QuestionJune 1, 2025

8. At a particular temperature, K=4.38 for the reaction SO_(2)(g)+NO_(2)(g)rightarrow SO_(3)(g)+NO(g) If all four gases had initial concentrations of 1.0 M, calculate the equilibrium concentration for all species. (5 points) 9. The Ksp for silver chromate, Ag_(2)CrO_(4), is 9.0times 10^-12 a. Write the balanced chemical equation for this dissolving process AND the Ksp expression. (3 points) b. Calculate the solubility of silver chromate in pure water (3 points) c. Calculate the solubility of silver chromate in the presence of 0.0050 M potassium chromate solution. (3 points)

8. At a particular temperature, K=4.38 for the reaction SO_(2)(g)+NO_(2)(g)rightarrow SO_(3)(g)+NO(g) If all four gases had initial concentrations of 1.0 M, calculate the equilibrium concentration for all species. (5 points) 9. The Ksp for silver chromate, Ag_(2)CrO_(4), is 9.0times 10^-12 a. Write the balanced chemical equation for this dissolving process AND the Ksp expression. (3 points) b. Calculate the solubility of silver chromate in pure water (3 points) c. Calculate the solubility of silver chromate in the presence of 0.0050 M potassium chromate solution. (3 points)
8. At a particular temperature, K=4.38 for the reaction SO_(2)(g)+NO_(2)(g)rightarrow SO_(3)(g)+NO(g) If all four gases had initial concentrations of 1.0 M, calculate the equilibrium concentration for all species. (5 points) 9. The Ksp for silver chromate, Ag_(2)CrO_(4), is 9.0times 10^-12 a. Write the balanced chemical equation for this dissolving process AND the Ksp expression. (3 points) b. Calculate the solubility of silver chromate in pure water (3 points) c. Calculate the solubility of silver chromate in the presence of 0.0050 M potassium chromate solution. (3 points)

Solution
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Answer

1. Set up the equilibrium expression for the reaction ### For SO_{2}(g) + NO_{2}(g) \leftrightarrow SO_{3}(g) + NO(g), the equilibrium constant expression is K = \frac SO_3][NO SO_2][NO_2 . ## Step2: Define changes in concentration ### Let the change in concentration at equilibrium be x. Then, [SO_3] = [NO] = 1.0 + x and [SO_2] = [NO_2] = 1.0 - x. ## Step3: Substitute into the equilibrium expression ### 4.38 = \frac{(1.0 + x)^2}{(1.0 - x)^2}. ## Step4: Solve for x ### Solving gives x \approx 0.48. ## Step5: Calculate equilibrium concentrations ### [SO_3] = [NO] = 1.0 + 0.48 = 1.48 M; [SO_2] = [NO_2] = 1.0 - 0.48 = 0.52 M. # Answer: ### [SO_2] = 0.52 M, [NO_2] = 0.52 M, [SO_3] = 1.48 M, [NO] = 1.48 M # Explanation: ## Step1: Write the balanced chemical equation ### Ag_2CrO_4(s) \leftrightarrow 2Ag^+(aq) + CrO_4^{2-}(aq). ## Step2: Write the Ksp expression ### K_{sp} = [Ag^+]^2[CrO_4^{2-}]. ## Step3: Calculate solubility in pure water ### Let solubility be s. Then, [Ag^+] = 2s and [CrO_4^{2-}] = s. K_{sp} = (2s)^2(s) = 4s^3 = 9.0 \times 10^{-12}. ## Step4: Solve for s ### s = \sqrt[3]{\frac{9.0 \times 10^{-12}}{4}} \approx 1.3 \times 10^{-4} M. ## Step5: Calculate solubility in 0.0050 M potassium chromate ### [CrO_4^{2-}] = 0.0050 + s', where s' is the new solubility. K_{sp} = (2s')^2(0.0050) \approx 4s'^2(0.0050) = 9.0 \times 10^{-12}. ## Step6: Solve for s' ### s' \approx \sqrt{\frac{9.0 \times 10^{-12}}{4 \times 0.0050}} \approx 6.7 \times 10^{-6} M. # Answer: ### a. Ag_2CrO_4(s) \leftrightarrow 2Ag^+(aq) + CrO_4^{2-}(aq); K_{sp} = [Ag^+]^2[CrO_4^{2-}] ### b. 1.3 \times 10^{-4} M ### c. 6.7 \times 10^{-6} M
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