Animated Solution for Chemistry - Basic Concepts in Chemistry: The hardness of a water sample containing $10^{-3}{\text{ M MgSO}}_4$ expressed as ${\text{CaCO}}_3$ equivalents (in ppm) is...... (molar mass of ${\text{MgSO}}_4$ is $120.37\text{ g/mol}$)

Title: Decoding Water Hardness: The Calcium Carbonate Equivalent

Have you ever wondered what makes water "hard"? It's the presence of dissolved minerals, primarily calcium and magnesium ions. But when we measure hardness, we don't just list the concentrations of all these different salts. Instead, we use a clever trick: we express the total hardness as if it were all caused by a single substance—calcium carbonate (${\text{CaCO}}_3$). Let's dive into a classic JEE problem to see exactly how this works!

We are given a water sample containing magnesium sulfate (${\text{MgSO}}_4$) with a concentration of $10^{-3}\text{ M}$. Our goal is to find the hardness of this sample in parts per million (ppm), expressed as ${\text{CaCO}}_3$ equivalents.

First, let's understand what $10^{-3}\text{ M}$ means. It tells us that in exactly $1\text{ L}$ of this water sample, there are $10^{-3}\text{ moles}$ of ${\text{MgSO}}_4$. Since each molecule of ${\text{MgSO}}_4$ provides one ${\text{Mg}}^{2+}$ ion, we have $10^{-3}\text{ moles}$ of hardness-causing ${\text{Mg}}^{2+}$ ions.

To express hardness in ppm, we use the following formula:

Why do we use ${\text{CaCO}}_3$? It's not just a random choice! The molar mass of ${\text{CaCO}}_3$ is exactly $100\text{ g/mol}$. This beautiful, round number makes calculations incredibly simple and minimizes errors.

Since $1\text{ mole}$ of ${\text{Mg}}^{2+}$ causes the same amount of hardness as $1\text{ mole}$ of ${\text{CaCO}}_3$ (both are divalent cations), our $10^{-3}\text{ moles}$ of ${\text{MgSO}}_4$ is equivalent to $10^{-3}\text{ moles}$ of ${\text{CaCO}}_3$.

Now, let's convert these moles into mass.

So, the hardness in our $1\text{ L}$ sample is equivalent to $0.1\text{ g}$ of ${\text{CaCO}}_3$.

To use our ppm formula, we also need the mass of the water sample. For dilute aqueous solutions, the density is approximately $1\text{ g/mL}$. Therefore, $1\text{ L}$ (which is $1000\text{ mL}$) of water has a mass of $1000\text{ g}$.

Let's plug everything into the master equation:

This problem beautifully illustrates the power of standardizing units. By converting everything to ${\text{CaCO}}_3$ equivalents, we can easily compare and add the hardness from various different salts. Next time you drink a glass of water, you might just find yourself calculating its ppm!