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 tin(II) chloride, which would react with silver nitrate solution like this: SnCl2 (aq) + 2AgNO3 (aq) → 2AgCl (s) + SnNO32 (aq) The chemist adds 24.0m M silver nitrate solution to the sample until silver chloride stops forming. He then washes, dries, and weighs the precipitate. He finds he has collected 3.7mg of silver chloride. Calculate the concentration of tin(II) chloride contaminant in the original groundwater sample. Be sure your answer has the correct number of significant digits. express answer in mg/L

Answer 1

SnCl2

V = 250 mL sample

[AgNO3] = 25 mM = 25*10^-3 M

mass of AgCl = 3.7 mg = 3.7*10^-3 g

mol of AgCl = mass/MW = (3.7*10^-3)/(143.32 ) = 0.00002581 mol of AgCl

now...

2 mol of AgCl = 2 mol of AgNO3

then, 0.00002581 mol of AgNO3

1 mol of SnCl = 2 ml of AgNO3

mol of SnCl2 = 1/2*0.00002581 = 0.000012905 mol of SnCl2

[SnCl2] = mol /V = (0.000012905)/(250*10^-3) = 0.00005162 M = 0.00005162*10^3 mM = 0.052 mM

change to mg /L

MW of SnCl2 = 189.6 *0.052 mM = 9.8592 mg of SnCl2 / L