One way the U.S. Environmental Protection Agency (EPA) tests for chloride contam

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

CuCl2(aq)+ 2AgNO3(aq) 2AgCl(s)+ CuNO32(aq)

The chemist adds 39.0mM silver nitrate solution to the sample until silver chloride stops forming. He then washes, dries, and weighs the precipitate. He finds he has collected 8.4mg of silver chloride.

Calculate the concentration of copper(II) chloride contaminant in the original groundwater sample. Round your answer to 2 significant digits and in mg/L.

Explanation / Answer

Molar mass of AgCl = 108 + 35.45 = 143.45 gm/mol

Number of moles of AgCl = Mass/molar mass = 8 * 10^(-3)/143.45 = 5.5768 * 10^(-5) moles

One mole of CuCl2 will give 2 moles of AgCl(s)

Hence number of CuCl2 in the solution = 1/2 * 5.5768 * 10^(-5) = 2.7884 * 10^(-5) moles

Number of moles of CuCl2 in the solution = Volume of solution * Molarity

2.7884 * 10^(-5) moles = 250/1000 * Molarity

Molarity = 1.153 * 10^(-4) moles

Hence the molarity of the CuCl2 solution will be 1.153 * 10^(-4) M

Molarity in 2 significant digits will be 1.2 * 10^(-4) M

Molar mass of CuCl2 = 63.5 + 2(35.45) = 134.4 gm/mol

Concentration in mg/L = 1.2 * 10^(-4) mol/L * 134.4 gm/mol * 10^(3) mg/gm

=> 16.128 mg/L = 16 mg/L

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