Question

A water contains the following dissolved ions.
$\left[N{a}^{+}\right]=56mg/L;\left[C{a}^{2+}\right]=40mg/L;$
$\left[M{g}^{2+}\right]=30mg/L;\left[A{l}^{3+}\right]=3mg/L;$
$\left[HC{O}_3^{-}\right]=190mg/L;\left[C{l}^{-}\right]=165mg/L;$
Water pH is 7.
Atomic weights:
$Ca=40,Mg=24,Al=27,H=1,C=12,O=16,Na=23,Cl=35.5$

The total hardness of the sample in mg/L as $CaC{O}_3$ is

• A. $225$
• B. $450$
• C. $242$
• D. $484$

Answer 1

The correct option is C $242$

Total hardness is due to multivalent metallic cations $(C{a}^{2+},M{g}^{2+},A{l}^{3+},F{e}^{2+},M{n}^{2+},S{r}^{2+}..etc)asCaC{O}_3$

Total hardness = conc. of $C{a}^{2+}$
$\times \frac{\text{Eq.wt.of}CaC{O}_3}{\text{Eq.wt.of}C{a}^{2+}}+\text{conc.of}M{g}^{2+}$
$\times \frac{\text{Eq.wt.of}CaC{O}_3}{\text{Eq.wt.of}M{g}^{2+}}+\text{conc.of}A{l}^{3+}\times$
$\frac{\text{Eq.wt.of}CaC{O}_3}{\text{Eq.wt.of}A{l}^{3+}}$

$=40\times \frac{50}{20}+30\times \frac{50}{12}+3\times \frac{50}{\left(27/3\right)}$

$=241.66mg/LasCaC{O}_3$

$=242mg/LasCaC{O}_3$