Problem 36 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

(a) One litre of a sample of hard water contains \(1 \mathrm{mg}\) of \(\mathrm{CaCl}_{2}\) and 1 \(\mathrm{mg}\) of \(\mathrm{MgCl}_{2}\). Find the total hardness in terms of parts of \(\mathrm{CaCO}_{3}\) per \(10^{6}\) parts of water by weight. (b) A sample of hard water contains \(20 \mathrm{mg}\) of \(\mathrm{Ca}^{++}\)ions per litre. How many milli-equivalent of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) would be required to soften 1 litre of the sample? (c) \(1 \mathrm{~g}\) of \(\mathrm{Mg}\) is burnt in a closed vessel which contains \(0.5 \mathrm{~g}\) of \(\mathrm{O}_{2}\). (i) Which reactant is left in excess? (ii) Find the weight of the excess reactants? (iii) How may milliliters of \(0.5 \mathrm{~N} \mathrm{H}_{2} \mathrm{SO}_{4}\) will dissolve the residue in the vessel.

Step-by-Step Solution

Answer

(a) 1.952 mg/L as CaCO3. (b) 1 meq/L of Na2CO3. (c) (i) O2 is limiting, (ii) 0.240 g of Mg in excess, (iii) 31.2 mL of 0.5 N H2SO4 needed.

Step 1: Calculate Total Hardness for CaCl2

To find the hardness due to calcium chloride (\( \mathrm{CaCl}_2 \)), consider that 1 mg of \( \mathrm{CaCl}_2 \) contains calcium. The molar mass of \( \mathrm{CaCl}_2 \) is 110.98 g/mol, and for \( \mathrm{CaCO}_3 \), it is 100.09 g/mol.Hardness conversion is based on calcium carbonate, requiring calculation of equivalent calcium as \( \mathrm{CaCO}_3 \): \[ \left( \frac{1 \text{ mg of } \mathrm{CaCl}_2}{110.98 \text{ g/mol}} \right) \times \left( \frac{100.09 \text{ g/mol of } \mathrm{CaCO}_3}{1 \text{ mg of } \mathrm{CaCO}_3} \right) = 0.901 \text{ mg/L as } \mathrm{CaCO}_3 \].

Step 2: Calculate Total Hardness for MgCl2

Similarly, calculate the hardness due to magnesium chloride (\( \mathrm{MgCl}_2 \)). The molar mass of \( \mathrm{MgCl}_2 \) is 95.21 g/mol.Hardness conversion:\[ \left( \frac{1 \text{ mg of } \mathrm{MgCl}_2}{95.21 \text{ g/mol}} \right) \times \left( \frac{100.09 \text{ g/mol of } \mathrm{CaCO}_3}{1 \text{ mg of } \mathrm{CaCO}_3} \right) = 1.051 \text{ mg/L as } \mathrm{CaCO}_3 \].

Step 3: Total Hardness Calculation

Add the hardness from \( \mathrm{CaCl}_2 \) and \( \mathrm{MgCl}_2 \):\[ 0.901 \text{ mg/L as } \mathrm{CaCO}_3 + 1.051 \text{ mg/L as } \mathrm{CaCO}_3 = 1.952 \text{ mg/L as } \mathrm{CaCO}_3 \].

Step 4: Calculate Milli-equivalents for Softening

To soften the water, calculate the milli-equivalents needed. Given 20 mg of \( \mathrm{Ca}^{++} \) ions:Milli-equivalents = \( \frac{\text{mass (mg)}}{\text{equivalent mass (mg/meq)}} \).Equivalent mass of \( \mathrm{Ca} \) is half of its molar mass (20.04 mg/meq).\[ \text{Milli-equivalents for } \mathrm{Ca} = \frac{20 \text{ mg}}{20.04 \text{ mg/meq}} = 1.0 \text{ meq/L} \].One meq of \( \mathrm{Ca} \) is replaced by one meq of \( \mathrm{Na}_2\mathrm{CO}_3 \), so 1 meq/L of \( \mathrm{Na}_2\mathrm{CO}_3 \) is required.

Step 5: Determine Limiting Reactant for Mg and O2

For part (c), use stoichiometry of magnesium oxide formation. Reaction is \( 2\mathrm{Mg} + \mathrm{O}_2 \rightarrow 2\mathrm{MgO} \).Molar masses: \( \mathrm{Mg} = 24.305 \) g/mol, \( \mathrm{O}_2 = 32 \) g/mol.Mg moles: \( \frac{1 \text{ g}}{24.305 \text{ g/mol}} = 0.0411 \text{ mol} \).O2 moles: \( \frac{0.5 \text{ g}}{32 \text{ g/mol}} = 0.0156 \text{ mol} \).Since \( 2 \) mol of Mg requires \( 1 \) mol of O2, O2 is the limiting reactant.

