Question

$R{H}_2$ (ion exchange resin) can replace $C{a}^{2+}$ in hard water
$R{H}_2+C{a}^{2+}⟶RCa+2H^{+}$
$1L$ of hard water after passing through $R{H}_2$ has $pH2$. Hence, hardness in ppm of $C{a}^{2+}$ is:

• A. 200
• B. 100
• C. 50
• D. 125

Answer 1

Answer: (A)

$R{H}_2+C{a}^{2+}⟶RCa+2H^{+}$

Energy $C{a}^{2+}$ ion is replaced by two $H^{+}$ ions $pH=2$

$\therefore \left[H^{+}\right]=10^{-2}M$

$\therefore \left[C{a}^{2+}\right]=\frac{10^{-2}}{2}M=\frac{10^{-2}\times 40}{2}g{L}^{-1}$

$\therefore ppm$ of $C{a}^{2+}=\frac{10^{-2}\times 40\times 10^3}{2\times 10^6}C{a}^{2+}=200$