Question

the chemist adds 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 of silver chloride. calculate the concentration of nickel(ii) chloride contaminant in the original groundwater sample. round your answer to significant digits.

Answer 1

The concentration of iron(II) chloride contaminant in the original groundwater sample is 8.96 10^-4 M. Rounding this answer to 2 significant digits gives us a final answer of 8.96 x 10^-4 M.

To solve for the concentration of iron(II) chloride contaminant in the original groundwater sample, we can use the following steps:

1. Calculate the moles of silver chloride precipitate:

6.4 mg AgCl * (1 mol AgCl / 143.32 g AgCl) = 4.47 * 10^-5 mol AgCl

2. Use the balanced chemical equation to determine the moles of iron(II) chloride that reacted with the silver nitrate:

FeCl2 (aq) + 2 AgNO3(aq) → AgCl(s) + Fe(NO3)3(aq)

From the balanced equation, we can see that 1 mole of iron(II) chloride reacts with 2 moles of silver nitrate. Therefore, the moles of iron(II) chloride that reacted is:

4.47 * 10^-5 mol AgCl * (1 mol FeCl2 / 2 mol AgCl) = 2.24 * 10^-5 mol FeCl2

3. Calculate the concentration of iron(II) chloride in the original groundwater sample:

Molarity of FeCl2 = moles of FeCl2 / volume of groundwater sample

The volume of the groundwater sample is 250 mL, which is equal to 0.250 L. Therefore, the molarity of iron(II) chloride in the groundwater sample is: Molarity of FeCl2 = 2.24 * 10^-5 mol FeCl2 / 0.250 L = 8.96 * 10^-4 M .