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: \(Pb(NO_3)_2 (aq) + Na_2SO_4 (aq) \rightarrow PbSO_4 (s) + 2NaNO_3 (aq)\). The limiting reactant in the reaction is lead(II) nitrate. After the reaction is complete, the concentrations of the ions in the solution are: \(Pb^{2+}\): 0 M, \(NO_3^-\): 0.0724 M, \(Na^+\): 0.127 M, and \(SO_4^{2-}\): 0.0638 M.

Step-by-step solution

Write the balanced chemical equation for the reaction

The reaction between lead(II) nitrate and sodium sulfate can be represented as: \(Pb(NO_3)_2 (aq) + Na_2SO_4 (aq) \rightarrow PbSO_4 (s) + 2NaNO_3 (aq)\) This tells us that one mole of lead(II) nitrate reacts with one mole of sodium sulfate to produce one mole of lead(II) sulfate and two moles of sodium nitrate.

Calculate the moles of lead(II) nitrate and sodium sulfate

We can now use the mass of lead(II) nitrate and the volume and concentration of sodium sulfate to find the initial moles of the reactants. Moles of lead(II) nitrate: Molar mass of Pb(NO₃)₂ = 331.21 g/mol Mass of Pb(NO₃)₂ = 1.50 g Moles of Pb(NO₃)₂ = \(1.50~g/(331.21~g/mol)\) = 0.00453 moles Moles of sodium sulfate: Concentration of Na₂SO₄ = 0.100 M Volume of Na₂SO₄ = 125 mL = 0.125 L Moles of Na₂SO₄ = (0.100 M) * (0.125 L) = 0.0125 moles

Identify the limiting reactant

To determine the limiting reactant, we will use stoichiometry to determine which reactant will be completely consumed first. \(moles_{initial} (Pb(NO_3)_2) / mol/mol_{rxn} (Pb(NO_3)_2) = 0.00453 / 1 = 0.00453\) \(moles_{initial} (Na_2SO_4) / mol/mol_{rxn} (Na_2SO_4) = 0.0125 / 1 = 0.0125\) Since the ratio for lead(II) nitrate is smaller, it is the limiting reactant.

Calculate the concentrations of the ions after the reaction

As the reaction reaches completion, the limiting reactant will be completely consumed, and the amount of sodium sulfate left can be determined using stoichiometry: Moles of Na₂SO₄ remaining = Initial moles - moles reacted with Pb(NO₃)₂ = 0.0125 moles - 0.00453 moles = 0.00797 moles Now, we can determine the concentrations of the ions after the reaction: Concentration of \(Pb^{2+}\) = 0 M (completely consumed) Concentration of \(NO_3^-\) = \((2~mol~NaNO_3 / 1~mol~Pb(NO_3)_2) * 0.00453~moles~Pb(NO_3)_2 / 0.125~L = 0.0724~M\) Concentration of \(Na^+\) = \((2~mol~NaNO_3 / 1~mol~Na_2SO_4) * 0.00797~moles~Na_2SO_4 / 0.125~L = 0.127~M\) Concentration of \(SO_4^{2-}\) = 0.00797 moles remaining / 0.125 L = 0.0638 M The final concentrations of all ions in the solution are as follows: \(Pb^{2+}\): 0 M \(NO_3^-\): 0.0724 M \(Na^+\): 0.127 M \(SO_4^{2-}\): 0.0638 M