Problem 49 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Upon mixing \(45.0 \mathrm{~mL}\). of \(0.25 \mathrm{M}\) lead nitrate solution with \(25.0 \mathrm{~mL}\) of \(0.10 \mathrm{M}\) chromic sulphate solution, precipitation of lead sulphate takes place. How many moles of lead sulphate are formed? Also, calculate the molar concentrations of the species left behind in the final solution. Assume that lead sulphate is completely insoluble. [1993 - 3 Marks]

Step-by-Step Solution

Answer

0.00375 moles of lead sulphate are formed. Remaining Pb(NO₃)₂ is 0.107 M and Cr(NO₃)₃ is 0.0536 M.

Step 1: Calculate Initial Moles of Reactants

First, we calculate the moles of lead nitrate and chromic sulphate. The formula to calculate moles is: \( ext{moles} = ext{volume in L} \times ext{molarity} \). For lead nitrate:\(45.0 \, \text{ml} = 0.045 \, \text{L} \)\(0.045 \, \text{L} \times 0.25 \, \text{M} = 0.01125 \, \text{moles} \)For chromic sulphate:\(25.0 \, \text{ml} = 0.025 \, \text{L} \)\(0.025 \, \text{L} \times 0.10 \, \text{M} = 0.0025 \, \text{moles} \)

Step 2: Determine the Limiting Reactant

The reaction is:\(\text{3Pb(NO}_3)_2 + \text{2Cr}_2(\text{SO}_4)_3 \rightarrow \text{3PbSO}_4 + \text{2Cr(NO}_3)_3\)We must determine which reactant limits the formation of lead sulphate. For every 3 moles of Pb(NO₃)₂, we need 2 moles of Cr₂(SO₄)₃. Calculate the required moles of Cr₂(SO₄)₃ for 0.01125 moles of Pb(NO₃)₂:\(0.01125 \, \text{moles Pb(NO}_3)_2 \times \frac{2}{3} = 0.0075 \, \text{moles Cr}_2(\text{SO}_4)_3\)Since we only have 0.0025 moles of Cr₂(SO₄)₃, Cr₂(SO₄)₃ is the limiting reactant.

Step 3: Calculate Moles of Lead Sulphate Formed

Since Cr₂(SO₄)₃ is the limiting reactant, we use its amount to calculate the moles of PbSO₄ produced.From the stoichiometry of the reaction, 2 moles of Cr₂(SO₄)₃ yield 3 moles of PbSO₄.\(0.0025 \, \text{moles Cr}_2(\text{SO}_4)_3 \times \frac{3}{2} = 0.00375 \, \text{moles of PbSO}_4\)

Step 4: Calculate Remaining Concentrations

The total volume of the solution is:\(45.0 \, \text{ml} + 25.0 \, \text{ml} = 70.0 \, \text{ml} = 0.070 \, \text{L}\)a) For Pb(NO₃)₂, initially 0.01125 moles, 0.00375 moles react (in ratio 1:1 to PbSO₄), thus remaining is:\(0.01125 - 0.00375 = 0.0075 \, \text{moles}\)Concentration:\(\frac{0.0075}{0.070} = 0.107 \, \text{M}\)b) Cr(NO₃)₃ concentration starts at 0, gains all from reacted Cr₂(SO₄)₃ (0.0025 moles to 0.00375 moles):\(\frac{0.00375}{0.070} = 0.0536 \, \text{M}\)

Key Concepts

Limiting Reactant

In chemical reactions, the limiting reactant determines how much product can be formed. It is the reactant that will be completely used up first, limiting the extent of the reaction. To identify the limiting reactant, you need to compare the mole ratio of the reactants used in the equation to the ratio calculated from the initial amounts of each reactant.

For instance, in the reaction between lead nitrate and chromic sulphate to form lead sulphate, you can see from the chemical equation: \[3 \text{Pb(NO}_3)_2 + 2 \text{Cr}_2(\text{SO}_4)_3 \rightarrow 3 \text{PbSO}_4 + 2 \text{Cr(NO}_3)_3\]that 3 moles of lead nitrate react with 2 moles of chromic sulphate.

• If you have fewer moles of Cr₂(SO₄)₃ than needed (relative to lead nitrate), it will run out first, making it the limiting reactant.

Determining the limiting reactant in a reaction helps predict the maximum amount of product that can form, which is valuable for efficient resource use and process planning.

Moles Calculation

Understanding moles is fundamental in stoichiometry. A mole is a unit that measures the amount of substance, and it plays a central role in chemical calculations.

To find the number of moles of a reactant or product, you can use the formula:\[\text{moles} = \text{volume in L} \times \text{molarity}\]

• For example, if you have 0.045 L of a 0.25 M lead nitrate solution, the moles would be calculated as: \[0.045 \, \text{L} \times 0.25 \, \text{M} = 0.01125 \, \text{moles} \]

Moles connect the macroscopic measurable quantities (like volume and mass) with molecular quantities, thus enabling you to predict how much product a reaction can produce and how much reactant is needed. This is particularly important when dealing with reactions involving gases, solutions, and solid precipitates.

Molar Concentration

Molar concentration, often simply called "molarity," is a way to express the concentration of a solution. It tells you how many moles of a substance are present in one liter of the solution.

The formula used is:\[\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\]

• In our previous example, for remaining lead nitrate, having 0.0075 moles in a total solution volume of 0.070 L would result in a molarity of: \[\frac{0.0075}{0.070} = 0.107 \, \text{M}\]

Molarity is critical in preparing solutions to required strengths and in performing dilution and neutralization calculations. It allows for the comparison of concentrations of different solutions and provides information essential for quantitatively describing reactions in a chemical equation.

Precipitation Reaction

A precipitation reaction results in the formation of an insoluble product, or precipitate, when two solutions are mixed.

In the context of this exercise, when solutions of lead nitrate and chromic sulphate are combined, an insoluble compound, lead sulphate, forms and precipitates out of the solution. The reaction can be written as:\[\text{Pb}^{2+} + \text{SO}_4^{2-} \rightarrow \text{PbSO}_4(s)\]

• This type of reaction is useful for removing ions from solutions and can be utilized in various applications, such as water purification and qualitative analysis in chemistry laboratories.
• Understanding how to predict whether a precipitate will form is crucial, typically achieved through solubility rules and product solubility.

The solid that forms includes the ions that were present in the reacting solutions initially. The amount of solid precipitate formed can be related back to the limiting reactant, showing the interplay between chemical quantities and the physical observation of a solid forming in the reaction mixture.