Question

Suppose an EPA chemist tests a 200.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) + FeNO32 (aq) The chemist adds 14.0m 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. Be sure your answer has the correct number of significant digits.

Answer 1

To find the concentration of iron(II) chloride in the groundwater sample, 6.9 mg of AgCl is converted to moles and then used to calculate the moles of FeCl2. The calculated moles of FeCl2 are then divided by the sample volume to get the concentration of 0.120 mM FeCl2.

Step-by-step explanation:

To calculate the concentration of iron(II) chloride contaminant in the groundwater sample, we first need to use the given data on the amount of silver chloride precipitated. The balanced equation for the reaction between iron(II) chloride and silver nitrate is: FeCl2 (aq) + 2AgNO3 (aq) → 2AgCl (s) + Fe(NO3)2 (aq). Since 6.9 mg of AgCl was collected, we can convert this mass to moles and use stoichiometry to find the moles of FeCl2.

First, we calculate the moles of AgCl: 6.9 mg AgCl × (1 g / 1000 mg) × (1 mol / 143.32 g) = 4.817 × 10-5 mol AgCl. Next, we use the molar ratio from the balanced equation to find the moles of FeCl2: 4.817 × 10-5 mol AgCl × (1 mol FeCl2 / 2 mol AgCl) = 2.4085 × 10-5 mol FeCl2.

Lastly, to find the concentration of FeCl2 in the original 200 mL sample, we divide the moles of FeCl2 by the volume of the sample in liters: 2.4085 × 10-5 mol FeCl2 / 0.200 L = 1.204 × 10-4 M FeCl2, which is 0.120 mM FeCl2. This is the concentration of iron(II) chloride in the groundwater sample with the correct number of significant digits.