Question

Calculate the solubility (in grams per liter) of silver chloride in the following. (a) pure water (b) \(0.025 \mathrm{M} \mathrm{BaCl}_{2}\) (c) \(0.17 \mathrm{M} \mathrm{AgNO}_{3}\)

Short answer

Question: Calculate the solubility of silver chloride (AgCl) in the following solutions: (a) pure water, (b) 0.025 M BaCl2, and (c) 0.17 M AgNO3. Express the solubility in grams per liter. Answer: The solubility of AgCl in the given solutions are: (a) pure water: 0.00192 g/L, (b) 0.025 M BaCl2: \(1.03 \times 10^{-6}\) g/L, and (c) 0.17 M AgNO3: \(1.52 \times 10^{-7}\) g/L.

Step-by-step solution

Write the balanced chemical equation for the dissociation of AgCl in water

The balanced chemical equation for the dissolution of silver chloride in water is: AgCl(s) \(\rightleftharpoons\) Ag+(aq) + Cl-(aq)

Write the solubility product expression for the balanced equation

The solubility product expression for AgCl is: Ksp = [Ag+][Cl-]

Find the Ksp value of AgCl

From a reference source, the Ksp value for AgCl is \(1.8 \times 10^{-10}\).

Calculate the solubility of AgCl in each solution

(a) For pure water, the concentration of both Ag+ and Cl- ions is equal to the solubility of AgCl (s). So, Ksp = s² Solving for s, we get: s = \(\sqrt{1.8 \times 10^{-10}}\) s = \(1.34 \times 10^{-5} \mathrm{M}\) To convert the solubility to grams per liter, multiply the molar solubility by the molar mass of AgCl: Solubility (g/L) = \((1.34 \times 10^{-5} \mathrm{M}) \times (143.3 \mathrm{g/mol})\) Solubility (g/L) ≈ 0.00192 g/L (b) For \(0.025 \mathrm{M} \mathrm{BaCl}_{2}\), the concentration of Cl- ions from BaCl2 is 0.025 M. Let s be the solubility of AgCl to form Ag+ and Cl-. Therefore, the total Cl- concentration will be s + 0.025 M, and the Ag+ concentration will be s. So, Ksp = s(s + 0.025) Solving for s, we get: s ≈ \(7.2 \times 10^{-9} \mathrm{M}\) Solubility (g/L) = \((7.2 \times 10^{-9} \mathrm{M}) \times (143.3 \mathrm{g/mol})\) Solubility (g/L) ≈ \(1.03 \times 10^{-6}\) g/L (c) For \(0.17 \mathrm{M} \mathrm{AgNO}_{3}\), the concentration of Ag+ ions from AgNO3 is 0.17 M. Since Ag+ ions are already present, adding more AgCl will not significantly increase the concentration of Ag+ ions (s will be very small compared to 0.17 M). Therefore, we can assume the Ag+ concentration to be approximately 0.17 M. As before, let s be the amount of added AgCl. So, Ksp = (0.17)(s) Solving for s, we get: s ≈ \(1.06 \times 10^{-9} \mathrm{M}\) Solubility (g/L) = \((1.06 \times 10^{-9} \mathrm{M}) \times (143.3 \mathrm{g/mol})\) Solubility (g/L) ≈ \(1.52 \times 10^{-7}\) g/L In summary, the solubility of AgCl in the given solutions are: (a) pure water: 0.00192 g/L (b) \(0.025 \mathrm{M} \mathrm{BaCl}_{2}\): \(1.03 \times 10^{-6}\) g/L (c) \(0.17 \mathrm{M} \mathrm{AgNO}_{3}\): \(1.52 \times 10^{-7}\) g/L