Question

6.80 mL of a 1.30 M solution of lead(II) nitrate was combined with 3.20 mL of a 5.0 M

solution of ammonium sulfide.

1. Write the chemical equation for this reaction and balance it.

2. What is the total concentration of all soluble ions after these two solutions are mixed

and allowed to react completely?

3. A solid product precipitated out of this reaction, and was collected. If 2.01 g of this

product were collected, what is the percent yield?

Answer 1

1) The reaction will be

Here, PbS will precipitate out.

2)

Total Volume = 6.80 mL + 3.20 mL = 10.00 mL

milli moles of Pb(NO₃)₂ = Volume * Molarity = 6.80 mL * 1.30 M = 8.84 mmmol

milli moles of (NH₄)₂S = Volume * Molarity = 3.20 mL * 5.0 M = 16.0 mmmol

Concentration of NO₃^-

1 mmol of Pb(NO₃)₂ will give 2 mmol of NO₃^- which will remain soluble.

Thus, mmol of NO₃^- = 2 * 8.84 mmol = 17.68 mmol

Concentration of NO₃^- = Moles/Volume = 17.68 mmol / 10.00 mL = 1.768 M

Concentration of NH₄⁺

1 mmol of (NH₄)₂Swill give 2 mmol of NH₄⁺ which will remain soluble.

Thus, mmol of NH₄⁺ = 2 * 16.00 mmol = 32.00 mmol

Concentration of NH₄⁺ = Moles/Volume = 32.00 mmol/10.00 mL = 3.200 M

Concentration of Pb²⁺

1 mmol of Pb(NO₃)₂ will give 1 mmol of Pb²⁺

Thus, mmol of Pb²⁺= 8.84 mmol

But all of it will react with 8.84 mmol S^2- to form precipitate and none of it will remain.

Concentration of Pb²⁺ =0 M

Concentration of S^2-

1 mmol of (NH₄)₂Swill give 1 mmol of S^2- which will remain soluble.

Thus, mmol of S^2- = 16.00 mmol

But 8.84 mmol will react with Pb²⁺.

Moles remaining = 16.00 mmol - 8.84 mmol = 7.16 mmol

Concentration of S^2-  = Moles/Volume = 7.16 mmol/10.00 mL = 0.716 M

Total concentration of all soluble ions = 1.768 M + 3.200 M + 0.716 M = 5.684 M = 5.68 M (3 sig figure)

c)

Theoretical moles of PbS produced = 8.84 mmol = 8.84 * 10^-3 mol

Molar Mass of PbS = 239.3 g/mol

Theoretical Mass = (8.84 * 10^-3 mol) * (239.3 g/mol) = 2.115 g

Actual Mass = 2.01 g

% Yield = Actual Mass / Theoretical Mass * 100 = 2.01 g / 2.115 g * 100= 95.03%