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 non-carbonate hardness of the sample in $mg/L$ as $CaC{O}_3$ is

• A. $225$
• B. $156$
• C. zero
• D. $86$

Answer 1

The correct option is D $86$

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

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

$=241.66mg/LasCaC{O}_3$

$AlkalinityisduetoC{O}_3^{-}andHC{O}_3^{-}ofC{a}^{2+}andM{g}^{2+}$

Alkalinity = conc. of $HC{O}_3^{-}\times \frac{\text{Eq.wt.of}CaC{O}_3}{\text{Eq.wt.of}HC{O}_3^{-}}$

$=190\times \frac{50}{61}$

$=155.73mg/LasCaC{O}_3$

Carbonate Hardness = Minimum of total hardness and alkalinity
$=155.73mg/LasCaC{O}_3$

Non carbonate hardness =Total hardness - Carbonate hardness
$=241.56-155.73$
$=85.93$
$=86mg/LasCaC{O}_3$