Question

The lead nitrate, \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\), in \(25 \mathrm{~mL}\) of a \(0.15 \mathrm{M}\) solution reacts with all of the aluminium sulphate, \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\), in \(20 \mathrm{~mL}\) of solution. What is the molar concentration of the \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} ?\) \(3 \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q) \longrightarrow 3 \mathrm{PbSO}_{4}(s)+2 \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}(a q)\) (a) \(6.25 \times 10^{-2} M\) (b) \(2.421 \times 10^{-2} M\) (c) \(0.1875 M\) (d) None of these

Short answer

The molar concentration of the Al2(SO4)3 is (b) \(2.421 \times 10^{-2} M\).

Step-by-step solution

- Calculate moles of Pb(NO3)2

First, calculate the number of moles of lead nitrate using the molarity formula. Molarity (M) is moles of solute divided by liters of solution. Given the molarity of Pb(NO3)2 is 0.15 M and the volume is 25 mL (0.025 L), use the formula: Moles of Pb(NO3)2 = Molarity * Volume = 0.15 M * 0.025 L.

- Determine the stoichiometry of the reaction

The balanced chemical equation indicates that 3 moles of Pb(NO3)2 react with 1 mole of Al2(SO4)3. Based on the stoichiometry, the ratio of Pb(NO3)2 to Al2(SO4)3 is 3:1.

- Calculate moles of Al2(SO4)3

Use the stoichiometry from Step 2 to find the moles of Al2(SO4)3 that will react with the moles of Pb(NO3)2. Since the ratio is 3:1, divide the moles of Pb(NO3)2 by 3 to get the moles of Al2(SO4)3.

- Calculate the molarity of Al2(SO4)3

Finally, use the moles of Al2(SO4)3 and the volume of the Al2(SO4)3 solution to calculate its molarity. The volume of Al2(SO4)3 solution is 20 mL (0.020 L). Molarity = Moles of solute / Liters of solution.

Key concepts

Stoichiometry

Stoichiometry is the core concept of quantitative relationships between the reactants and products in a chemical reaction. It's a bit like a recipe for chemistry, where it tells you how much of each substance is needed to react completely.

Referring to our exercise, stoichiometry comes into play when we notice that the balanced chemical equation shows a 3:1 ratio between lead nitrate, \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\), and aluminium sulfate, \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\). This ratio is essential to calculate the moles of aluminium sulfate reacting with a given amount of lead nitrate.

When improving understanding of stoichiometry, ensure to emphasize:

• How to interpret the coefficients in a balanced chemical equation.
• Understanding that these coefficients represent mole ratios of reactants and products.
• Applying these ratios to find the unknown quantity in a chemical reaction.

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, turn into different substances, called products. In the context of our exercise, the reaction between lead nitrate and aluminium sulfate forms lead sulfate and aluminum nitrate.

While exploring chemical reactions, it's crucial to comprehend:

• The law of conservation of mass, which states that matter is neither created nor destroyed in a chemical reaction.
• How to write and balance chemical equations to reflect this law.
• Identifying the states of matter for each compound (like (aq) for aqueous or (s) for solid) in the reaction and how they can influence the reaction’s dynamics.

Molar Concentration

Molar concentration, often referred to as molarity, describes the amount of a substance in a given volume of solution. It is a key concept for understanding the concentration of solutions in chemistry.

In our exercise, molarity is used to determine the concentration of lead nitrate and to find the concentration of aluminium sulfate. By using the provided molarity and volume of lead nitrate, students can calculate the moles of lead nitrate and, consequently, the moles of aluminium sulfate by employing stoichiometry.

To improve clarity on molarity:

• Clarify that molarity is calculated as moles of solute per liter of solution (\(M = \frac{\text{moles}}{\text{liters}}\)).
• Emphasize the conversion of milliliters to liters, as this is a common source of confusion.
• Show how to apply the molarity concept in stoichiometry calculations to find out the concentration of other reactants or products in a reaction.