Question

A sample of hard water contains $0.005mole$ of calcium chloride $perlitre$. What is the minimum concentration of sodium sulphate which must be added for removing the ${Ca}^{2+}$ ions from this water sample? $(K_{sp}$ of $Ca{SO}_4=2.4\times {10}^{-5})$

• A. $4.8\times {10}^{-2}$
• B. $4.8\times {10}^{-3}$
• C. $2.4\times {10}^{-2}$
• D. $2.4\times {10}^{-3}$

Answer 1

The correct option is B $4.8\times {10}^{-3}$

Hard water contains $0.005$ mole of ${CaCl}_2$ per litre.

$\therefore$ $\left[{Ca}^{2+}\right]=0.005M$

now, $\left[{Ca}^{2+}\right]\times \left[{SO}_4^{2-}\right]=K_{sp}=2.4\times {10}^{-5}$

$\Rightarrow \left[{SO}_4^{2-}\right]=\frac{2.4\times {10}^{-5}}{0.005}=4.8\times {10}^{-3}M$

$\therefore$ The minimum concentration of ${Na}_2{SO}_4$ must be $4.8\times {10}^{-3}$ mole per litre.