Question

A solution prepared by mixing $200ml$ of $0.025M$ $CaC{l}_2$ and $400ml$ of $0.15M$ $N{a}_{2}S{O}_4$? Given $K_{sp}$ of $CaS{O}_4$ = $2.4\times {10}^{-5}$ Then

• A. will get precipitate
• B. No precipitate is formed
• C. get ready to precipitate
• D. NaCl will get precipitate

Answer 1

The correct option is A

will get precipitate

The equation for the equilibrium is

$CaS{O}_4\left(s\right)$ $\rightleftharpoons$ $C{a}^{2+}\left(aq\right)+S{O}_4^{2-}\left(aq\right)$

and the solubility product expression is

$K_{sp}$ = $\left[C{a}^{2+}\right]\left[S{O}_4^{2-}\right]$ = $2.4\times {10}^{-5}$

If we assume that the volumes of the solutions that are mixed are additive, the final solution will have a volume of 600 ml. This total volume contains the equivalent of 200 ml of $CaC{l}_2$, so the concentration of $C{a}^{2+}$ ions is

$\left[C{a}^{2+}\right]$ = $\left(\frac{200mlCaC{l}_{2}solution}{600mltotalvolume}\right)$$\times \left(0.15M\right)$

= $8.33\times {10}^{-3}M$

and the concentration of $S{O}_4^{2-}$ ion is:

$\left[S{O}_4^{2-}\right]$ = $\left(\frac{400mlN{a}_{2}S{O}_{4}solution}{600mltotalvolume}\right)\times \left(0.15M\right)$

= $0.1M$

The ion product is

$\left[C{a}^{2+}\right]\left[S{O}_4^{2-}\right]$ = $(8.33\times {10}^{-3})(1.0\times {10}^{-1})$

= $8.33\times {10}^{-4}$

Which is larger than $K_{sp}$, so $CaS{O}_4$ should precipitate from the solution.