In a lab experiment you mix 0.20 m lead(ii) nitrate and 0.10 m sodium sulfate. in the reaction, you added 20 drops of each solution (approximately 0.90 ml of each solution) and observed a precipitate. what are the concentration(s) of all species remaining in the solution? you may assume that the compound precipitating is very insoluble and drops out of solution completely.

Pb(NO₃)₂ = 0.05 M

PbSO₄ = 0.05 M

NaNO₃= 0.1 M

Explanation:

The ions in lead(II) nitrate are Pb⁺² and NO₃⁻, so its formula is Pb(NO₃)₂. The ions of sodium sulfate are Na⁺ and SO₄²⁻, so its formula is Na₂SO₄. The reaction will be:

Pb(NO₃)₂(aq) + Na₂SO₄(aq)→ PbSO₄(s) + 2NaNO₃(aq)

For the sulfates, the lead sulfate is insoluble. The salts formed by elements from group 1 are solubles (such as sodium). So the precipitated formed is PbSO₄, and because it is insoluble, the reaction can be considered irreversible.

The number of moles of each reactant is the concentration multiplied by the volume in L (0.90 mL = 0.0009 L):

Pb(NO₃)₂ = 0.20x0.0009 = 0.00018 mol

Na₂SO₄ = 0.1x0.0009 = 0.00009 mol

The stoichiometry is 1 mol of Pb(NO₃)₂ from 1 mol of Na₂SO₄, so Pb(NO₃)₂ is in excess (by double) and Na₂SO₄ is the limiting reactant and it's all consumed.  

The stoichiometry calculus must be done with the limiting reactant. So, for Pb(NO₃)₂, the number of moles resulting will be:

Pb(NO₃)₂ = 0.00018  - 0.00009 = 0.00009 mol

For PbSO₄:

1 mol of Na₂SO₄ 1 mol of PbSO₄

PbSO₄= 0.00009 mol

For NaNO₃:

1 mol of Na₂SO₄ 2 mol of NaNO₃

0.00009 mol   x

By a simple direct three rule:

x = 0.00018 mol of NaNO₃

The final volume is = 0.90 + 0.9 = 1.8 mL = 0.0018 L, and the concentrations are the number of mol divided by the volume:

Pb(NO₃)₂ = 0.00009/0.0018  = 0.05 M

PbSO₄ = 0.00009/0.0018  = 0.05 M

NaNO₃= 0.00018/0.0018 = 0.1 M

i think   the answer is c hope i'm right