Question

Results of a water sample analysis are as follows.

Cation	Concentration $\left(mg/L\right)$	Equivalent weight
$N{a}^{+}$	$40$	$23$
$M{g}^{2+}$	$10$	$12.2$
$C{a}^{2+}$	$55$	$20$
$K^{+}$	$2$	$39$

(Milliequivalent weight of $CaC{O}_3=50mg/meq$). Hardness of the water sample in mg/L as $CaC{O}_3$ is

• A. 358
• B. 89.5
• C. 179
• D. 44.8

Answer 1

The correct option is C 179

Hardness $=\left[M{g}^{2+}\right]\times \frac{\text{Eq.wt.of}CaC{O}_3}{\text{Eq.wt.of}\left[M{g}^{2+}\right]}+\left[C{a}^{2+}\right]\times \frac{\text{Eq.wt.of}CaC{O}_3}{\text{Eq.wt.of}\left[C{a}^{2+}\right]}$

$=10\times \frac{50}{12.2}+\frac{55\times 50}{20}=178.48$

$\simeq 179mg/LasCaC{O}_3$