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 nickel(II) chloride, which would react with silver nitrate solution like this: NiCl2 (aq) + 2AgNO3 (aq) -> 2AgCl (s) + Ni(NO3)2 (aq) The chemist adds 74.0 mM silver nitrate solution to the sample until silver chloride stops forming. She then washes, dries, and weighs the precipitate. She finds she has collected 8.3 of silver chloride. Calculate the concentration of nickel(II) chloride contaminant in the original groundwater sample. Show your work!

Answer 1

15.0g/L is the concentration of nickel(II) chloride contaminant in the original groundwater sample.

Step-by-step explanation:

Based on the reaction:

NiCl2 (aq) + 2AgNO3 (aq) -> 2AgCl (s) + Ni(NO3)2 (aq)

Where 1 mole of NiCl₂ reacts producing 2 moles of AgCl.

To solve this problem, we need to convert mass of AgCl to moles to know the moles of NiCl₂ that reacts. With these moles and the volume of the sample (250mL = 0.250L), we can determine the molar concentration of the contaminant in the sample

Moles AgCl:

8.3g of AgCl were collected. In moles (Molar mass AgCl: 143.32g/mol):

8.3g AgCl * (1mol / 143.32g) = 0.05791 moles AgCl

Moles NiCl₂:

As 2 moles of AgCl are produced from 1 mole of NiCl₂. Moles of NiCl₂ are:

0.05791 moles AgCl * (1 mole NiCl₂ / 2 moles AgCl) = 0.02896 moles NiCl₂

Molar concentration:

0.02896 moles NiCl₂ / 0.250L =

0.1158M

In g/L (Molar mass NiCl₂: 129.6g/mol):

0.1158 mol / L * (129.6g / mol) =

15.0g/L is the concentration of nickel(II) chloride contaminant in the original groundwater sample.