Principles 2 Week 10

.

Week 10 Day 1 Monday October 24, 2022 Part I Solubility Product K_sp: Solids in water and solubility
Textbook Readings 17.5: Solubility Equilibria and the Solubility Product Constant 17.6: Precipitation	Course Lectures 16.1 pdf Video* Solubility and K_sp
Objectives 1. Correctly write solubility equilibrium reactions with the solids always appearing on the reactant side 2. Write the Law of Mass Action expressions for solubility equilibria. 3. Calculate the molar and gram solubilities for salts in pure water given K_sp values. 4. Given molar solubility values, determine K_sp 5. Use K_sp values to compare solubilites of solids.	Solubility Equilibria: Ksp and molar solubility
Homework Problems Answer these questions. Click and drag over the space below for answers. 36.1 Write the chemical equation showing how PbBr₂ dissociates and write the K_sp Law of Mass Action expression. 36.2 Write the chemical equation showing how Bi₂S₃ dissociates and write the K_sp Law of Mass Action expression. 36.3 a. Calculate the molar solubility (S) of silver chloride (AgCl) in distilled water by solving the following ICE problem. Report the value of S with the correct units. AgCl_(s) ↔ Ag⁺_(aq) + Cl^-_(aq) K_sp = 1.77 x 10^-10. I ~ 0.00 M 0.00 M C + S + S E S S b. Convert the molar solubility (mol_AgCl/L) you calculated in part "a" into gram solubility (g_AgCl/L) using the AgCl molar mass: 143.320689 g/mol 36.4 Calculate the molar AND gram solubility tin (II) hydroxide (Sn(OH)₂) in distilled water given that K_sp =5.45 x 10^-27. 36.5 Calculate the gram AND molar solubility of bismuth sulfide (Bi₂S₃ ) in distilled water given K_sp = 1.82 x 10^-99. Answers: Click and drag in the space below 1. PbBr_2(s) <---> Pb²⁺_(aq) + 2 Br^-_(aq) K_sp = [Pb²⁺][Br^-]² 2. Bi₂S_3(s) <---> 2 Bi³⁺_(aq) + 3 S^2-_(aq) K_sp= [Bi³⁺]²[S^2-]³ 3. a. Molar solubility = 1.33 x 10^-5mole/Liter b. 1.91 x 10^-3g/L 4. Molar solubility = 1.11 x 10^-9 mole/Liter Gram Solubility = 1.69 x 10^-7 g/L 5. Molar solubility = 7.00 x 10^-21 mole/Liter Gram Solubility = 3.60 x 10^-18 g/L

.
.

Week 10 Day 2 Tuesday October 25, 2022 PART 2 Solubility Product K_sp: Solubility in Solutions The "COMMON ION" effect
Textbook Readings 17.6: Precipitation	Course Lectures 16.2 pdf Video* Solubility and Common Ions
Objectives 1. Describe what is meant by a "Common Ion" 2. Use pre-existing common ion concentrations to determine new molar and gram solubilities for sparingly soluble salts 3. Describe why a pre-existing common ion concentration reduces the solubility of a sparingly soluble salt. 4. Convert molar solubility (moles/L) into gram solubility (g/L) using the molar mass of the solute.	Solubility Equlibria: Common Ion Effect
Homework Problems 37.1 In the Le Chatelier Principle lab, you studied the following colorful complex-ion equilibrium that took place in a methanol solution: CoCl₄^2-_(methanol) + 6 H₂O_(methanol) ↔ Co(H₂O)₆²⁺_(methanol) + 4 Cl^-_(methanol) One of the "stresses" you applied to this equilibrium was small amounts of HCl. Why was additional HCl a "stress" and specifically what was the "Common Ion" in this case? 37.2 Calculate the molar solubility (S) and gram solubility (μg/L) of PbSO₄ in distilled water by solving the following ICE problem: PbSO_4(s) ↔ Pb²⁺_(aq) + SO₄^2-_(aq) K_sp = 1.96 x 10^-8 I ~ 0.00 M 0.00 M C + S + S E S S 37.3 Calculate the molar solubility (S) and gram solubility (μg/L) of PbSO₄ in 0.100 M Na₂SO₄ by solving the following ICE problem: PbSO_4(s) ↔ Pb²⁺_(aq) + SO₄^2-_(aq) K_sp = 1.96 x 10^-8 I ~ 0.00 M 0.100 M C + S + S E S S 37.4 Compare the solubilities you calculated in 37.2 and 37.3. Explain why PbSO₄ is so much more soluble in distilled water. 37.5 Calculate the molar and gram solubility (μg/L) of AgCl in a 0.050 M KCl solution? K_sp for AgCl is 1.00 x 10^-10 37.6 Calculate the molar and gram solubility (mg/L) of Ag₂CrO₄ in 0.010 M K₂CrO₄ solution. K_sp for Ag₂CrO₄ = 1.12 x 10^-12 Answers: Click and drag in the space below 37.1 HCl is a strong acid and completely ionizes. So, you were really adding H⁺ and Cl^- ions to the equilibrium mixture. H⁺ doesn't appear in the equilibrium reaction and in this case, has no effect. However, by adding HCl, you're also adding Cl^- ions and these do appear in the equilibrium reaction. Thus, Cl^- is considered a "common ion" (...an ion that your addition of HCl and the equilibrium reaction have in common.) The additional Cl^- ions cause the reaction to shift left in favor of the blue CoCl₄^2- reactant. 37.2 Molar Solubility = 1.40 x 10^-4 moles/Liter Gram Solubility = 42459 μg/L 37.3 Molar Solubility = 1.96 x 10^-7 moles/Liter Gram Solubility = 59.4 μg/L 37.4 PbSO₄ is more soluble in distilled water since significant amounts of both Pb²⁺ and SO₄^2- must be produced to reach equilibrium. (They both started at zero). However, in problem 37.3, SO₄^2- has a "head start" and SO₄^2- is the common ion. Consequently, the equilibrium in 37.3 doesn't have to shift as far to the right to reach equilibrium levels and less of the solid dissolves. 37.5 Molar Solubility = 2.00 x 10^-9 moles/Liter Gram Solubility = 0.287 μg/L 37.6 Molar Solubility = 5.29 x 10^-6 moles/Liter Gram Solubility = 1.76 mg/L

