Question

If 25.8 ml of an AgNO₃ solution is needed to precipitate all the Cl⁻ ions in a 785-mg sample of KCl (forming AgCl), what is the molarity of the AgNO₃ solution?

Answer 1

The molarity of the AgNO₃ solution used to precipitate all Cl⁻ ions in a 785-mg sample of KCl is 0.408 M. To find the molarity of the AgNO₃ solution, convert the mass of KCl to moles, use the stoichiometry of the reaction between AgNO₃ and KCl, which is 1:1, then divide the moles of AgNO₃ by the volume in liters to get the molarity.

Step-by-step explanation:

To calculate the molarity of the AgNO₃ solution, we first need to understand the reaction between AgNO₃ and KCl, which produces AgCl as a precipitate:

AgNO₃ (aq) + KCl (aq) → AgCl (s) + KNO₃ (aq)

According to the stoichiometry of this reaction, each mole of KCl will react with one mole of AgNO₃ to form one mole of AgCl. Now, let's convert the mass of KCl to moles:

1. First, determine the molar mass of KCl, which is 39.1 (K) + 35.45 (Cl) = 74.55 g/mol.
2. Next, convert the mass of KCl to moles by dividing the mass of the sample (785 mg, which is 0.785 g) by its molar mass: 0.785 g / 74.55 g/mol ≈ 0.01053 moles of KCl.
3. Since KCl and AgNO₃ react in a 1:1 ratio, there are also 0.01053 moles of AgNO₃.
4. Finally, calculate the molarity of AgNO₃ by dividing the number of moles of AgNO₃ by the volume of the solution in liters: 0.01053 moles / 0.0258 L = 0.408 M.

The molarity of the AgNO₃ solution is therefore 0.408 M.