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) + FeNO33(aq) The chemist adds 57.0mM silver nitrate solution to the sample until silver chloride stops forming. She then washes, dries, and weighs the precipitate. She finds she has collected 3.5mg of silver chloride. Calculate the concentration of iron(III) chloride contaminant in the original groundwater sample. Round your answer to 2 significant digits.

Answer 1

Answer:

4.1 × 10⁻⁵ M

Explanation:

Let's consider the following balanced equation.

FeCl₃(aq) + 3 AgNO₃(aq) → 3 AgCl(s) + Fe(NO₃)₃(aq)

We can establish the following relations:

• The molar mass of AgCl is 143.32 g/mol.
• The molar ratio of AgCl to FeCl₃ is 3:1.

When 3.5 mg of AgCl are collected, the moles of of FeCl₃ that reacted are:

$3.5 \times 10^{-3} gAgCl.\frac{1molAgCl}{143.32gAgCl} .\frac{1molFeCl_{3}}{3molAgCl} =8.1 \times 10^{-6} molFeCl_{3}$

The sample has a volume of 200 mL (0.200 L). The molar concentration of FeCl₃ is:

$\frac{8.1 \times 10^{-6} molFeCl_{3}}{0.200L} =4.1 \times 10^{-5}M$