Question

$80mL$ of water sample contains $0.162gCa(HC{O}_3{)}_2$ and $0.146gMg(HC{O}_3{)}_2.$ The hardness of water expressed in term of $CaC{O}_3$ is:
(Given : molar mass of $Ca\left(HC{O}_3{)}_{2}andMg\right(HC{O}_3{)}_2$ are 162 and 146 g/mol respectively. )

• A. $2.5\times {10}^{3}ppm$
• B. $1.5\times {10}^{3}ppm$
• C. $5\times {10}^{4}ppm$
• D. $6.2\times {10}^{4}ppm$

Answer 1

The correct option is A $2.5\times {10}^{3}ppm$

The hardness of water is calculated in terms of $CaC{O}_3.$
$\text{Equivalents of}CaC{O}_3=\text{Equivalents of}Ca(HC{O}_3{)}_2+\text{Equivalents of}Mg(HC{O}_3{)}_2$
$\frac{W_{CaC{O}_3}}{100}\times 2=\frac{0.162}{162}\times 2+\frac{0.146}{146}\times 2$
$\therefore W_{CaC{O}_3}\approx 0.2g$

We know, density of water $=1g/mL$
$W_{H_{2}O}=\left(1g/mL\right)\times \left(80mL\right)=80g$

$\text{Hardness}=\frac{\text{Mass of}CaC{O}_3}{\text{Mass of hard water}}\times {10}^{6}ppm$
$\therefore \text{Hardness}=\frac{0.2}{80}\times {10}^{6}ppm$
$\text{Hardness}\approx 2.5\times {10}^{3}ppm$


Theory:
Hardness is generally represented in terms of ppm of $CaC{O}_3$
Hardness of Water = $\frac{MassofCaC{O}_3}{Massofhardwater}\times {10}^6$
If there is no information about the salt given, then the hardness of water is calculated in terms of $CaC{O}_3.$

Note :- Hardness of water can also be represented in terms of other salts like $CaC{l}_2,MgC{l}_2,CaS{O}_4$, etc.