{-6}=0.1\times \left[S{O}_4^{2-}\right]$

$\left[S{O}_4^{2-}\right]={10}^{-5}$ M
3. For ${Ag}_{2}S{O}_4$

$K_{sp}{Ag}_{2}S{O}_4=\left[A{g}^{+}{]}^2\right[S{O}_4^{2-}]$

${10}^{-5}=(0.1{)}^2\times [S{O}_4^{2-}]$

$\left[S{O}_4^{2-}\right]={10}^{-3}$ M
Thus, precipitation of $BaS{O}_4$ will require least value of $\left[S{O}_4^{2-}\right]$.