Why we take #"CaCO"_3# as standard to determine the hardness of water?

1 Answer
Mar 8, 2018

Answer:

There are two reasons.

Explanation:

1. #"CaCO"_3# causes most of the hardness

Water passing over limestone deposits dissolves some of the minerals.

The concentrations of #"Ca"^"2+", "Mg"^"2+"#, and #"HCO"_3^"-"# ions present in the water increase greatly and cause the water to become "hard."

#"Ca"# and #"Mg"# often occur together, so it is convenient to have a single number that represents the total concentration of both ions.

2. The calculation is easy

We measure the concentrations of individual ions #"Ca"^"2+"# and #"Mg"^"2+"#.

We usually express the hardness of water in "ppm" (parts per million or milligrams per kilogram of water or milligrams per litre).

The molar mass of #"CaCO"_3"# is 100 g/mol, so it is easy to manipulate the numbers.

Thus,

#M_text(CaCO₃)/M_text(Ca) = 100/40 = 2.5#

#M_text(CaCO₃)/M_text(Mg) = 100/24.3 = 4.1#

Thus we can express the total hardness of water as

#["CaCO"_3] = 2.5["Ca"^"2+"] + 4.1["Mg"^"2+"]#

Example

If #["Ca"^"2+"] = "30 mg/L"# and #["Mg"^"2+"] = "3 mg/L"#,

The total hardness expressed as #"CaCO"_3# is

#["CaCO"_3] = "2.5 × 30 mg/mL+ 4.1 × 3 mg/L = 75 mg/L + 12 mg/L"#

# "= 87 mg/L"#