Step 6: Calculate Excess Magnesium

From Step 5, \( 0.0156 \) mol of O2 reacts completely with \( 2 \times 0.0156 = 0.0312 \) mol of Mg.Excess Mg moles: \( 0.0411 - 0.0312 = 0.0099 \) mol.Weight of excess Mg: \( 0.0099 \text{ mol} \times 24.305 \text{ g/mol} = 0.240 \text{ g} \).

Step 7: Calculate H2SO4 Volume Required

For dissolving MgO, reaction is \( \mathrm{MgO} + \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{MgSO}_4 + \mathrm{H}_2\mathrm{O} \).Moles of Mg burnt = Moles of MgO formed = \( 0.0156 \) mol.\( 0.0156 \) mol of \( \mathrm{H}_2\mathrm{SO}_4 \) required (1:1 ratio).1 N solution has 1 eq/L; 0.5 N has \( 0.5 \) eq/L.Volume of 0.5 N \( \mathrm{H}_2\mathrm{SO}_4 \) = \( \frac{0.0156 \text{ mol}}{0.5 \text{ eq/L}} = 0.0312 \text{ liters} = 31.2 \text{ mL} \).

Key Concepts

Chemical Stoichiometry

Chemical stoichiometry is like the recipe of a chemical reaction. It helps us understand how much of each ingredient (or reactant) is needed to make the desired amount of product. Imagine you have a recipe for a cake. If you know how much flour and sugar you need, you can bake a cake with the same taste every time. Similarly, stoichiometry tells us how to mix chemicals in the exact proportions for a reaction to occur without leftovers.

### Chemical Reactions with Stoichiometry
Let's consider the reaction between magnesium and oxygen to form magnesium oxide: - **Reaction**: \(2\mathrm{Mg} + \mathrm{O}_2 \rightarrow 2\mathrm{MgO}\)
This means 2 moles of magnesium react with 1 mole of oxygen to produce 2 moles of magnesium oxide.

Stoichiometry helps us calculate which reactant will run out first (called the limiting reactant), and how much product can be created. If you have more of one reactant than needed, it becomes the excess reactant and will be left over after the reaction.

### Practical Example
If we have 1 gram of magnesium and 0.5 grams of oxygen in a container, stoichiometry allows us to calculate:- How many moles of each reactant are present.- Which reactant is limiting based on the stoichiometry of the reaction.- How much magnesium oxide can be produced.- How much excess reactant remains.

Hardness Calculation

Water hardness is a measure of the concentration of calcium and magnesium ions in water. These ions can interfere with soap's ability to lather and cause scaling in pipes, boilers, and appliances. Calculating water hardness often involves converting the amounts of these ions into equivalent concentrations of calcium carbonate (\(\mathrm{CaCO}_3\)). This is because \(\mathrm{CaCO}_3\) serves as a standard reference for measuring hardness.

### Understanding Water Hardness- **Total Hardness Calculation**: You convert the masses of \(\mathrm{CaCl}_2\) and \(\mathrm{MgCl}_2\) into \(\mathrm{CaCO}_3\) equivalents to add them up and find the total hardness.- **Conversion Example**: If your water sample has 1 mg of \(\mathrm{CaCl}_2\) and 1 mg of \(\mathrm{MgCl}_2\), you'd convert these into their \(\mathrm{CaCO}_3\) equivalents and sum them to get the total hardness.

By using the molecular weights and the conversion factor, we calculate how much hypothetical calcium carbonate would be present if it were fully converted from these chlorides. This provides a standardized measure for comparing the hardness levels of different water samples.

Ionic Equivalents

Ionic equivalents are used in chemistry to compare different ions based on their quantifiable effects or reactions. It's particularly important in reactions involving ionic compounds, like the treatment of hard water or softening of water, where equivalent weights help precisely measure the substances needed. This concept works by taking into account the valence, or charge, of the ions involved.

### Calculating Ionic Equivalents
To calculate ionic equivalents, we use the formula:
\[ \text{meq} = \frac{\text{mass (mg)}}{\text{equivalent weight (mg/meq)}} \]
Equivalent weight is determined by dividing the molar mass of the ion by its valency (charge). For example, the equivalent weight of \(\mathrm{Ca}^{++}\) is half its molar mass since its charge is +2.

### Application in Water Softening
Suppose you have a sample containing calcium ions, and you want to soften the water using \(\mathrm{Na}_2\mathrm{CO}_3\). If your water has 20 mg/L of \(\mathrm{Ca}^{++}\) ions, and the equivalent weight of calcium is roughly 20.04 mg/meq, the number of milli-equivalents needed will be 1 meq/L. One milli-equivalent of \(\mathrm{Ca}^{++}\) is replaced by one milli-equivalent of \(\mathrm{Na}_2\mathrm{CO}_3\), providing the necessary measurements for softening the water effectively.