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 250. mL sample of groundwater known to be contaminated with iron(II) chloride, which would react with silver nitrate solution like this: FeCl2(aa) + 2 AgNO3(aa) -2 AgCl(s) Fe(NO3),(aq) The chemist adds 56.0 mM silver nitrate solution to the sample until silver chloride stops forming. He then washes, dries, and weighs the precipitate. He finds he has collected 7.3 mg of silver chloride Calculate the concentration of iron(II) chloride contaminant in the original groundwater sample. Round your answer to 2 significant digits mg 0%

Answer 1

The balanced chemical reaction is -

FeCl2 + 2AgNO3 -> 2AgCl + Fe(NO3)2

Chemist got 7.3 mg of AgCl after reaction which is 7.3/143.3 millimoles of AgCl = 0.0509 millimoles

From the chemical reaction we know that the ratio of FeCl2 and AgCl must be 1/2

So, 0.0509 millimoles of AgCl means 0.0509/2 millimoles = 0.0254 of FeCl2 must be present in 250ml sample so, in 1 L sample will contain 0.0254*4 millimoles of FeCl2 = 0.1018 millimoles

Molar mass of FeCl2 = 126.75. So, weight of 0.1018 millimoles = 0.1018*126.75 = 12.913 mg per litre of groundwater sample.