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 50.0m 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 2.6mg of silver chloride. Calculate the concentration of cadmium chloride contaminant in the original groundwater sample. Round your answer to 2 significant digits.

Answer 1

ans)

from above data that

CdCl₂ (aq) + 2 AgNO₃ (aq) --> 2 AgCl (s) + cd(NO₃)₂ (aq)

Molar mass of AgCl = 143.32 g/mol

Moles of AgCl formed = 2.6 mg x 10^-3 / 143.32

= 1.81 x 10^-5

Moles of AgNO3 reacted = 1.81 x 10^-5

Volume of AgNO3 added, V1 = Moles / Concentration

= 1.81 x 10^-5 / 50 x 10^-3 = 3.62 x 10^-4 L

Original sample volume, Vo = 200 mL = 0.2 L

Total volume of solution, Vo + V1 = 0.2004 L

Moles of cdCl2 reacted = Moles of AgCl formed / 2 (by stoichiometry)

= 0.905 x 10^-5

Molar mass of cdCl2 = 183.82 g/mol

Mass of cdCl2 in original sample = 183.82*0.905 x 10^-5 = 0.00166 g

Concentration of cdCl2 in original sample = 0.00166 g * 1000 mg/g / 0.2 L

= 8.30 mg/L