Question
Calculate the temporary and permanent hardness of water sample having the following constituents per litre.
Ca(HCO3)2 = 162 mg, MgCl2 = 95 mg, NaCl = 585 mg, Mg(HCO3)2 = 73 mg, CaSO4 = 136 mg
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Solution
Temporary hardness is due to Ca(HCO3)2 and Mg(HCO3)2.
$n_{eq}CaCO_{3} = n_{eq}Ca\left( HCO_{3} \right)_{2} + n_{eq}Mg\left( HCO_{3} \right)_{2}$
Or $\frac{w}{100} \times 2 = \frac{162 \times 10^{- 3}}{162} \times 2 + \frac{73 \times 10^{- 3}}{146} \times 2 \Rightarrow w = 150 \times 10^{- 3}g$
∴ Temporary hardness = $\frac{150 \times 10^{- 3}}{1000} \times 10^{6} = 150\text{ ppm}$
Permanent hardness is due to MgCl2 and CaSO4.
$n_{eq}CaCO_{3} = n_{eq}MgCl_{2} + n_{eq}CaSO_{4}$
Or $\frac{w}{100} \times 2 = \frac{95 \times 10^{- 3}}{95} \times 2 + \frac{136 \times 10^{- 3}}{136} \times 2 \Rightarrow w = 200 \times 10^{- 3}g$
∴ Permanent hardness = $\frac{200 \times 10^{- 3}}{1000} \times 10^{6} = 200\text{ ppm}$
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