Chemistry > Stoichiometry > 11.0 Hardness of Water

  Stoichiometry
    1.0 The Mole
    2.0 The Limiting Reagent
    3.0 Gravimetric Analysis
    4.0 Volumetric Analysis
    5.0 Calculation of n-factor
    6.0 Redox Reactions
    7.0 Titration
    8.0 Iodimetric and Iodometric Titrations
    9.0 Volume strength peroxide solution
    10.0 Percentage Labeling of Oleum
    11.0 Hardness of Water

11.2 Permanent Hardness
It is due to the presence of dissolved $CaC{l_2},{\text{ }}CaS{O_4},{\text{ }}MgC{l_2}$ and $MgS{O_4}$ in water. A known volume of hard water is taken and an excess of known equivalents of $N{a_2}C{O_3}$ are added in it. $N{a_2}C{O_3}$ reacts with $C{a^{ + + }}$ and $M{g^{ + + }}$ forming precipitates of $CaC{O_3}$ and $MgC{O_3}.$

These precipitates are filtered off. The filtrate is titrated with a strong acid $\left( {HCl{\text{ }}or{\text{ }}{H_2}S{O_4}} \right).$ Knowing the equivalents of $N{a_2}C{O_3}$ added and left unreacted, the equivalents of $N{a_2}C{O_3}$ consumed by hard water is known.

The equivalents of $N{a_2}C{O_3}$ consumed is equal to the total equivalents of $C{a^{ + + }}$ and $M{g^{ + + }}$ ions present in hard water.

Hardness of water is represented in ppm (mg/litre) of $CaC{O_3}$ i.e. milli grams of $CaC{O_3}$ present per litre of hard water. But hard water does not contain $CaC{O_3}$ Hard water contains $CaC{l_2},{\text{ }}MgC{l_2},{\text{ }}Ca{\left( {HC{O_3}} \right)_2}$ etc.

One mole ${\text{ }}CaC{l_2}$ $ \equiv $ one mole $CaC{O_3}$

or
$111g{\text{ }}CaC{l_2}$ $ \equiv $ $100g{\text{ }}CaC{O_3}$
Similarly,

\[120g{\text{ }}MgS{O_4} \equiv 100g{\text{ }}CaC{O_3}\]

Thus mass of $CaC{O_3}$ corresponding to the mass of $CaC{l_2},{\text{ }}MgS{O_4}$ etc., present in hard water is calculated. Milligrams of $CaC{O_3}$ per litre of hard water is called hardness of water in ppm.
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