Question

When 38.0 mL of 0.1250 M H2SO4 is added to 100. mL of a solution of Pb(NO3)2, a precipitate of PbSO4 forms. The PbSO4 is then filtered from the solution, dried, and weighed. If the recovered PbSO4 is found to have a mass of 0.0471 g, what was the concentration of nitrate ions in the original solution, assuming that all of Pb2+ has been precipitated?

a) 1.55 × 10–3 M
b) 3.11 × 10–3 M
c) 1.55 × 10–4 M
d) 6.20 × 10–3 M
e) 3.10 × 10–4 M

Answer 1

b) 3.11x10⁻³ M

Step-by-step explanation:

We have:

38 ml H2SO4 0.1250 M

100 ml Pb(NO₃)₂

The chemical equation for this case is:

H₂SO₄ (ac) + Pb(NO₃)₂ (ac) → PbSO₄(s) + 2HNO₃(ac)

Sentence says that, once the solution of H₂SO₄ was added to the Pb(NO₃)₂ solution, we obtain a precipitate of PbSO₄ that finally weights 0.0471 g

First, we should obtain the molecular mass of this precipitate:

Pb= 207 g/mol

S= 32 g/mol

O= 16 g/mol

These values are obtained from Periodic Table, then:

Molecular mass of PbSO₄= 303 g

303 g PbSO₄=1 mol

0.0471 g PbSO₄= 1.55x10⁻⁴ mol

According to the stoichiometry of the above reaction, with 2 moles of NO₃⁻, we get 1 mol of PbSO₄, so:

1 mol PbSO₄=2 moles NO₃⁻

1.55x10⁻⁴ mol PbSO₄=3.11x10⁻⁴ moles NO₃⁻

Finally, we must obtain the concentration of these ions, in 1 Lt (1000 ml):

100 ml=3.11x10⁻⁴ mol NO₃⁻

1000 ml (1Lt) = 3.11x10⁻³ M NO₃⁻