Problem 8 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The hardness of a water sample (in terms of equivalents of \(\mathrm{CaCO}_{3}\) ) containing \(10^{-3} \mathrm{M} \mathrm{CaSO}_{4}\) is: (molar mass of \(\mathrm{CaSO}_{4}=136 \mathrm{~g} \mathrm{~mol}^{-1}\) ) [Main Jan. 12, 2019 (I)] (a) \(10 \mathrm{ppm}\) (b) \(50 \mathrm{ppm}\) (c) \(90 \mathrm{ppm}\) (d) \(100 \mathrm{ppm}\)

Step-by-Step Solution

Answer

(d) 100 ppm

Step 1: Calculate the mass of CaSO4 per liter

The concentration of \( \mathrm{CaSO_4} \) is given as \( 10^{-3} \mathrm{~M} \). This means there is \( 10^{-3} \) moles of \( \mathrm{CaSO_4} \) per liter of solution. Using the molar mass of \( \mathrm{CaSO_4} \) which is \( 136 \mathrm{~g/mol} \), we can calculate the mass: \( 10^{-3} \times 136 \mathrm{~g} = 0.136 \mathrm{~g} \).

Step 2: Convert grams to milligrams

We need to convert \( 0.136 \mathrm{~g} \) to milligrams because hardness is usually expressed in ppm (parts per million), which is equivalent to mg/L. \( 0.136 \mathrm{~g} = 136 \mathrm{~mg} \).

Step 3: Equivalents of CaCO3

Since the hardness is measured in terms of equivalent \( \mathrm{CaCO_3} \), we must convert the mass of \( \mathrm{CaSO_4} \) to the equivalent \( \mathrm{CaCO_3} \). The molar mass of \( \mathrm{CaCO_3} \) is \( 100 \mathrm{~g/mol} \). Therefore, \( 136 \mathrm{~mg} \text{ of } \mathrm{CaSO_4} \) is equivalent to \( 100 \mathrm{~mg} \text{ of } \mathrm{CaCO_3} \) because each mole of \( \mathrm{CaSO_4} \) releases the same number of calcium ions as a mole of \( \mathrm{CaCO_3} \).

Step 4: Calculate ppm of CaCO3

Since \( 136 \mathrm{~mg} \mathrm{CaSO_4} \) is equivalent to \( 100 \mathrm{~mg} \mathrm{CaCO_3} \), the water hardness is equivalent to \( 100 \mathrm{~ppm} \) \( \mathrm{CaCO_3} \).

Key Concepts

Calcium Carbonate Equivalence

In water chemistry, calcium carbonate equivalence is an important concept used to express water hardness. Water hardness is often measured in terms of calcium carbonate (\(\mathrm{CaCO}_3\)) equivalents.
This standardized measure allows for the comparison of water samples in terms of neutralizing power.

• Hardness as Equivalent of \(\mathrm{CaCO}_3\): It's based on the fact that different compounds like \(\mathrm{CaSO}_4\) and \(\mathrm{CaCO}_3\) release calcium ions in water.
  
• Reason for Using \(\mathrm{CaCO}_3\): Calcium carbonate is selected because of its common occurrence in natural water sources and its typical impact on water quality.

In our case, the water sample contains \(10^{-3}\) M of \(\mathrm{CaSO}_4\). Since \(\mathrm{CaSO}_4\) and \(\mathrm{CaCO}_3\) release the same number of \(\mathrm{Ca}^{2+}\) ions, 136 mg of \(\mathrm{CaSO}_4\) was converted to its \(\mathrm{CaCO}_3\) equivalent, calculated as 100 mg. This equivalence helps us to communicate and compare hardness regardless of specific compounds involved.
It's crucial to understand how different substances can be expressed in terms of \(\mathrm{CaCO}_3\) equivalents to assess their impact on water hardness accurately.

PPM Calculation

PPM, or "parts per million," is a unit of concentration that describes the amount of a substance in a million parts of a solution. It is effectively used to measure the concentration of substances in water, often representing water hardness.
How It Works:

• If a solution has 1 mg of a solute in 1 liter of water, that concentration is 1 ppm.
  
• With our example, we converted 0.136 g \(\mathrm{CaSO}_4\) to 136 mg/L—directly giving its ppm since 1 liter is approx. 1,000,000 mg of water.

This conversion helps standardize the reporting of water quality and allows for the simple comparison of results.
To find the ppm as \(\mathrm{CaCO}_3\), we equate the mass of \(\mathrm{CaSO}_4\) to its \(\mathrm{CaCO}_3\) equivalent since they both contribute equally to the water's hardness. Thus, we find that the water has a hardness of 100 ppm as \(\mathrm{CaCO}_3\). Understanding ppm is fundamental in evaluating water samples for municipal and industrial usage.

Molar Mass Conversion

Molar mass conversion is a fundamental step in chemical calculations as it allows us to bridge the quantity of substance (moles) with weight mass (grams).
Basics of Molar Mass:

• The molar mass is the mass in grams of one mole of a substance, based on its atomic constituents.
  
• \(\mathrm{CaSO}_4\) has a molar mass of 136 g/mol, meaning 1 mole of \(\mathrm{CaSO}_4\) weighs 136 g.

In our example, the concentration of \(\mathrm{CaSO}_4\) was given in molarity \((10^{-3} \mathrm{M})\). By multiplying this by its molar mass, we calculate the mass of \(\mathrm{CaSO}_4\) in 1 liter of solution (0.136 g).
Further, by converting this mass into milligrams (136 mg), we facilitate its expression in ppm.
This swift conversion between moles and grams and subsequently to milligrams aids in interpreting and communicating concentrations correctly in practical terms.