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 cadmium chloride, which would react with silver nitrate solution like this:

CdCl2

(aq)

+

2AgNO3

(aq)

→

2AgCl

(s)

+

CdNO32

(aq)

The chemist adds

58.0m

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

8.1mg

of silver chloride.

Calculate the concentration of cadmium chloride contaminant in the original groundwater sample. Be sure your answer has the correct number of significant digits.

ANSWER IN mg/L

Answer 1

The reaction:

CdCl2(aq)+2AgNO3(aq)→2AgCl(s)+CdNO32(aq)

V = 200 mL of water;

58.0 mM of AgNO3 solution ...

mass = 8.1 mg of AgCl, find [CdCl2]

mol of AgCl = mass/MW = (8.1*10^-3)/(143.32 ) = 0.0000565 mol of AgCl

2 mol of AgCl = 1 mol of CdCl2

0.0000565 mol --> 1/2*0.0000565   = 0.00002825 mol of CdCl2

change to mass

mass = mol*MW = 0.00002825 *183.32 = 0.005178 g

0.005178*10^3 mg = 5.18 mg

This is in 200 mL = 0.2 L

C = mass/V = 5.18/0.2 = 25.9 mg/L