Question

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

Answer 1

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"`