When 38.0 mL of 0.1250 M H 2 SO 4 is added to 100. mL of a solution of Pb(NO 3 )

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?

Question 12 options:

1.55 × 10–3 M

3.11 × 10–3 M

1.55 × 10–4 M

6.20 × 10–3 M

3.10 × 10–4 M

1.55 × 10–3 M

3.11 × 10–3 M

1.55 × 10–4 M

6.20 × 10–3 M

3.10 × 10–4 M

Explanation / Answer

we know that

moles = mass / molar mass

also

molar mass of PbS04 = 303 g/mol

so

moles of PbS04 = 0.0471 / 303

moles of PbS04 = 1.55 x 10-4

given all the Pb+2 has been precipitated

so

this moles of Pb+2 have come from Pb(N03)2

now

Pb(N03)2 --> Pb+2 + 2 N03-

we can see that

moles of N03- = 2 x moles of Pb+2

so

moles of N03- = 2 x 1.55 x 10-4

moles of N03- = 3.1 x 10-4

now

we know that

conc = moles / volume (L)

so

[N03-] = 3.1 x 10-4 / 100 x 10-3

[N03-] = 3.1 x 10-3

so

the concentration of nitrate ions is 3.1 x 10-3 M

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