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Friday, April 5

Friday, April 5 . Work on review problems –solubility Work in groups of 3 on review problems Come up and write answer on board when you think you have the correct answer. Calculate the molar solubility of. Molar Solubility Answers. Ag 3 PO 4 ; x = 1.8 x 10 -5 M

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Friday, April 5

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  1. Friday, April 5 • Work on review problems –solubility • Work in groups of 3 on review problems • Come up and write answer on board when you think you have the correct answer

  2. Calculate the molar solubility of

  3. Molar Solubility Answers • Ag3PO4; x = 1.8 x 10-5 M • Al(OH)3; x = 1.0 x 10-9 M • PbCl2; x = 1.6 x 10-2 M • Ag2CrO4; x = 8.7 x 10-5 M

  4. Solubility continued The magnesium and calcium ions present in seawater ([Mg2+] = 0.059 M and [Ca2+] = 0.011 M) can be separated by selective precipitation with KOH. Calculate the [OH-] that would separate the metal ions. Kspof Mg(OH)2= 6.3 x 10-10; Kspof Ca(OH)2= 6.5 x 10-6 Ans: [OH-]: 1.03 x 10-4 M

  5. Solubility questions LEFT side: The water solubility of SrF2 is 0.011 g/ 100 mL. Calculate the solubility product constant. Ksp = 2 x 10-9 RIGHT side: The water solubility of Ag2S is 3.12 x 10-15 g/ 100 mL. Calculate the solubility product constant. Ksp = 8 x 10-48

  6. Conceptual solubility question Explain how the molar solubility of CaF2 will differ in a solution containing 0.100 M NaF compared to in a solution of pure water? Common ion effect Molar solubility of CaF2 in NaF is lower than in pure water.

  7. Buffer questions Calculate the pH of a buffer that consists of 0.45 M of C6H5COOH and 0.25 M of C6H5COONa. Ka= 6.3 x 10-5. Ans: pH = 3.94

  8. Buffer questions Now assume that the pH increased to 5.10, what are the concentrations of the buffer components? i.e.) [C6H5COOH], [C6H5COONa]? Ans: [A] = 0.078 M; [B] = 0.622 M Check: Does this make sense? Yes, an increase in pH means we have more base in solution.

  9. Buffer questions Find the pH of a buffer that consists of 0.25 M NH3 and 0.15 M NH4Cl (p­Kb of NH3 = 4.75). Approach: First find pKa. Then sub pKa into Henderson Hasselbalch equation. Ans: pH = 9.47

  10. More Buffer questions What is the ratio of [Pr-]/ [HPr] buffer that has a pH of 5.44 (Ka of HPr = 1.3 x 10-5)? Ans: [Pr-]/[HPr]= 3.55 This means need ~3.6 times more [Pr-] than [HPr] to get to pH of 5.44.

  11. Last Buffer question Last Buffer Question A buffer containing 0.2000 M of acid, HA, and 0.1500 M of its conjugate base, A-, has pH of 3.35. What is the pH after 0.0015 mol of NaOH is added to 0.5000 L of this solution? Approach: First find pKa. Set up equation showing the neutralization reaction with ICE table. Use Henderson Hasselbalch equation to solve for pH. Ans: pH = 3.37

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