Question

What mass of silver chloride can be prepared by the reaction of \(100.0 \mathrm{~mL}\) of \(0.20 \mathrm{M}\) silver nitrate with \(100.0 \mathrm{~mL}\) of \(0.15 \mathrm{M}\) calcium chloride? Calculate the concentrations of each ion remaining in solution after precipitation is complete.

Short answer

0.860 g of silver chloride (AgCl) can be prepared from these initial concentrations, and after precipitation is complete, the concentration of Ag+ ions in the solution is 0.0050 M and the concentration of NO3- ions is 0.020 M.

Step-by-step solution

Write the balanced chemical equation

The balanced chemical equation for this reaction can be given as: \(2AgNO_3(aq) + CaCl_2(aq) \rightarrow 2AgCl(s) + Ca(NO_3)_2(aq)\)

Convert volumes to moles using molarity

Using the given volumes and molarities, we can calculate the initial moles of silver nitrate and calcium chloride: Moles of silver nitrate (AgNO3) = \(Volume \times Molarity = 100.0 mL \times 0.20 M = 0.020 L \times 0.20 \frac{mol}{L} = 0.0040 mol\) Moles of calcium chloride (CaCl2) = \(Volume \times Molarity = 100.0 mL \times 0.15 M = 0.020 L \times 0.15 \frac{mol}{L} = 0.0030 mol\)

Calculate moles of silver chloride formed and convert to mass

From the balanced equation, we can see that 2 moles of silver nitrate react with 1 mole of calcium chloride. Ratio of moles = 0.0040 mol AgNO3 / 0.0030 mol CaCl2 = 1.33 Since the ratio is greater than 1, AgNO3 is in excess and CaCl2 is the limiting reagent. All the CaCl2 will be consumed in the reaction. Moles of AgCl formed = 0.0030 mol CaCl2 × (2 mol AgCl / 1 mol CaCl2) = 0.0060 mol AgCl Mass of AgCl = Moles of AgCl × Molar mass of AgCl Mass of AgCl = 0.0060 mol × 143.32 g/mol = 0.85992 g (rounded to 0.860 g)

Calculate concentrations of remaining ions after precipitation

After the precipitation is complete, the remaining ions in the solution will be Ag+ from the excess silver nitrate and the NO3- ions from both the silver and calcium nitrate. Moles of remaining AgNO3 = 0.0040 mol AgNO3 - 0.0030 mol AgNO3 (since 1 mol of CaCl2 reacts with 2 mol of AgNO3) = 0.0010 mol AgNO3 Total moles of NO3- ions = 0.0010 mol (from remaining AgNO3) + 0.0030 mol (from the reaction of CaCl2) = 0.0040 mol The total volume of the mixture = 100.0 mL (AgNO3) + 100.0 mL (CaCl2) = 200.0 mL (0.200 L) Concentration of remaining Ag+ ions = (\(moles \div volume ) = \frac{0.0010 mol}{0.200 L} = 0.0050 M\) Concentration of remaining NO3- ions = (\(moles \div volume) = \frac{0.0040 mol}{0.200 L} = 0.020 M\) In summary, 0.860 g of silver chloride (AgCl) can be prepared from these initial concentrations, and after precipitation is complete, the concentration of Ag+ ions in the solution is 0.0050 M and the concentration of NO3- ions is 0.020 M.