Question

One litre of a sample of a hard water contains 1 mg of CaCl₂ and 1 mg of MgCl₂. Find the total hardness in terms of parts of CaCO₃ per 10⁶ parts of water by weight.

Answer 1

The relevant mass-mass relationships are

\(\underset{40 +2 \times 35.5\\=111mg}{CaCl_2} ≡ \underset{40 + 12 + 48\\=100mg}{CaCO_3}\)

\(\underset{24 + 2 \times35.5 \\= 95 mg}{MgCl_2} ≡ \underset{40 + 12 + 48\\=100mg}{CaCO_3}\)

Thus, 111 mg of CaCl₂ = 100 mg of CaCO₃ 

1 mg of CaCl₂ = \(\frac{100}{111}\) x 1 mg of CaCO₃ = 0.9 mg of CaCO₃

Similarly, 95 mg of MgCl₂ = 100 mg of CaCO₃ 

1 mg of MgCl₂ = \(\frac{100}{95}\) x 1 mg of CaCO₃ = 1.053 mg of CaCO₃. 

Total wt. of CaCO₃ in 1 litre (i.e., 1000 g) of water = 0.9 + 1.053 mg = 1.953 mg 

Also total wt. of CaCO₃ in 10⁶ mg of water = 1.953 mg

Hence total hardness in terms of CaCO₃ per 10⁶ parts by weight = 1.953 mg