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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.

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    $\begingroup$ HINT - ppm in this case is done on a weight basis. $\endgroup$ – MaxW Apr 26 at 6:24
  • $\begingroup$ Hardness is usually expressed in mmol/L (1=equivalent of 56 mg CaO/L) or in dGH ( 1 dGH=56 mg CaO/L) $\endgroup$ – Poutnik Apr 26 at 10:14
  • $\begingroup$ 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. $\endgroup$ – M. Farooq Apr 26 at 12:32

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