1 / 37

Topic 10 – Part II

Topic 10 – Part II. Applying Equilibrium Buffers Titrations Solubility Product. The Common Ion Effect. When the salt with the anion of a weak acid is added to that acid, It reverses the dissociation of the acid. Lowers the percent dissociation of the acid.

lajos
Télécharger la présentation

Topic 10 – Part II

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Topic 10 – Part II • Applying Equilibrium • Buffers • Titrations • Solubility Product

  2. The Common Ion Effect • When the salt with the anion of a weak acid is added to that acid, • It reverses the dissociation of the acid. • Lowers the percent dissociation of the acid. • The same principle applies to salts with the cation of a weak base. • The calculations are the same as last chapter.

  3. HF (aq) H+ (aq) + F- (aq) • If NaF is added to the solution • [F-] increases and the equilibrium shifts left • This makes the solution less acidic • Soln of NaF + HF less acidic than soln of HF

  4. Buffered solutions • A solution that resists a change in pH. • Either a weak acid and its salt or a weak base and its salt. • We can make a buffer of any pH by varying the concentrations of these solutions. • Same calculations as before.

  5. 15.2 – The pH of a Buffered Solution • A buffered solution contains 0.50 M acetic acid (HC2H3O2, Ka = 1.8 x 10-5) and 0.50 M sodium acetate (NaC2H3O2). • Calculate the pH of the solution.

  6. Adding a strong acid or base • Do the stoichiometry first. • A strong base will grab protons from the weak acid reducing [HA]0 • A strong acid will add its proton to the anion of the salt reducing [A-]0 • Then do the equilibrium problem.

  7. 15.3 – pH Changes in Buffered Solutions • Calculate the change in pH that occurs when 0.010 mol solid NaOH is added to 1.0 L of the buffered solution in sample exercise 15.2.

  8. Buffers – How do they work? • A buffered solution contains large quantities of a weak acid (HA) and its conjugate base (A-) from the salt. • When OH- ions are added they pull off the H+ from the HA • OH- + HA A- + H2O • OH- cannot accumulate. It reacts with HA to produce A-

  9. General equation • Ka = [H+] [A-] [HA] • so [H+] = Ka [HA] [A-] • The [H+] depends on the ratio [HA]/[A-] • taking the negative log of both sides • pH = -log(Ka [HA]/[A-]) • pH = -log(Ka)-log([HA]/[A-]) • pH = pKa + log([A-]/[HA])

  10. This is called the Henderson-Hasselbach equation • pH = pKa + log([A-]/[HA]) • pH = pKa + log(base/acid)

  11. 15.4 – The pH of a Buffered Solution II • Calculate the pH of a solution containing 0.75 M lactic acid (Ka = 1.4 x 10-4) and 0.25 M sodium lactate. 15.5 – The pH of a Buffered Solution III • A buffered solution contains 0.25 M NH3 (Kb = 1.8 x 10-5) and 0.40 M NH4Cl. Calculate the pH of this solution.

  12. 15.6 – Adding Strong Acid to a Buffered Solution I • What would the pH be if 0.10 mol of HCl is added to 1.0 L of the buffered solution from Ex. 15.5.

  13. Buffering Capacity • The pH of a buffered solution is determined by the ratio [A-]/[HA]. • As long as this doesn’t change much the pH won’t change much. • The more concentrated these two are the more H+ and OH- the solution will be able to absorb. • Larger concentrations bigger buffer capacity.

  14. 15.7 – Adding Strong Acid to a Buffered Solution II • Calculate the change in pH that occurs when 0.010 mol of HCl(g) is added to 1.0L of each of the following: • 5.00 M HAc and 5.00 M NaAc • 0.050 M HAc and 0.050 M NaAc • Ka= 1.8x10-5

  15. Buffer capacity • The best buffers have a ratio [A-]/[HA] = 1 • This is most resistant to change • True when [A-] = [HA] • Make pH = pKa (since log1=0)

  16. 15.8 – Preparing a Buffer • A chemist needs a solution buffered at pH 4.30 and can choose from the following acids (and their sodium salts): • a. chloroacetic acid (Ka = 1.35 x 10-3) • b. propanoic acid (Ka = 1.3 x 10-5) • c. benzoic acid (Ka = 6.4 x 10-5) • d. hypochlorous acid (Ka = 3.5 x 10-8) • Caclulate the ratio [HA] / [A-] required for each to yield a pH of 4.30. Which system will work best?

  17. Titrations • Millimole (mmol) = 1/1000 mol • Molarity = mmol/mL = mol/L • Makes calculations easier because we will rarely add Liters of solution. • Adding a solution of known concentration until the substance being tested is consumed. • This is called the equivalence point. • Graph of pH vs. mL is a titration curve.

