Question

Calculate hardness of water in terms of $\ce{CaCO_3}$.
Given $\ce{Ca(HCO3)2}$ = 0.81 gm, $\ce{Mg(HCO3)2}$ = 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)

$\ce{gmeq. of \; CaCO3 = gmeq. of \; Ca(HCO3) + gmeq. of \; Mg(HCO3)}$
$\text{mole} \times n.f = \frac{0.81}{162}\times2 + \frac{0.73}{146}\times2$
$ \text{mole} \; \text{of} \; CaCO_{3} = \frac{1}{200} \times2 + \frac{1}{200} \times2$
$ \text{mole} \text{of} CaCO_{3} = \frac{1}{200} + \frac{1}{200} = \frac{2}{200} = \frac{1}{ 100} \text{mole} $
Hardness $ = \frac{Wt. \text{of} \; CaCO_{3} \; \text{in} \; gm }{ Wt. of \; \text{water} \; \text{in} \; gm } \times10^{6} $
$ = \frac{\frac{1}{100} \times100}{100} \times10^{6} = 10^{4} ppm $