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? NaF(s) Na + (aq) + F - (aq) H 2 O(l)

4. Step 1: Identify the major species HF, Na +, F - and H 2 O HF(aq) H + (aq) + F - (aq) HF(aq) H + (aq) + F - (aq) + F - (aq) + Na + A solution contains HF (ka = 7.2 x 10 -4 ) and its salt (NaF).

5. Previously the equilibrium concentration of H + in a 1.0M HF solution was determined to be 2.7 x 10 -2 M, 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 -4 M 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 Weak Acid Formula of the acid Example of a salt of the weak acid Hydrofluoric HF KF – Potassium fluoride Formic HCOOH KHCOO – Potassium formate Benzoic C 6 H 5 COOH NaC 6 H 5 COO – Sodium benzoate Acetic CH 3 COOH NaH 3 COO – Sodium acetate Carbonic H 2 CO 3 NaHCO 3 - Sodium bicarbonate Propanoic HC 3 H 5 O 2 NaC 3 H 5 O 2 - Sodium propanoate Hydrocyanic HCN KCN - potassium cyanide 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, HNO 3 ) Base Formula of the base Example of a salt of the weak acid Ammonia NH 3 NH 4 Cl - ammonium chloride Methylamine CH 3 NH 2 CH 3 NH 3 Cl – methylammonium chloride Ethylamine C 2 H 5 NH 2 C 2 H 5 NH 3 NO 3 - ethylammonium nitrate Aniline C 6 H 5 NH 2 C 6 H 5 NH 3 Cl – aniline hydrochloride Pyridine C 5 H 5 N C 5 H 5 NHCl – pyridine hydrochloride

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 (HC 3 H 5 O 3 ). pH = 3.38

10. Give this a try! A buffered solution contains 0.50 M acetic acid (HC 2 H 3 O 2, Ka = 1.8 x 10 -5 ) and 0.50 M sodium acetate (NaC 2 H 3 O 2 ). 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. 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 A buffered solution contains 0.50 M acetic acid (HC 2 H 3 O 2, Ka = 1.8 x 10 -5 ) and 0.50 M sodium acetate (NaC 2 H 3 O 2 ). pH = 4.74

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

13. Preparing/Choosing a Buffer 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 ) I want a solution buffered @ pH = 4.30 What information can we get from pH? [H + ] = 5.0 x 10 -5 M

14. Ka = [A - ] [H + ] [HA] [H+] = Ka [HA] [A - ] = 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 (CCl 4 ) and benzoic acid (HC 7 H 5 O 2 ), 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 HC 7 H 5 O 2 in the original sample that Klaudia consumed.

18. Sample is titrated with OH - ions HC 7 H 5 O 2 (aq) + OH - (aq) H 2 O(l) + C 7 H 5 O 2 - (aq) Hint #1: Determine the number of moles of OH - required to react with all the HC 7 H 5 O 2. Answer: 1.637 x 10 -3 mol OH - Hint #2: Determine grams of the acid! Answer: 0.1997 g of HC 7 H 5 O

19. The mass percent of HC 7 H 5 O 2 in the original sample is: 0.1997 g 0.3518 g X 100 =56.77% WAKE UP Umang!

20. Titration of an Unbuffered Solution A solution that is 0.10 M CH 3 COOH is titrated with 0.10 M NaOH

21. Titration of a Buffered Solution A solution that is 0.10 M CH 3 COOH and 0.10 M NaCH 3 COO is titrated with 0.10 M NaOH

22. Comparing Results Buffered Unbuffered

23. Unbuffered Buffered  In what ways are the graphs different?  In what ways are the graphs similar?

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

25. Weak Acid/Strong Base Titration A solution that is 0.10 M CH 3 COOH 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 NH 3 Endpoint is below pH 7

28. Selection of Indicators

29. Some Acid-Base Indicators Indicator pH Range in which Color Change Occurs Color Change as pH Increases Crystal violet Thymol blue Orange IV Methyl orange Bromcresol green Methyl red Chlorophenol red Bromthymol blue Phenol red Neutral red Thymol blue Phenolphthalein Thymolphthalein Alizarin yellow Indigo carmine 0.0 - 1.6 1.2 - 2.8 1.4 - 2.8 3.2 - 4.4 3.8 - 5.4 4.8 - 6.2 5.2 - 6.8 6.0 - 7.6 6.6 - 8.0 6.8 - 8.0 8.0 - 9.6 8.2 - 10.0 9.4 - 10.6 10.1 - 12.0 11.4 - 13.0 yellow to blue red to yellow red to yellow red to yellow yellow to blue red to yellow yellow to red yellow to blue yellow to red red to amber yellow to blue colourless to pink colourless to blue yellow to blue blue to yellow

30. pH Indicators and their ranges

31. Solving Solubility Problems For the salt AgI at 25  C, K sp = 1.5 x 10 -16 AgI(s)  Ag + (aq) + I - (aq) I C E O O +x x x 1.5 x 10 -16 = x 2 x = solubility of AgI in mol/L = 1.2 x 10 -8 M

32. Solving Solubility Problems For the salt PbCl 2 at 25  C, K sp = 1.6 x 10 -5 PbCl 2 (s)  Pb 2+ (aq) + 2Cl - (aq) I C E O O +x +2x x 2x 1.6 x 10 -5 = (x)(2x) 2 = 4x 3 x = solubility of PbCl 2 in mol/L = 1.6 x 10 -2 M

33. Solving Solubility with a Common Ion For the salt AgI at 25  C, K sp = 1.5 x 10 -16 What is its solubility in 0.05 M NaI? AgI(s)  Ag + (aq) + I - (aq) I C E 0.05 O +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