Question

Calculate hardness of water in terms of $\mathrm{CaCO}{}_3$.
Given $\mathrm{Ca}(\mathrm{HCO}{}_3){}_2$ = 0.81 gm, $\mathrm{Mg}(\mathrm{HCO}{}_3){}_2$ = 0.73 gm, Volume of solution = 100 ml

• A. $10^{2}ppm$
• B. $10^{3}ppm$
• C. $10^{4}ppm$
• D. $10^{5}ppm$

Answer 1

Answer: (C)

$\mathrm{gmeq}\cdot \mathrm{of}\mathrm{CaCO}{}_3=\mathrm{gmeq}\cdot \mathrm{of}\mathrm{Ca}(\mathrm{HCO}{}_3)+\mathrm{gmeq}\cdot \mathrm{of}\mathrm{Mg}(\mathrm{HCO}{}_3)$
$\text{mole}\times n.f=\frac{0.81}{162}\times 2+\frac{0.73}{146}\times 2$
$\text{mole}\text{of}CaC{O}_3=\frac{1}{200}\times 2+\frac{1}{200}\times 2$
$\text{mole}\text{of}CaC{O}_3=\frac{1}{200}+\frac{1}{200}=\frac{2}{200}=\frac{1}{100}\text{mole}$
Hardness $=\frac{Wt.\text{of}CaC{O}_3\text{in}gm}{Wt.of\text{water}\text{in}gm}\times 10^6$
$=\frac{\frac{1}{100}\times 100}{100}\times 10^6=10^{4}ppm$