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 sample of groundwater known to be contaminated with iron(III) chloride, which would react with silver nitrate solution like this: (aq) (aq) (s) (aq) The chemist adds M silver nitrate solution to the sample until silver chloride stops forming. She then washes, dries, and weighs the precipitate. She finds she has collected 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 1

3.8x10⁻⁵ M

Step-by-step explanation:

By the reaction given, the stoichiometry between FeCl3 and AgNO3 and AgCl is 1:3:3. The molar mass of AgCl is 143.32 g/mol, so the number of moles that was formed is:

n = mass/molar mass

n = 0.0036g/143.32g/mol

n = 2.51x10⁻⁵ mol

Thus, the number of moles of FeCl3 that had precipitated was:

1 mol of FeCl3 ------------ 3 moles of AgCl

x -------------2.51x10⁻⁵ mol

By a simple direct three rule:

x = 8.37x10⁻⁶ mol

The precipitation is a reversible reaction, and, for the AgCl, is has an equilibrium constant Kps = 1.8x10⁻¹⁰, then, the solubility, S, (the maximum concentration of the ions that will not precipitate) can be calculated:

S² = Kps

S = √1.8x10⁻¹⁰

S = 1.34x10⁻⁵ mol/L

The volume of the sample was 250 mL = 0.250 L, so the number of moles of Cl⁻ that was not precipitated is:

n = 1.34x10⁻⁵ mol/L*0.250 L

n = 3.35x10⁻⁶ mol

Because of the stoichiometry:

1 mol of FeCl3 ------ 3 moles of Cl-

y ------ 3.35x10⁻⁶ mol

By a simple direct three rule:

y = 1.12x10⁻⁶ mol of FeCl3

The total number of moles of FeCl3 is then:

n = x + y

n = 9.5x10⁻⁵ mol

The concentration is the number of moles dived by the volume:

9.5x10⁻⁵ /0.250 = 3.8x10⁻⁵ M