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.mLsample 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 39.0mM silver nitrate solution to the sample until silver chloride stops forming. She then washes, dries, and weighs the precipitate. She finds she has collected 7.4mg of silver chloride. Calculate the concentration of cadmium chloride contaminant in the original groundwater sample. Round your answer to 2 significant digits.

Answer 1

CdCl₂ (aq) + 2AgNO₃ (aq) → 2AgCl (s) + Cd(NO₃)₂ (aq)

Mole ratio of AgCl to CdCl₂ is 2:1

Mass of AgCl = 7.4 mg

Molar Mass of AgCl = 143.32 g/mol

Moles of AgCl = Mass of AgCl / Molar Mass of AgCl

= 7.4*10^-3/143.32 = 5.16*10^-5 mol

Moles of CdCl₂ = (1/2)*Moles of AgCl

= (1/2)*5.16*10^-5 = 2.58*10^-5 mol

Volume of original groundwater sample = 250 mL = 0.25 L

Concentration of CdCl₂ = Moles of CdCl₂ / Volume of original groundwater sample

= 2.58*10^-5 / 0.25 = 1.032*10^-5 M

= 1.0*10^-5 M [Rounding off to 2 significant digits]

Therefore, the concentration of CdCl₂ in the original groundwater sample is 1.0*10^-5 M.