Question 4.14: To determine the concentration of chloride ions in a 100.0 m...
To determine the concentration of chloride ions in a 100.0 mL sample of groundwater, a chemist adds a large enough volume of a solution of AgNO_{3} to the sample to precipitate all the Cl^{-} as AgCl. The mass of the resulting AgCl precipitate is 71.7 mg. What is the chloride concentration in milligrams of Cl^{-} per liter of groundwater?
Collect and Organize We are given the sample volume, 100.0 mL, and the mass of AgCl formed, 71.7 mg. Our task is to determine the chloride concentration in milligrams of Cl^{-} per liter of groundwater.
Analyze We need a balanced chemical equation for the reaction. Although we can use the molecular, overall ionic, or net ionic equation, we choose the net ionic equation because it contains only those species involved in the precipitation reaction. The net ionic equation is
Ag^{+}(aq) + Cl^{-}(aq) → AgCl(s)
It shows that 1 mole of Cl^{-} ions reacts with 1 mole of Ag^{+} ions to produce 1 mole of AgCl. The molar mass of Cl^{-} is 35.45 g/mol, and that of AgCl is 143.32 g/mol. We use the methods developed in Chapter 3 to determine the mass of chloride ion in 100.0 mL of groundwater and then use this mass to calculate the concentration we need.
Solve Because our molar masses are in grams per mole, a logical first step is to convert the AgCl(s) mass to grams:
71.7 \sout{mg} \times \frac{10^{-3} g}{1 \sout{mg}}=0.0717 g
The mass of Cl^{-} ions in the 100.0 mL sample of groundwater is therefore
0.0717 \sout{g AgCl(s)} \times \frac{1 \sout{mol AgCl(s)}}{143.32 \sout{g AgCl(s)}} \times \frac{1 \sout{mol Cl^{-}(aq)}}{1 \sout{mol AgCl(s)}} \times \frac{35.45 g Cl^{-}(aq)}{1 \sout{mol Cl^{-}(aq)}}
= 0.0177 g Cl^{-}(aq)= 17.7 mg Cl^{-}(aq)
The Cl^{-} ion concentration in milligrams per liter of water is
\frac{17.7 mg Cl^{-}}{100.0 \sout{mL}} \times \frac{1000 \sout{mL}}{1 L} =177 mg Cl^{-}/L groundwater
An alternative way to solve this problem is to determine the mass of Cl^{-} ions in 1 mg of AgCl(s):
\frac{1 \sout{mol Cl^{-}}}{1 \sout{mol AgCl}} \times \frac{35.45 g Cl^{-}/\sout{mol Cl^{-}}}{143.32 g AgCl/ \sout{mol AgCl}} = \frac{0.2473 g Cl^{-}}{g AgCl} = \frac{0.2473 mg Cl^{-}}{mg AgCl}
Using this factor with the mass of AgCl(s) precipitated, we get
71.7 \sout{mg AgCl} \times \frac{0.2743 mg Cl^{-}}{1 \sout{mg AgCl}} =17.7 mg Cl^{-} precipitated as AgCl(s)
From here, the problem proceeds as before.
Think About It In Sample Exercise 4.1 we noted that the chloride concentration of drinking water should not exceed 250 ppm. Since mg/L is equivalent to ppm for dilute solutions, the concentration of Cl^{-} in the sample for this exercise is 177 ppm, which meets the guidelines for drinking water.