A sample of hard water contains 0.005 mole of CaCl₂ per litre. What is the minimum concentration of ${Na}_{2} {SO}_{4}$ (in mole/ L)which must be exceeded for removing ${Ca}^{2 +}$ ions from this water sample? The solubility product of CaSO₄at 25° C is 2.4 x 10^-5 • (Give your answer after multiplying with 10000 and· excluding any decimal places)

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answer is 48.

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Detailed Solution

Here, we know that,

$[{Ca}^{2 +}] \times [{SO}_{4}^{2 -}] = K_{sp}$

Put the values,

$\begin{array}{l} 5 \times 10^{- 3} \times [{SO}_{4}^{2 -}] = 2.4 \times 10^{- 5} \\ \Rightarrow [{SO}_{4}^{2 -}] = 48 \times 10^{- 3} \end{array}$

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