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

CuCl2 (aq) + 2AgNO3 (aq) --> 2AgCl (s) + Cu(NO3)2 (aq)

The chemist adds 38.0 mM silver nitrate solution to the sample until silver chloride stops forming. She then washes, dries, and weighs the precipitate. She finds she has collected 7.2 mg of silver chloride. Calculate the concentration of copper(II) chloride contaminant in the original groundwater sample. Be sure your answer has the correct number of significant digits. ( anserer in mg/L)

Answer 1

Mass of silver chloride obtained = 7.2 mg = 0.0072 g.

Molar mass of silver chloride = 143.3 g/mol

Moles of silver chloride obtained = mass / molar mass = 0.072 / 143.3 = 0.0000502 mol

From the balanced equation,

2 mol of silver chloride is obtained from 1 mol of copper(II) chloride

Then, 0.0000502 mol of silver chloride is obtained from 0.0000251 mol of copper(II) chloride.

Mass of copper(II) chloride = moles * molar mass = 0.0000251 * 134.4 = 0.00338 g. = 3.38 mg.

Volume of copper(II) chloride used = 200. mL = 0.200 L

Therefore concentration of sopper(II) chlorid ein the origianl contaminant sample = mass / volume

C = 3.38 / 0.200

C = 16.9 mg/L