Question

$R{H}_2$ (ion exchange resin) can replace $C{a}^{2+}$ ions in hard water as: $R{H}_2+C{a}^{2+}\to RCa+2H^{+}$.

If $1L$ of hard water after passing through $R{H}_2$ has $pH=3$, then hardness in parts per million of $C{a}^{2+}$ is:

• A. 80
• B. 10
• C. 40
• D. 100

Answer 1

The correct option is A 80

The hydrogen ion concentration is calcuated from pH
$\left[H^{+}\right]={10}^{-pH}={10}^{-3}M$
As two $H^{+}$ ions are needed for 1 calcium ion, the concentration of Ca(II) is twice the concentration of hydrogen ions.
$\left[C{a}^{2+}\right]=2\left[H^{+}\right]=2\times {10}^{-3}M$
The molar mass of Ca is 40.
Hence, the ppm of Ca(II) $=\frac{40\times 2\times {10}^{-3}}{1000}\times {10}^6=80ppm$
Thus, the hardness of water in terms of Ca(II) is 80 ppm.