Question

The balanced equation for the precipitation reaction between lead ion and potassium chloride is shown below.

Pb2+(aq)+2KCl(aq)⟶PbCl2(s)+2K+(aq)

Lead(II) ion is highly poisonous. To determine the amount of lead ion, a water sample was treated with excess of potassium chloride. A precipitate of lead(II) chloride was formed, weighing 80.1μg. Determine the mass of lead (in micrograms) in the water sample.

Answer 1

Answer:

The balanced equation is

Pb^2+(aq)+2KCl(aq)⟶PbCl2(s)+2K^+(aq)

Given mass of precipitate, PbCl2=80.1 ug=80.1 x 10^-6 g

Since 1 ug=10^-6 g.

Molar mass of PbCl2=278.1 g/mol.

Moles of PbCl2=mass/molar mass=80.1 x 10^-6 g/278.1 g/mol=2.88x10^-7 mol.

From the balanced equation, moles of Pb^+2 =moles of PbCl2=2.8x10^-7 mol.

Molar mass of Pb^+2=207.2 g/mol.

Mass of Pb^+2 in water sample=moles x molar mass=(2.8 x 10^-7 mol x 207.2 g/mol)=5.968 x 10^-5 g =59.68 x 10^-6 g.

Mass of Pb^+2 in water sample=59.68 ug.

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