.
.

Week 10 Day 3 Wednesday October 26, 2022 Solubility Product: Precipitation Threshholds And Silver Recovery
Textbook Readings OS: 15.1 Precipitation and Dissolution LT: 17.6: Precipitation	Course Lectures 16.4 pdf Video* Precipitation Thresholds 16.5 pdf Video* Co-precipitation
Objectives 1. Perform dilution calculations to determine the new initial concentrations of reactant species 2. Use the balanced solubility chemical equilibrium reaction to determine the Law of Mass Action expression 3. Use the Law of Mass action and diluted reactant concentrations to determine a value of Q_sp	Predicting Precipitate Formation Predicting Precipitate Formation
4. Compare Qsp to K_sp: Q_sp < K_sp Reaction shifts right and solid dissolves. ...Solution is unsaturated Q_sp = K_sp No shift. Equilibrium. ...Solution is saturated. Q_sp > K_sp Reaction shifts left and precipitate forms. ... Solution is saturated 5. Given several different precipitation possibilities, predict if any will precipitate under the given conditions. 6. Use solubility principles to determine the percent recovery of aqueous metals in solution as precipitates.
Homework Problems 38.1 Consider the following solution preparations and ionic concentrations for AgCl. Calculate Q_sp , compare it to K_sp and predict if solid AgCl will precipiate. Useful Information: AgCl_(s) ↔ Ag⁺_(aq) + Cl^-_(aq) K_sp = 1.77 x 10^-10 Q_sp = [Ag⁺][Cl^-] a. [Ag⁺] = 2.50 x 10^-5 M [Cl^-] = 2.50 x 10^-5 M b. [Ag⁺] = 1.33 x 10^-13 M [Cl^-] = 8.90 x 10^-5 M c. [Ag⁺] = 5.00 x 10^-5 M [Cl^-] = 2.00 x 10^-6 M 38.2 1250.0 mL of 0.100 M AgNO₃ are mixed with 1.50 mL of 0.0050 M NaCl? a. Perform dilution calculations and determine the initial concentrations of the Ag⁺ and Cl^- ions. b. Use the concentrations you determined in part "b" and calculate a value for Q_sp. c. Compare your value of Q_sp to K_sp and determine whether a precipitate should form under these conditions. 38.3 Given a 1.35 x 10^-6 M AgNO₃ solution, what is the minimum chloride ion concentration required to begin the AgCl precipitation formation? 38.4 Silver, if present in a waste solution, is worth recovering (Silver is expensive!) Given 500.0 mL of a waste solution where [Ag⁺] = 0.00330 M a. How many moles and grams of silver metal are present in this waste solution? b. Concentrated HCl is added in small amounts to increase [Cl^-]. What is the minimum [Cl^-] required for precipitation of AgCl to begin? Note: For this problem and those that follow, we'll assume the volume contributed by the additional HCl is insignificant. c. If enough HCl is added to the solution for the concentration of chloride ion to be 0.0000100 M, what is the concentration of silver ions still remaining in solution? d. Use your answer to part "c" and the total solution volume (500.0 mL) to determine the moles and grams of silver metal still present in this waste solution. e. What percent of the original silver was precipitated in this procedure? 38.5 Small amounts of concentrated HCl are added to 750.0 mL of a waste solution that is known to be 0.155 M in silver metal ions. If the [Cl^-] levels are increased to 4.5 x 10^-7 M, what percent of the original silver is precipitated? (Assume the added volume of the HCl is insignificant) Click and drag the region below for correct answers 38.1 a. Q_sp = 6.25 x 10^-10 Q_sp > K_sp Reaction shifts left. Solid precip forms Solution is Saturated b. Q_sp = 1.18 x 10^-17 Q_sp < K_sp Reaction shifts right. Any solid AgCl present dissolves (solubility increases) Solution is initially unsaturated c. Q_sp = 1.00 x 10^-10 Q_sp = K_sp Reaction at equilibrium No visible change to any solid present Solution is saturated. 38.2 a. [Ag⁺] = 0.099₈₈ M [Cl^-] = 5.99₂₈ x 10^-6 M b. Q_sp = 5.98₅₆ x 10^-7 c. Q_sp > K_sp Reaction shifts left and precipitate is observed. 38.3 K_sp = [Ag⁺][Cl^-] 1.77x 10^-10 = (1.35 x 10^-6 ) [Cl^-] [Cl^-] = 1.3₁ x 10^-4 M 38.4 a. 0.00165₀ moles of Ag = 0.177₉ grams Ag⁺ in solution b. minimum [Cl^-] for precipitation to begin = 5.36 x 10^-8 M c. [Ag⁺] = 1.77 x 10^-5 M d. Assuming 500.0 mL total volume: 8.85 x 10^-6 moles Ag = 9.55 x 10^-4 grams Ag in solution e. 99.5 % of silver originally has been precipitated from solution. 38.5 99.8% of dissolved silver has been precipitated as solid AgCl

