Question

Ion concentrations obtained for a groundwater sample (having $pH=8.10$) are given below.

Ion	$C{a}^{2+}$	$M{g}^{2+}$	$N{a}^{+}$	$HC{O}_3^{-}$	$S{O}_4^{2-}$	$C{l}^{-}$
Ion concentration $\left(mg/L\right)$	$100$	$6$	$15$	$250$	$45$	$39$
Atomic weight	$Ca=40$	$Mg=24$	$Na=23$	$H=1$ $C=12$ $O=16$	$S=32$ $O=16$	$Cl=35.5$

Carbonate hardness mg/L $\left(CaC{O}_3\right)$ present in the above sample is

• A. $205$
• B. $250$
• C. $289$
• D. $275$

Answer 1

The correct option is A $205$

Total hardness in water is due to multivalent cations

Total hardness
$=\left[C{a}^{2+}\right]\times \frac{\text{Equivalent weight of}CaC{O}_3}{\text{Equivalent weight of}C{a}^{2+}}+\left[M{g}^{2+}\right]\times \frac{\text{Equivalent weight of}CaC{O}_3}{\text{Equivalent weight of}M{g}^{2+}}$

$=100\times \frac{50}{20}+6\times \frac{50}{12}$

$=275mg/LasCaC{O}_3$

Carbonate hardness is equal to total harndess or alkalinity whichever is less.

Alkalinity is caused by $C{O}_3^{2-},HC{O}_3^{-},O{H}^{-}$

Hence, alkainity
$=\left[HC{O}_3^{-}\right]\times \frac{\text{Equivalent weight of}CaC{O}_3}{\text{Equivalent weight of}HC{O}_3^{-}}$
$=250\times \frac{50}{61}=205mg/L$

Hence, carbonate hardness $=min\{275,205\}mg/L$

$=205mg/LasCaC{O}_3$