Question

On adding 0.1 M solution each of $\left[A{g}^{+}\right],\left[B{a}^{+2}\right],\left[C{a}^{+2}\right]$ in a $N{a}_{2}S{O}_4$, solution, species first precipitated is:

$[K_{sp},BaS{O}_4={10}^{-11}M,K_{sp}CaS{O}_4={10}^{-6},K_{sp}A{g}_{2}S{O}_4={10}^{-5}]$

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

Answer 1

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

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

$\therefore \left[S{O}_4^{2-}\right]$ needed for precipitation of $A{g}_{2}S{O}_4>\frac{K_{s}p}{[A{g}^{+}{]}^2}$

= $\frac{{10}^{-5}}{01.\times 0.1}={10}^{-3}M$

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

$S{O}_4^{2-}$ needed for precipitation of $BaS{O}_4>\frac{K_{sp}}{\left[B{a}^{2+}\right]}$

= $\frac{{10}^{-11}}{0.1}={10}^{-10}M$

For $CaS{O}_4$;

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

$\therefore \left[S{O}_4^{2-}\right]$ needed for precipitation of $CaS{O}_4>\frac{K_{sp}}{\left[C{a}^{2+}\right]}=\frac{{10}^{-6}}{0.1}={10}^{-5}M$

Thus, minimum $\left[S{O}_4^{2-}\right]$ is required for precipitation of $BaS{O}_4$ and hence it is precipitated out first.