chemistry question

Question

Would a precipitate form if 255 mL of 0.00016 mol/L lead (II) nitrate, Pb(NO₃)₂(aq), was mixed with 456 mL of a 0.00023 mol/L solution of sodium sulfate, Na₂SO₄(aq)? Show your work.

Answer 1

Write the correctly balanced equation for the double replacement reaction taking place:

Pb(NO₃)₂(aq) + Na₂SO₄(aq) ==> PbSO₄(s) + 2NaNO₃(aq) ... balanced equation

From the solubility rules, we see that lead sulfate (PbSO₄) is insoluble, i.e. will precipitate. However, we must find the concentration of PbSO₄ and compare it to the Ksp for PbSO₄ (which we will look up in a table).

I found the Ksp for PbSO₄ to be 1.8x10^-8

To find the [PbSO₄] in the current reaction, we will consider the amounts of both reactants, find the limiting reactant, and then calculate the [PbSO₄] formed.

moles Pb(NO₃)₂ = 255 ml x 1 L/1000 ml x 0.00016 mol/L = 4.08x10^-5 mols

moles Na₂SO₄ = 456 mls x 1 L/1000 ml x 0.00023 mol/L = 1.05x10^-4 mols

Mole ratio in balanced equation is 1 mol Pb(NO₃)₂ : 1 mol Na₂SO₄

Therefore, Pb(NO₃)₂ is limiting and will dictate how much PbSO₄ can be formed.

Moles PbSO₄ formed = 4.08x10-5 mols Pb(NO₃)₂ x 1 mol PbSO₄/mol Pb(NO₃)₂ = 4.08x10^-5 mols PbSO₄

Total volume = 255 ml + 456 ml = 711 mls = 0.711 L

Final [PbSO₄] = 4.08x10^-5 mols / 0.711 L = 5.74x10^-5 M

This is considerably greater than the Ksp, and therefore a precipitate will form.