Animated Solution for Chemistry - Basic Concepts in Chemistry: The hardness of a water sample (in terms of equivalents of ${\text{CaCO}}_3$) containing $10^{-3}{\text{ M CaSO}}_4$ is (Molar mass of ${\text{CaSO}}_4=136{\text{ g mol}}^{-1}$)

The ${\text{CaCO}}_3$ Equivalent

Water hardness is a classic topic in chemistry that often trips students up, not because the math is hard, but because the conventions are a bit unique. When we talk about the "hardness" of water, we are primarily talking about the presence of dissolved calcium (${\text{Ca}}^{2+}$) and magnesium (${\text{Mg}}^{2+}$) ions.

However, instead of reporting the concentration of every single salt (like ${\text{CaSO}}_4$, ${\text{MgCl}}_2$, etc.), chemists use a universal standard: Calcium Carbonate (${\text{CaCO}}_3$) equivalents.

Imagine you have a beaker filled with a water sample. The problem states that this water contains calcium sulfate (${\text{CaSO}}_4$) with a concentration of $10^{-3}\text{ M}$.

This means that in exactly $1\text{ Litre}$ of this water, there are $10^{-3}\text{ moles}$ of ${\text{CaSO}}_4$.

Because ${\text{CaSO}}_4$ dissociates completely into ${\text{Ca}}^{2+}$ and ${\text{SO}}_4^{2-}$, we can confidently say that we have $10^{-3}\text{ moles}$ of ${\text{Ca}}^{2+}$ ions floating around in that litre of water.

Why ${\text{CaCO}}_3$?

Why do we convert everything to ${\text{CaCO}}_3$? Two reasons: 1. It is the most common precipitate (scale) formed by hard water. 2. Its molar mass is exactly $100\text{ g/mol}$, making calculations incredibly elegant.

Since $1\text{ mole}$ of ${\text{CaSO}}_4$ provides $1\text{ mole}$ of ${\text{Ca}}^{2+}$, and $1\text{ mole}$ of ${\text{CaCO}}_3$ also provides $1\text{ mole}$ of ${\text{Ca}}^{2+}$, they are directly equivalent in terms of hardness.

Therefore, $10^{-3}\text{ moles}$ of ${\text{CaSO}}_4$ is equivalent to $10^{-3}\text{ moles}$ of ${\text{CaCO}}_3$.

Now, we just need to find the mass of this equivalent ${\text{CaCO}}_3$.

So, we have $0.1\text{ grams}$ of ${\text{CaCO}}_3$ equivalent per litre of water. Let's convert this to milligrams:

We have $100\text{ mg}$ of ${\text{CaCO}}_3$ per Litre of solution.

Here is the golden rule for dilute aqueous solutions: $1\text{ ppm}$ (part per million) is exactly equal to $1\text{ mg/L}$.

Why? Because $1\text{ Litre}$ of water weighs approximately $1000\text{ grams}$, which is $1,000,000\text{ milligrams}$. So, $1\text{ mg}$ in $1,000,000\text{ mg}$ is exactly one part per million!

Therefore, our hardness is simply $100\text{ ppm}$.

Did you notice the molar mass of ${\text{CaSO}}_4$ ($136\text{ g/mol}$) given in the question? We never used it! This is a classic distractor designed to test if you truly understand the concept of equivalents or if you are just blindly plugging numbers into formulas. Always stay focused on the core principles!