Question

A sample of \(1.50 \mathrm{~g}\) of lead(II) nitrate is mixed with \(125 \mathrm{~mL}\) of \(0.100 \mathrm{M}\) sodium sulfate solution. (a) Write the chemical equation for the reaction that occurs. (b) Which is the limiting reactant in the reaction? (c) What are the concentrations of all ions that remain in solution after the reaction is complete?

Short answer

The balanced chemical equation for the reaction between lead(II) nitrate and sodium sulfate is: \(Pb(NO_3)_2 + Na_2SO_4 \longrightarrow PbSO_4 + 2NaNO_3\) The limiting reactant in this reaction is lead(II) nitrate. After the reaction is complete, the concentrations of the remaining ions in the solution are: Sulfate ions (SO_4^{2-}) = 0.064 M Sodium ions (Na^+) = 0.199 M Nitrate ions (NO_3^-) = 0.0725 M

Step-by-step solution

Write the chemical equation

Lead(II) nitrate reacts with sodium sulfate to produce lead(II) sulfate and sodium nitrate. The unbalanced chemical equation is as follows: \(Pb(NO_3)_2 + Na_2SO_4 \longrightarrow PbSO_4 + NaNO_3\) Now let's balance the chemical equation: \(Pb(NO_3)_2 + Na_2SO_4 \longrightarrow PbSO_4 + 2NaNO_3\)

Calculate moles of each reactant

First, we need to find the number of moles of lead(II) nitrate and sodium sulfate. Given, Mass of lead(II) nitrate = 1.50 g Molar mass of lead(II) nitrate = 331.21 g/mol Moles of lead(II) nitrate = \(\frac{mass}{molar~mass}\) = \(\frac{1.50~g}{331.21~g/mol} = 0.00453~mol\) For sodium sulfate, we have: Volume of sodium sulfate solution = 125 mL = 0.125 L Concentration of sodium sulfate solution = 0.100 M Moles of sodium sulfate = (concentration)(volume) = (0.100 M)(0.125 L) = 0.0125 mol

Identify the limiting reactant

To identify the limiting reactant, we will compare the mole ratio of the reactants with the stoichiometry of the balanced chemical equation. Mole ratio of reactants (Pb(NO_3)_2 : Na_2SO_4) = \(\frac{0.00453~mol}{0.0125~mol} = 0.362\) Stoichiometric ratio from the balanced equation (Pb(NO_3)_2 : Na_2SO_4) = 1:1 Since the mole ratio of the reactants (0.362) is less than the stoichiometric ratio (1), lead(II) nitrate is the limiting reactant.

Calculate the concentrations of all ions remaining in the solution

Now we need to find the moles of ions that are consumed and the moles of ions that are produced in the reaction: Moles of lead(II) nitrate consumed = moles of lead(II) sulfate produced = 0.00453 mol Moles of sodium sulfate consumed = 0.00453 mol (from 1:1 stoichiometry) Moles of sodium sulfate remaining = 0.0125 mol - 0.00453 mol = 0.00797 mol Moles of sodium ions produced (from 2 moles of sodium nitrate) = 2 x 0.00453 mol = 0.00906 mol Moles of sodium ions remaining from sodium sulfate = 2 x 0.00797 mol = 0.0159 mol Total moles of sodium ions remaining = 0.0159 mol + 0.00906 mol = 0.0249 mol Moles of nitrate ions remaining (from produced sodium nitrate) = 2 x 0.00453 mol = 0.00906 mol Now, calculate the concentrations of remaining ions: The total volume of the solution = volume of sodium sulfate solution = 0.125 L Concentration of remaining sulfate ions = \(\frac{moles ~of ~sulfate~ ions}{total~ volume}\) = \(\frac{0.00797~mol}{0.125~L} = 0.064~M\) Concentration of remaining sodium ions = \(\frac{moles ~of ~sodium~ ions}{total~ volume}\) = \(\frac{0.0249~mol}{0.125~L} = 0.199~M\) Concentration of remaining nitrate ions = \(\frac{moles ~of ~nitrate~ ions}{total~ volume}\) = \(\frac{0.00906~mol}{0.125~L} = 0.0725~M\) The concentrations of all ions remaining in the solution after the reaction is complete are: Sulfate ions (SO_4^{2-}) = 0.064 M Sodium ions (Na^+) = 0.199 M Nitrate ions (NO_3^-) = 0.0725 M

Key concepts

Limiting Reactant

Understanding the concept of the limiting reactant is fundamental in stoichiometry, which deals with the quantities of reactants and products in chemical reactions. The limiting reactant is essentially the 'bottleneck' of the reaction, determining the maximum amount of product that can be formed.

Consider a real-life analogy: If you're making sandwiches, and you have 8 slices of bread but only 3 slices of cheese, the cheese is the limiting reactant because it runs out first, limiting the number of sandwiches you can make. Similarly, in a chemical reaction, the reactant that gets used up first limits the extent of the reaction and dictates the amount of products formed.

In the given exercise, we compared the actual mole ratio of reactants with the stoichiometric ratio from the balanced chemical equation. The limiting reactant was determined to be the one present in a smaller stoichiometric amount, which in this case was lead(II) nitrate. Hence, it governs the formation of products and the calculation of the remaining concentrations of ions in solution.

Mole Concept

The mole concept is a cornerstone of chemical stoichiometry and allows us to count atoms, molecules, and ions in bulk matter by weighing it. One mole is Avogadro's number (\(6.022 \times 10^{23}\)) of particles and this number is analogous to a 'dozen' but for atoms and molecules.

In the context of our problem, the mole concept helps to translate the mass of lead(II) nitrate and the volume and molarity of sodium sulfate solution into moles, which can then be used in stoichiometric calculations. Moles offer a way to scale reactions observed at the atomic level up to quantities that can be observed and measured in the laboratory. For instance, by knowing the molar mass of lead(II) nitrate, we could convert the mass given into moles to find out how much of it would react with sodium sulfate.

Chemical Reaction Equation

A chemical reaction equation represents the chemical changes that occur during a reaction. It illustrates the conversion of reactants to products through chemical formulas. Properly balancing this equation is crucial as it obeys the Law of Conservation of Mass; the number of atoms for each element must be the same on both sides of the equation.

In our chemical reaction, lead(II) nitrate reacts with sodium sulfate to form lead(II) sulfate and sodium nitrate. The balanced equation clearly indicates not only the substance's change but also the stoichiometric ratios, which are essential when identifying the limiting reactant and subsequently predicting the amount of each substance after the reaction has proceeded to completion.

Ionic Concentration

The ionic concentration in a solution refers to the amount of a particular ion present per unit volume of solution. It is usually expressed in moles per liter (Molarity, M). In the realm of stoichiometry, it's important for calculating how ions in solution react with each other to form products.

After the limiting reactant is used up, the remaining ions' concentrations must be calculated based on the amount of reactant left unreacted and the volume of the solution (now including both reactants and products after the reaction). In our exercise, by understanding the reaction stoichiometry and the change in quantity of each ion, we calculated the final concentrations of the sulfate, sodium, and nitrate ions.