Question

The analysis of a water sample produces the following results

$$\begin{array}{ccc}\text{Ion}& \text{Concentration}mg/l& \text{Atomic Weight}\\ C{a}^{2+}& \text{20.0}& \text{60}\\ M{g}^{2+}& \text{12.2}& \text{36.6}\\ N{a}^{+}& \text{23.0}& \text{92}\\ K^{+}& \text{39.1}& \text{78.2}\\ S{O}_4^{2-}& \text{48.0}& \text{72}\\ HC{O}_3^{-}& \text{61.0}& \text{122}\end{array}$$The total hardness (in mg/L as $CaC{O}_3$) of the water sample is

• A. 300

Answer 1

The correct option is A 300
Hardness will be due to multivalent cations only
$C{a}^{2+}=60mg/L\to$
equivalent weight =20 (given)

$M{g}^{2+}=36.6mg/L\to$
equivalent wight $=12.2$ (given)

milliequivalent of
$C{a}^{2+}=\frac{60}{20}=3$
$M{g}^{2+}=\frac{36.6}{12.2}=3$

$Totalhardnessmg/Las\left(CaC{O}_3\right)$
$=$ (milliequivalent of $C{a}^{2+}+$ milliequivalent of
$m{g}^{2+})\times$ equivalent weight of $CaC{O}_3$

$=(3+3)\times \frac{100}{2}$

$=300mg/LasCaC{O}_3$