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)+ Cu(NO3)2(aq).The chemist adds 81.0 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 7.7 mg of silver chloride. Calculate the concentration of copper(II) chloride contaminant in the original groundwater sample. Round your answer to 2 significant digits

Answer 1

Answer:

1.1 × 10⁻⁴ M

Explanation:

Let's consider the following double displacement reaction.

CuCl₂(aq) + 2 AgNO₃(aq) → 2 AgCl(s)+ Cu(NO₃)₂(aq)

We can establish the following relations:

• The molar mass of AgCl is 143.32 g/mol.
• The molar ratio of AgCl to CuCl₂ is 2:1

The moles of CuCl₂ that reacted to produce 7.7 mg of AgCl are:

$7.7 \times 10^{-3} gAgCl.\frac{1molAgCl}{143.32gAgCl} .\frac{1molCuCl_{2}}{2molAgCl} =2.7 \times 10^{-5}molCuCl_{2}$

The molarity of CuCl₂ is:

$M=\frac{2.7 \times 10^{-5}molCuCl_{2}}{0.250L} =1.1\times 10^{-4} M$