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(III) chloride, which would react with silver nitrate solution like this:

FeCl₃(aq)+3AgNO₃(aq) -> 3AgCl(s)+Fe(NO₃)₃(aq)

The chemist adds 34.0mM silver nitrate solution to the sample until silver chloride stops forming. She then washes, dries, and weighs the precipitate. She finds she has collected 6.0mg of silver chloride.

Calculate the concentration of iron(III) chloride contaminant in the original groundwater sample. Be sure your answer has the correct number of significant digits.

Answer ? mg/L

Answer 1

silver in mmoles = 6mg/143.32g/mol = 0.04186 mmol { 143.32g/mol is molar mass of AgCl}

FeCl₃(aq)+3AgNO₃(aq) -> 3AgCl(s)+Fe(NO₃)₃(aq)

1 mole FeCl3 needs 3 mol AgNO3,which forms 3 mol AgCl

mmoles of FeCl3 = 1/3 x mmoles of AgCl = 0.013955 mmol

molarity = mmoles / volume in mL = 0.013955 mmol / 200 mL = 6.97739 x 10^-5 M

in terms of mg/mL = 6.97 x 10^-5 mol /L = 162.2 g/mol x 6.97739 x 10^-5 mol / L = 1.1317 x 10^-2 mg/mL

{162.2 g/mol is molar mass of FeCl3 }

3 significant figures minimum in the given data

1.13 x 10^-2 mg/L

can also be written as : 0.0113 mg/L