Question

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) + 3AgNO3(aq) ⟶ 3AgCl(s) + FeNO3(aq)The chemist adds 82.0 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 2.5mg of silver chloride.Calculate the concentration of iron(III) chloride contaminant in the original groundwater sample.

Answer 1

`2.326* 10^(-5) M` the concentration of iron(III) chloride contaminant in the original groundwater sample.

Step-by-step explanation:

`FeCl_3(aq) + 3AgNO_3(aq)\rightarrow 3AgCl(s) + Fe(NO_3)_3(aq)`

Mass of silver chloride = 2.5 mg = 0.0025 g

1 mg = 0.001 g

Moles of silver chloride =
`(0.0025 g)/(143.32 g/mol)=1.744* 10^(-5) mol`

According to reaction, 3 moles of silver chloride are obtained from 1 mole of ferric chloride, then
`1.744* 10^(-5) mol` od silver chloride will be obtained from ;

`(1)/(3)* 1.744* 10^(-5) mol=5.814* 10^(-6) mol` of ferric chloride

Volume of the ground water sample = 250 mL= 0.250 L

1 mL = 0.001 L

Concentration of iron(III) chloride =

`[FeCl_3]=(5.814* 10^(-6) mol)/(0.250 L)=2.326* 10^(-5) M`

`2.326* 10^(-5) M` the concentration of iron(III) chloride contaminant in the original groundwater sample.