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

Mar 8, 2018

There are two reasons.

#### Explanation:

1. ${\text{CaCO}}_{3}$ causes most of the hardness

Water passing over limestone deposits dissolves some of the minerals.

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

$\text{Ca}$ and $\text{Mg}$ 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 $\text{Ca"^"2+}$ and $\text{Mg"^"2+}$.

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 $\text{CaCO"_3}$ 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

$\left[\text{CaCO"_3] = 2.5["Ca"^"2+"] + 4.1["Mg"^"2+}\right]$

Example

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

The total hardness expressed as ${\text{CaCO}}_{3}$ is

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

$\text{= 87 mg/L}$