Question

2) When 38.0 mL of 0.1250 MH,SO, is added to 100 mL of a solution of Pbly, 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 iodide ions in the original solution?

Answer 1

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 PbI2

now

PbI2 --> Pb+2 + 2 I-

we can see that

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

so

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

moles of I- = 3.1 x 10-4

now

we know that

conc = moles / volume (L)

so

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

[I-] = 3.1 x 10-3

so

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