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 AgNO3AgNO_{3} to the sample to precipitate all the Cl Cl^{-} as AgCl. The mass of the resulting AgCl precipitate is 71.7 mg. What is the chloride concentration in milligrams of ClCl^{-} per liter of groundwater?

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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 ClCl^{-} 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)Ag^{+}(aq) + Cl^{-}(aq) → AgCl(s)

It shows that 1 mole of ClCl^{-} ions reacts with 1 mole of Ag+Ag^{+} ions to produce 1 mole of AgCl. The molar mass of ClCl^{-} 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 mg×103 g1 mg=0.0717 g71.7  \sout{mg} \times \frac{10^{-3}  g}{1  \sout{mg}}=0.0717  g

The mass of ClCl^{-} ions in the 100.0 mL sample of groundwater is therefore

0.0717 g AgCl(s)×1 mol AgCl(s)143.32 g AgCl(s)×1 mol Cl(aq)1 mol AgCl(s)×35.45 g Cl(aq)1 mol Cl(aq)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 mgCl(aq)= 0.0177  g  Cl^{-}(aq)= 17.7  mg Cl^{-}(aq)

The ClCl^{-} ion concentration in milligrams per liter of water is

17.7 mgCl100.0 mL×1000 mL1 L=177 mg Cl/L groundwater\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 ClCl^{-} ions in 1 mg of AgCl(s):

1 mol Cl1 mol AgCl×35.45 g Cl/mol Cl143.32 g AgCl/mol AgCl=0.2473 g Clg AgCl=0.2473 mg Clmg AgCl\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 mg AgCl×0.2743 mg Cl1 mg AgCl=17.7 mg Cl precipitated as AgCl(s)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 ClCl^{-} in the sample for this exercise is 177 ppm, which meets the guidelines for drinking water.

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