.

Week 10 Day 4 Thursday October 27, 2022 Selective-Precipitation
Textbook Reading Selective Precipitation	Course Lectures 16.5 pdf Video* Co-precipitation
Objectives 1. Use the associated K_sp values to determine the order solids precipitate from a solution that contains a mixture of metal cations 2. Determine anion threshold concentrations required for all possible solids to ppt. 3. For the first ion to precipitate, determine it's percent precipitation when a second ion begins to ppt.	Selective Precipitation (Dr. Dave)
Objectives 1. Use the associated K_sp values to determine the order solids precipitate from a solution that contains a mixture of metal cations 2. Determine anion threshold concentration required for a solid to precipitate. Homework Problems 39.1 A solution contains both Cl⁻ ions (0.050 M) and I⁻ ions (0.050 M). Next a concentrated Cu⁺ solution is slowly added to to precipitate CuCl and CuI sequentially. a. Use the K_sp values below to determine the copper ion thresholds required to precipitate CuCl and CuI. CuCl K_sp = 1.0 × 10^-6 CuI K_sp = 5.1 × 10^-12 b. Use your answers from part "a" to determine which of the two precipitates forms first. c. How can the K_sp values be used to arrange ions in order of precipitation and how does this relate to the idea of "solubility?" 39.2 A solution contains equal concentrations of of I^-, Br^-, Cl^- and CN^-. If concentrated, aqueous AgNO₃ is slowly added, what is the order of precipitation? AgI K_sp = 8.0 x 10^-17 AgCl K_sp = 1.77 x 10^-10 AgCN K_sp = 1.6 x 10^-14 AgBr K_sp = 7.7 x 10^-13 39.3 (Source) Concentrated BaCl₂ is slowly added to a solution that is 1.0 x 10^-4 M in sodium sulfate (Na₂SO₄) and 1.0 x 10^-4 M in sodium selenate, Na₂SeO₄. a. Which anion precipitates out of solution first? b. What percentage of the first ion has precipitated when the second anion begins to precipitate? BaSO₄ K_sp = 1.1 x 10^-10 BaSeO₄ K_sp = 2.8 x 10^-11. (Assume no volume change upon the addition of BaCl₂) 39.4 (Source) A solution containing Ni²⁺(0.10 M) and Cu²⁺ (0.10 M) are separated using selective precipitation by the addition of solid Na₂CO₃. How much of the first precipitated ion (in %) remains at the point when the second ion begins to precipitate? NiCO₃ K_sp = 1.4 x 10^-7 CuCO₃ K_sp = 2.5 x 10^-10 (Assume no volume change upon the addition of Na₂CO₃) Answers: Click and drag in the space below 39.1 a. CuCl(s) [Cu⁺]_threshold = 2.0 x 10^-5 M CuI(s) [Cu⁺]_threshold = 1.02 x 10^-10 M b. The CuI threshold is lower and therefore reached first. CuI precipitates before CuCl c. As long as the solid salts are of the same form, K_sp values can be used to predict the order of precipitation. In this case, both salts are 1 : 1 salts (i.e. 1 cation to 1 anion) and the salt with the small K_sp value will precipitate first (CuI). You would need to calculate solubility values when comparing 1:1 salts with other salts having different cation to anion ratios. For example comparing 1:1 CuOH to 1 : 2 Ca(OH)₂ would require an ICE solubility problem for each salt before deciding which precipitated out of solution first. 39.2 These are all 1:1 salts and their K_sp values can be used to order them from least soluble (first to precipitate) to most soluble (last to precipitate. First to ppt (least soluble) AgI ... AgCN ... AgBr ... AgCl Last to ppt (most soluble) 39.3 & 39.4 Solutions available here.

.
.
.
.
.
.
.