Question

A $200$ g sample of hard water is passed through the column of cation exchange resin in which $H^{+}$ is exchanged by $C{a}^{2+}$. The outlet water of column required $50$ mL of $0.1MNaOH$ for complete neutralization. What is the hardness of $C{a}^{2+}$ ion in ppm?

• A. $250$ ppm
• B. $500$ ppm
• C. $750$ ppm
• D. $1000$ ppm

Answer 1

The correct option is B $500$ ppm

Milli-moles of $H^{+}$ present in outlet water = Milli-moles of $O{H}^{-}=50\times 0.1=5$
Milli-moles of $C{a}^{2+}$ present in hard water$=\frac{5}{2}$$=2.5$ ($1C{a}^{2+}$ replaced by $2H^{+}$)
Number of mg of $C{a}^{2+}$ ion$=2.5\times 40=100$
$200$ g sample hard water contain $100$ mg of $C{a}^{2+}$.
$\therefore {10}^6$ g sample hard water contain $\frac{100}{200}\times {10}^6\times {10}^{-3}=500$ ppm.