Question

The concentration of lead ions (Pb2+) in a sample of polluted water that also contains nitrate ions (NO3-) is determined by adding sodium sulfate (Na2SO4) to exactly 500mL of the water. Calculate the molar concentration of Pb2+ if 0.0335g of Na2SO4 was needed for the complete precipitation of Pb2+ ions as PbSO4.

Answer 1

Given weight of Na₂SO₄ = 0.0335 g

& molar mass of Na₂SO₄ =142 g/mol

Now calculate number of mole of Na₂SO₄ = given mass / molar mass

num of mole of Na₂SO₄ = 0.0335 /142 =0.000236 mole

so num of mole of Na₂SO₄ is equal to num of mole of pb²⁺ ions when precipitation completed.

So we have moles of pb²⁺ =0.000236 mole

Given volume of water =500 mL (i.e. =0.500 L)

Molar concentration of pb²⁺ = num of mole / volume of water in liter

Molar concentration of pb²⁺ = 0.000236 / 0.500 =0.000472 M

Molar concentration of pb²⁺ =0.000472 M