  18. Strong acid with Strong Base • Do the stoichiometry. • There is no equilibrium . • They both dissociate completely. • The titration of 50.0 mL of 0.200 M HNO3 with 0.100 M NaOH • Calculate the pH after the following ml of NaOH are added: • A. no NaOH • B. 10ml of NaOH • C. 20ml of NaOH • D. 50ml of NaOH • E. 100ml of NaOH • F. 150ml of NaOH

  19. Weak acid with Strong base • There is an equilibrium. • Do stoichiometry. • Then do equilibrium. • Titrate 50.0 mL of 0.10 M HC2H3O2 (Ka = 1.8 x 10-5) with 0.10 M NaOH. Calculate the pH at the following points. • A. No NaOH • B. 10ml of NaOH • C. 25ml of NaOH • D. 40ml of NaOH • E. 50ml of NaOH • F. 60ml of NaOH

  20. Titration Curves • Strong acid with strong Base • Equivalence at pH 7 7 pH mL of Base added

  21. Weak acid with strong Base • Equivalence at pH >7 >7 pH mL of Base added

  22. Strong base with strong acid • Equivalence at pH 7 7 pH mL of Base added

  23. Weak base with strong acid • Equivalence at pH <7 <7 pH mL of Base added

  24. Summary: Titration of a Weak Acid with a Strong Base • At 0 ml base find pH with ICE and Ka • Weak acid before equivalence point • Stoichiometry first • Then Henderson-Hasselbach • Weak acid at ½ equivalence pt pH=pKa • Weak acid at equivalence point find pH with ICE and Kb • Weak base after equivalence - leftover strong base.

  25. Indicators • Weak acids that change color when they become bases. • weak acid written HIn • Weak base • HIn H+ + In- clear red • Equilibrium is controlled by pH • End point - when the indicator changes color.

  26. Indicators • Since it is an equilibrium the color change is gradual. • It is noticeable when the ratio of [In-]/[HIn] or [HIn]/[In-] is 1/10 • Since the Indicator is a weak acid, it has a Ka. • pH the indicator changes at is. • pH=pKa +log([In-]/[HIn]) = pKa +log(1/10) • pH=pKa - 1 on the way up (adding a base)

  27. Indicators • pH=pKa + log([HIn]/[In-]) = pKa + log(10) • pH=pKa+1 on the way down (adding an acid) • For strong acid base titrations, the indicator color change is sharp and a wide range of indicators , might be suitable. • For titrations of weak acids choose the indicator with a pKa as close to equivalence point as possible.

  28. Solubility Equilibria(Equilibria between solids and solutions) • All dissolving is an equilibrium. • If there is not much solid it will all dissolve. • As more solid is added the solution will become saturated. • Solid dissolved • The solid will precipitate as fast as it dissolves . • Equilibrium

  29. Consider the following equilibrium • CaF2 (s) Ca2+ (aq) + 2F- (aq) • Ultimately this reaches equilibrium • Solubility Product Constant (Ksp) is • Ksp = [Ca2+] [F-]2 • Called the solubility product.

  30. Solubility is not the same as solubility product • Solubility product is an equilibrium constant. • it doesn’t change except with temperature. • Solubility is an equilibrium position for how much can dissolve. • A common ion can change this.

  31. 15.12 - Calculating Ksp from Solubility I • Copper (I) Bromide has a measured solubility of 2.0 x 10-4 mol/L at 25oC. • Calculate its Ksp value

  32. 15.13 - Calculating Ksp from Solubility II • Calculate the Ksp value for bismuth sulfide (Bi2S3), which has a solubility of 1.0 x 10-15 mol/L at 25oC.

  33. 15.14 - Calculating Solubility from Ksp • The Ksp value for copper (II) iodate, Cu(IO3)2, is 1.4 x 10-7 at 25oC. Calculate its solubility at 25oC.

  34. Relative solubilities • Ksp will only allow us to compare the solubility of solids the that fall apart into the same number of ions. • The bigger the Ksp of those the more soluble. • If they fall apart into different number of pieces you have to do the math.

  35. Common Ion Effect • If we try to dissolve the solid in a solution with either the cation or anion already present less will dissolve. • 15.15 - Solubility and Common Ions • Calculate the solubility of solid CaF2 (Ksp = 4.0 x 10-11) in a 0.025 M NaF

  36. pH and solubility • OH- can be a common ion. • More soluble in acid or less in base. • Mg(OH)2 (s) Mg2+ (aq) + 2OH- (aq) • For other anions if they come from a weak acid they are more soluble in a acidic solution than in water. • H+ + PO43- HPO42- • Ag3PO4(s) 3Ag+(aq) + PO43-(aq)

  37. Precipitation and Qualitative Analysis • Ion Product, Q =[M+]a[N-]b • If Q>Ksp a precipitate forms. • If Q<Ksp No precipitate. • If Q = Ksp equilibrium. • 15.17 – Determining Precipitation Conditions • A solution is prepared by mixing 150.0 mL of 1.00 x 10-2 M Mg(NO3)2 and 250.0 mL of 1.00 x 10-1 M NaF. Calculate the concentrations of Mg2+ and F- at equilibrium with solid MgF2 (Ksp = 6.4 x 10-9).

More Related