Question

A sample of hard water contains 0.005 moles of $CaC{l}_2$ per litre. What is the minimum concentration Of $N{a}_{2}S{O}_4$ which must be added for removing $C{a}^{2+}$ ions from the water sample?

[$K_{sp}$ for $CaS{O}_4$ is $2.4\times {10}^{-5}$ at ${25}^{\circ }C$]

• A. $2.8\times 0^{-5}M$
• B. $4.8\times {10}^{-3}M$
• C. $6.1\times {10}^{-3}M$
• D. $4.1\times {10}^{-5}M$

Answer 1

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

The concentration of calcium ions is $0.005M$.

The solubility product of $CaS{O}_4$ is $2.4\times {10}^{-5}$

The expression for the solubility product of calcium sulphate is as shown below.

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

Substitute values in the above expression:

$2.4\times {10}^{-5}=0.005\left[S{O}_4^{2-}\right]$

Hence, $\left[S{O}_4^{2-}\right]=4.8\times {10}^{-3}$.

Therefore, option B is correct.