How do you solve this

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(II) chloride, which would react with silver nitrate solution like this:

FeCl2(aq)+2AgNO3(aq) →2AgCl(s)+Fe(NO3)2(aq)

The chemist adds 29.0m

M silver nitrate solution to the sample until silver chloride stops forming. She then washes, dries, and weighs the precipitate. She finds she has collected 6.9mg

 of silver chloride.Calculate the concentration of iron(II) chloride contaminant in the original groundwater sample. Round your answer to 2 significant digits.

Answer 1

So for this question, the first step is to determine what information is necessary to solve the problem and what is not. The 29.0mM concentration of the silver nitrate solution is not important since it is added in excess to precipitate out all of the silver chloride. That being said, we now have to convert the grams of precipitate to moles. 6.9 mg of AgCl is equivalent to 0.000048 mol (4.8e-5 mol) of AgCl.

Looking at the reaction formula, there is a ratio of 2 moles of AgCl for every one mole of FeCl2. This means that for every two moles of AgCl precipitated, there will be one mole of FeCl2 from the initial groundwater. This means that the moles of AgCl must be divided by 2 in order to find the moles of FeCl2, giving us a value of 2.4e-5 mol of FeCl2.

Lastly, to find the concentration in molarity, we divide this value by original 0.25L of sample, since molarity is moles/liters of solution. This gives us a final answer of 9.6e-5 M (0.000096 M) concentration of FeCl2 in the contaminated groundwater.

Hope this helps!