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Aqueous Equilibria

Aqueous Equilibria. Solubility Equilibrium. For substances that are not appreciably soluble, we use the solubility product, K sp , as the equilibrium expression for that process. HgCl 2(s)  Hg 2+ ( aq ) + 2 Cl - ( aq ) K sp = [Hg 2+ ][Cl - ] 2

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Aqueous Equilibria

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  1. Aqueous Equilibria

  2. Solubility Equilibrium • For substances that are not appreciably soluble, we use the solubility product, Ksp, as the equilibrium expression for that process. • HgCl2(s) Hg2+(aq) + 2 Cl-(aq) • Ksp = [Hg2+][Cl-]2 • Notice that the compound is not included because it’s a pure solid.

  3. Calculating Ksp from solubility • The molar solubility of Ag2CrO4 is 1.3x10-4 M. What is its Ksp? • Write the dissolution equation • Use stoichiometry to find the concentration of the ions. • Plug into Ksp and solve. • Ag2CrO4(s) 2 Ag+(aq) + CrO42-(aq) 1.3x10-4 2.6x10-4 1.3x10-4

  4. Finding Solubility from Ksp • The Ksp for CaF2 is 3.9x10-11. What is it’s solubility in grams per liter? • Write the dissociation reaction. • Use stoichiometry to determine the change in concentrations in terms of “x”. • Plug into the Ksp expression and solve for x. Then convert to desired units. CaF2(s) Ca2+(aq) + 2F-(aq) -x +x +2x

  5. Factors that affect solubility

  6. Common ion effect • Having one of the ions in a salt already present in solution decreases its solubility. • Le Chatlier’s argument – Adding a product shifts equilibrium to the left, making it less soluble. • Solve common ion problems just like Ksp problems… just add the amount of the ion present to the expression. • ICE tables may help

  7. pH and solubility • Salts with anions that come from weak acids are more soluble in acidic solution. • The F- can react as a base with excess H+, forming HF. • As F- reacts with H+, it reduces the concentration of F- in solution, shifting the first reaction to the right. PbF2(s) Pb2+(aq) + 2F-(aq) F- + H+ HF

  8. Buffers and titrations

  9. Definitions • Buffer – a solution containing a weak acid/base and its conjugate salt. • Buffer capacity – the amount of acid/base a buffer can absorb before the pH starts to appreciably change. • Titration – the slow addition of one substance to another until the reaction reaches stoichiometric completion. • Titrant – the substance being slowly added to a reaction mixture. • Analyte – the substance to which titrant is being added. • Equivalence point – when a reaction is stoichiometrically completed. • End point – the point at which an indicator changes color.

  10. Action of a buffer • Imagine we have an ammonia buffer (NH3 and NH4Cl). • As we add acid to the buffer, the NH3 will react with it to form NH4+. • As we add base to the buffer, the NH4+ will react with it to form NH3 • The change in pH of a buffer solution on the addition of acid/base is negligible, as long as you have not reacted all of the appropriate component. NH3 + HF  NH4+ + F- NH4+ + OH- NH3 + H2O

  11. Ph of a buffer • Henderson-Hasselbalch Equation • What is the pH of a buffer that is 0.12M lactic acid and 0.10M sodium lactate? Ka = 1.4x10-4

  12. Common buffers • Carbonate buffer – H2CO3 and HCO3- • Acetate buffer – HC2H3O2and C2H3O2- • Phosphate buffer – H2PO4- and HPO42- • You can use any soluble cation for the salts.

  13. Strong acid – strong base titration pH Calculations Initial pH  calculate from [acid] Before equivalence  calculate excess H+. At equivalence  pH = 7. After equivalence  calculate excess OH- Remember to add the volume of titrant added to the volume of analyte before calculating concentrations!

  14. Strong-Strong Calculations • 40 mL of 0.10M HCl was titrated with 0.10M NaOH. Calculate the pH at the following points: • 0 mL of titrant added • 10 mL of titrant added • 60 mL of titrant added There are 0.003 moles of H+ in excess. Divide by 50 mL to get [H+] = .06M. pH = 1.22 There are .002 moles of OH- in excess. Divide by 100 mL to get [OH-] = .02M. pH = 12.30

  15. Weak Acid – strong Base titration pH Calculations Initial  use ICE tables and Ka. Before equivalence  You have a buffer. Calculate [acid] and [base] using stoichiometry, then plug into HH equation. At equivalence  Use Kb and ICE tables. After equivalence  Calculate excess [OH-]. pH at ½ equivalence = pKa

  16. Weak-Strong Calculations (1 of 4) • 25 mL of 0.10M acetic acid (Ka = 1.8x10-5) is titrated with 0.10M sodium hydroxide. Calculate the pH after: • 0 mL of titrant added Since [HAc] >> Ka, we can ignore -x

  17. Weak Strong Calculations (2 of 4) • 15 mL of titrant added 25 mL of 0.1M HAc has 2.5x10-3 mole of acid. 15 mL of 0.10M NaOH has 1.5x10-3 mole base. We will form 1.5x10-3 mole of Ac-, with 1x10-3 mole of HAc excess. This is a buffer. Divide both amounts of moles by combined volume (40 mL), and plug into HH equation

  18. Weak Strong Calculations (3 of 4) • At equivalence All of the HAc has been turned into Ac-. We do a Kb problem for Ac-. The pH at equivalence for any weak acid/strong base combo will be greater than 7!

  19. Weak Strong Calculations (4 of 4) • 35 mL of titrant added 2.5x10-3 moles of acid and 3.5x10-3 moles of base. This leaves 1x10-3 moles of base in excess.

  20. Polyprotic acids • You will get an equivalence for each acidic proton. • Don’t worry about doing math here.

  21. Indicators • Weak acids with two forms: HIn and In-. • HIn and In- have different colors. • At pH < pKa of the indicator, HIn form predominates. • Once pH > pKa, In- predominates. • The point when an indicator changes color is called endpoint. • If the indicator is selected properly, endpoint is usually equivalence + 1 drop.

  22. phenopthalein • Phenopthalein (HPheno) is a weak acid with a pKa of 9.17. • HPheno is colorless • Pheno- is pink • At pH values less than 9.17, the colorless HPheno is dominant. • Above 9.17, the pink Pheno- ion is dominant. • If titrating anything with a strong base, the pH of the solution will be > pKaat endpoint.

  23. Common indicators Look at the equivalence pH for your titration to select the proper indicator.

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