Effect of Structure on Acid-Base Properties

1. Effect of Structure on Acid-Base Properties

2. Applications of Aqueous Equilibria

3. Common Ion Effect A solution contains HF (ka = 7.2 x 10-4) and its salt (NaF). What effect does the presence of the dissolved salt have on the dissociation equilibrium? H2O(l) NaF(s) Na+(aq) + F-(aq)

4. A solution contains HF (ka = 7.2 x 10-4) and its salt (NaF). Step 1: Identify the major species HF, Na+, F- and H2O HF(aq) H+(aq) + F-(aq) HF(aq) H+(aq) + F-(aq) + F-(aq) + Na+

5. Previously the equilibrium concentration of H+ in a 1.0M HF solution was determined to be 2.7 x 10-2M, and the percent dissociation of HF = 2.7%. Calculate [H+] and the percent dissociation of HF in a solution containing 1.0M HF (Ka = 7.2 x 10-4) and 1.0M NaF. [H+] = 7.2 x 10-4M Percent Dissociation = 0.072% WOW!

6. Buffered Solutions A solution that resists a change in pH when either hydroxide ions or protons are added. Buffered solutions contain either: A weak acid and its salt A weak base and its salt

7. Acid/Salt Buffering Pairs The salt will contain the anion of the acid, and the cation of a strong base (NaOH, KOH)

8. Base/Salt Buffering Pairs The salt will contain the cation of the base, and the anion of a strong acid (HCl, HNO3)

9. Try This! Calculate the pH of a solution containing 0.75 M lactic acid (Ka = 1.4 x 10-4) and 0.25 M sodium lactate. Lactic acid (HC3H5O3). pH = 3.38

10. Give this a try! 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 this solution. pH = 4.74 Now….try this! Calculate the change in pH that occurs when 0.01 M NaOH is added to 1.0 L of the buffered solution above. Compare this pH change with that which occurs when 0.01 M NaOH is added to 1.0 L of water.

11. A buffered solution contains 0.50 M acetic acid (HC2H3O2, Ka = 1.8 x 10-5) and 0.50 M sodium acetate (NaC2H3O2). pH = 4.74 Do Now: Please complete the following: Calculate the change in pH that occurs when 0.01 M NaOH is added to 1.0 L of the buffered solution above. Compare this pH change with that which occurs when 0.01 M NaOH is added to 1.0 L of water. Objective: Discuss buffers, titrations, Ksp and intermolecular forces of attraction. Homework: Study hard for Monday’s BIG DAY! 042916

12. Best approach: 2 steps: • Stoichiometry • Equilibrium WHAT???

13. Preparing/Choosing a Buffer I want a solution buffered @ pH = 4.30 chloroacetic acid (Ka = 1.35 x 10-3) propanoic acid (Ka = 1.3 x 10-5) benzoic acid (Ka = 6.4 x 10-5) hypochlorous acid (Ka = 3.5 x 10-8) What information can we get from pH? [H+] = 5.0 x 10-5M

14. [H+] = 5.0 x 10-5M [HA] Ka = [A-] [H+] [H+] = Ka = [A-] [HA] chloroacetic acid (Ka = 1.35 x 10-3) propanoic acid (Ka = 1.3 x 10-5) benzoic acid (Ka = 6.4 x 10-5) hypochlorous acid (Ka = 3.5 x 10-8)

15. TITRATION CURVES We can determine the amount of a certain substance by performing a technique called “titration.” Analyte

16. 3 Criteria Must Be Met for a Successful Titration: • Exact reaction between titrant and analyte MUST be known. • Equivalence point MUST be marked accurately. • Volume of titrant required to reach the equivalence point MUST be accurate. Note: When analyte is an acid or base…titrant is a strong base or acid, respectively.

17. Please solve: Klaudia has “snapped” and decides to consume the contents of a flask containing carbon tetrachloride (CCl4) and benzoic acid (HC7H5O2), a weak acid that has one acidic hydrogen atom per molecule. Prior to her “shocking” act, a sample of this solution weighing 0.3518 g was shaken with water, and the resulting aqueous solution required 10.59 mL of 0.1546 M NaOH for neutralization Calculate the mass percent of HC7H5O2 in the original sample that Klaudia consumed.

18. Sample is titrated with OH- ions HC7H5O2(aq) + OH-(aq) H2O(l) + C7H5O2-(aq) Hint #1: Determine the number of moles of OH- required to react with all the HC7H5O2. Answer: 1.637 x 10-3 mol OH- Hint #2: Determine grams of the acid! Answer: 0.1997 g of HC7H5O

19. The mass percent of HC7H5O2 in the original sample is: 0.1997 g X 100 = 56.77% 0.3518 g WAKE UP Umang!

20. Titration of an Unbuffered Solution A solution that is 0.10 M CH3COOH is titrated with 0.10 M NaOH

21. Titration of a Buffered Solution A solution that is 0.10 M CH3COOH and 0.10 M NaCH3COO is titrated with 0.10 M NaOH

22. Comparing Results Buffered Unbuffered

23. Comparing Results Unbuffered Buffered • In what ways are the graphs different? • In what ways are the graphs similar?

24. Strong Acid/Strong Base Titration Endpoint is at pH 7 A solution that is 0.10 M HCl is titrated with 0.10 M NaOH

25. Weak Acid/Strong Base Titration A solution that is 0.10 M CH3COOH is titrated with 0.10 M NaOH Endpoint is above pH 7

26. Strong Acid/Strong Base Titration A solution that is 0.10 M NaOH is titrated with 0.10 M HCl Endpoint is at pH 7 It is important to recognize that titration curves are not always increasing from left to right.

27. Strong Acid/Weak Base Titration A solution that is 0.10 M HCl is titrated with 0.10 M NH3 Endpoint is below pH 7

28. Selection of Indicators

29. Some Acid-Base Indicators

30. pH Indicators and theirranges

31. Solving Solubility Problems For the salt AgI at 25C, Ksp = 1.5 x 10-16 AgI(s)  Ag+(aq) + I-(aq) O O +x +x x x 1.5 x 10-16 = x2 x = solubility of AgI in mol/L = 1.2 x 10-8 M

32. Solving Solubility Problems For the salt PbCl2 at 25C, Ksp = 1.6 x 10-5 PbCl2(s)  Pb2+(aq) + 2Cl-(aq) O O +2x +x 2x x 1.6 x 10-5 = (x)(2x)2 = 4x3 x = solubility of PbCl2 in mol/L = 1.6 x 10-2 M

33. Solving Solubility with a Common Ion For the salt AgI at 25C, Ksp = 1.5 x 10-16 What is its solubility in 0.05 M NaI? AgI(s)  Ag+(aq) + I-(aq) 0.05 O 0.05+x +x 0.05+x x 1.5 x 10-16 = (x)(0.05+x)  (x)(0.05) x = solubility of AgI in mol/L = 3.0 x 10-15 M