# Degree of Hardness Numerical

To remove the temporary hardness from $$5\ \mathrm L$$ of $$\ce{H2O}$$ $$5.6\ \mathrm{mg}$$ of lime is required. Calculate the degree of hardness due to $$\ce{Ca(HCO3)2}$$ present in water.

My attempt:

$$\ce{Ca(HCO3)2 + CaO -> 2CaCO3 + H2O}$$ Since $$5.6\ \mathrm{mg}$$ of lime is required, amount of calcium bicarbonate present is $$16.2 \times 10^{-3}$$.

Molarity = $${16.2 \times 10^{-3}}/(162 \times 5)$$

$$= 0.002\ \mathrm M$$.

This is equivalent to the concentration of calcium carbonate.

ppm of calcium carbonate = $$0.002\times10^{-2}\times100\times10^6\times10^{-3} = 200$$

Therefore degree of hardness is $$200\ \mathrm{ppm}$$.

Is the solution correct? Also I am looking for an alternative method for solving the problem.

• HINT - ppm in this case is done on a weight basis. – MaxW Apr 26 '19 at 6:24
• Hardness is usually expressed in mmol/L (1=equivalent of 56 mg CaO/L) or in dGH ( 1 dGH=56 mg CaO/L) – Poutnik Apr 26 '19 at 10:14
• 1 dGH is most likely a German unit. Double check you answer with the following approach as well. The traditional unit is ppm CaCO3 calculated in terms of mg/L. Find out moles of CaO, from the equation see the 1:1 mol ratio between CaO and CaCO3. If x moles of lime were used, x moles of CaCO3 is present in 5 L of water. Convert moles of CaCO3 into mg CaCO3 and express it per liter. – M. Farooq Apr 26 '19 at 12:32