Question

A hard water sample has $131$ ppm $CaS{O}_4$. What fraction of the water must be evaporated in a container before solid $CaS{O}_4$ begins to deposit. $K_{sp}$ of $CaS{O}_4=9.0\times {10}^{-6}$

• A. $52$
• B. $68$
• C. $85$
• D. $42$

Answer 1

Answer: (B)

$68$

Maximum solubility of $CaS{O}_4$ in water
$S=\sqrt{K_{sp}}=3\times {10}^{-3}mol{L}^{-1}$

Let V litre of sample is taken, then $CaS{O}_4$ present
$=\frac{131\times V\times {10}^3}{{10}^6}g$
$[\because ppm=g$ of $CaS{O}_4$ in ${10}^{6}g$ of sample$]$

$=131\times {10}^{-3}Vg$
$=\frac{131\times {10}^{-3}\times V}{136}$ mole in V L

If water is evaporated on heating so that just precipitation of $CaS{O}_4$ occurs. Let $V_1$ L of water is left, then
$\frac{131\times {10}^{-3}\times V}{136}$ mol is present in $V_1$ L solution is equal to $3\times {10}^{-3}\times V_1$ mol.

$\therefore \frac{131\times {10}^{-3}\times V}{136}=3\times {10}^{-3}\times V_1\therefore V_1=0.32V$

Thus, volume evaporated $=V-0.32V=0.68V$ or 68% of water should be evaporated.