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 iron(III) chloride, which would react with silver nitrate solution like this: FeCl3(aq) + 3 AgNO3(aq) → 3 AgCl(s) + Fe(NO3),(aq) The chemist adds 61.0 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 3.2 mg of silver chloride. Calculate the concentration of iron(III) chloride contaminant in the original groundwater sample. Round your answer to 2 significant digits. x 6 ?

Answer 1

Molar mass of AgCl,

MM = 1*MM(Ag) + 1*MM(Cl)

= 1*107.9 + 1*35.45

= 143.35 g/mol

mass of AgCl = 3.2*10^-3 g

mol of AgCl = (mass)/(molar mass)

= 3.2*10^-3/1.434*10^2

= 2.232*10^-5 mol

According to balanced equation

mol of FeCl3 formed = (1/3)* moles of AgCl

= (1/3)*2.232*10^-5

= 7.441*10^-6 mol

Molar mass of FeCl3,

MM = 1*MM(Fe) + 3*MM(Cl)

= 1*55.85 + 3*35.45

= 162.2 g/mol

mass of FeCl3 = number of mol * molar mass

= 7.441*10^-6*1.622*10^2

= 1.207*10^-3 g

= 1.207 mg

Concentration = mass of FeCl3 / volume in L

= 1.207 mg / 0.250 L

= 4.83 mg/L

Answer: 4.8 mg/L