Question

One way the U.S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titrating a sample of silver nitrate solution. Any chloride anions in solution will combine with the silver cations to produce bright white silver chloride precipitate.

Suppose an EPA chemist tests a

200.mL

sample of groundwater known to be contaminated with iron(II) chloride, which would react with silver nitrate solution like this:

FeCl2

(aq)

+

2AgNO3

(aq)

→

2AgCl

(s)

+

FeNO32

(aq)

The chemist adds

65.0m

M silver nitrate solution to the sample until silver chloride stops forming. He then washes, dries, and weighs the precipitate. He finds he has collected

6.4mg

of silver chloride.

Calculate the concentration of iron(II) chloride contaminant in the original groundwater sample. Round your answer to

2

significant digits.

=mg/L

Answer 1

Given reaction is

FeCl₂ + 2 AgNO₃ = Fe(NO₃)₂ + 2 AgCl

1 mole 2 mole 1 mole 2 mole

6.4 mg of AgCl has been collected

Molar mass of AgCl = 143.32 g/mol

No. of moles of AgCl = Mass / Molar mass = 6.4 * 10^-3 g / 143.32 g/mol = 4.5 * 10^-3 moles

according to reaction stoicometry 1 mole of FeCl2 gives 2 moles of AgCl

so No. of moles of FeCl2 = 4.5 * 10^-5 moles /2 = 2.25 * 10^-5 moles of FeCl2

Molar mass of FeCl2 = 126.75 g/mol

Mass of FeCl2 = No. of moles * molar mass = 2.25 * 10^-5 moles * 126.75 g/mol = 28.5 * 10^-3 g = 28.5 mg

this is the amount of FeCl2 present in 200 ml = 0.2 L of groundwater

so Concentration = Mass / Volume in Litres = 28.52 mg / 0.2 L = 142.59 mg/L Answer