2. General properties ACIDS Taste sour Turn litmus React with active metals – Fe, Zn React with bases BASES Taste bitter Turn litmus Feel soapy or slippery (react with fats to make soap) React with acids blue to redred to blue

3. Acid Nomenclature Review Binary  Ternary An easy way to remember which goes with which… “In the cafeteria, you ATE something ICky”

4. Acid Nomenclature Flowchart

5. HBr (aq)HBr (aq) H 2 CO 3H 2 CO 3 H 2 SO 3H 2 SO 3  hydrobromic acid  carbonic acid  sulfurous acid Acid Nomenclature Review

6. Name ‘Em! HI (aq)HI (aq) HCl (aq)HCl (aq) H 2 SO 4H 2 SO 4 HNO 3HNO 3 H 3 PO 4H 3 PO 4

7. Definitions Acids – produce H + Bases - produce OH - Acids – donate H + Bases – accept H + Acids – accept e - pair Bases – donate e - pair Arrehenius Bronsted-Lowry Lewis only in water any solvent used in organic chemistry, wider range of substances

8. Examples Arrhenius Bronsted-Lowry Lewis HCl NaOH HClNH 3 :NH 3 BF 3 HCN The hydrogen ion in aqueous solution H + + H 2 O  H 3 O + (hydronium ion)

9. Neutralization In general: Acid + Base  Salt + Water All neutralization reactions are double replacement reactions. HCl + NaOH  NaCl + HOH HCl + Mg(OH) 2  H 2 SO 4 + NaHCO 3 

10. HA Let’s examine the behavior of an acid, HA, in aqueous solution. What happens to the HA molecules in solution?

11. HA H+H+ A-A- Strong Acid 100% dissociation of HA Would the solution be conductive?

12. HA H+H+ A-A- Weak Acid Partial dissociation of HA Would the solution be conductive?

13. HA H+H+ A-A- Weak Acid HA  H + + A - At any one time, only a fraction of the molecules are dissociated.

14. Strong and Weak Acids/Bases Strong acids/bases – 100% dissociation into ions HClNaOH HNO 3 KOH H 2 SO 4 Weak acids/bases – partial dissociation, both ions and molecules CH 3 COOHNH 3

15. pH 23456789101112 neutral @ 25 o C (H + ) = (OH - ) distilled water acidic (H + ) > (OH - ) basic or alkaline (H + ) < (OH - ) natural waters pH = 6.5 - 8.5 normal rain (CO 2 ) pH = 5.3 – 5.7 acid rain (NO x, SO x ) pH of 4.2 - 4.4 in Washington DC area 0-14 scale for the chemists fish populations drop off pH < 6 and to zero pH < 5

16. You are here! http://nadp.sws.uiuc.edu/isopleths pH of Rainwater across United States in 2001 Increasing acidity Why is the eastern US more acidic? air masses

17. What is acid rain? CO 2 (g) + H 2 O  H 2 CO 3  H + + HCO 3 - Dissolved carbon dioxide lowers the pH Atmospheric pollutants from combustion NO, NO 2 + H 2 O …  HNO 3 SO 2, SO 3 + H 2 O …  H 2 SO 4 both strong acids pH < 5.3

18. Effects of Acid Rain on Marble (calcium carbonate) George Washington: BEFORE George Washington: AFTER

19. pH 123456891011 The biological view in the human body gastric juice urine saliva cerebrospinal fluid blood pancreatic juice bile acidicbasic/alkaline 7 Tortora & Grabowski, Prin. of Anatomy & Physiology, 10 th ed., Wiley (2003)

20. Chapter 16.2 “Concentration of Solutions”

21. Concentration is... a measure of the amount of solute dissolved in a given quantity of solvent A concentrated solution has a large amount of solute A dilute solution has a small amount of solute –These are qualitative descriptions But, there are ways to express solution concentration quantitatively

22. Concentrated vs. Dilute

23. Molarity Molarity = moles of solute liters of solution Abbreviated with a capital M, such as 6.0 M This is the most widely used concentration unit used in chemistry.

24. - Page 481

25. Practice Problem A sample of 0.0341 mol iron(III) chloride, Fe Cl 3, was dissolved in water to give 25. 0 mL of solution. What is the molarity of the solution? Since molarity = Moles of FeCl 3 Liters of solution 0.0341 moles of FeCl 3 0.0250 Liters of solution Since molarity = = 1.36 M FeCl 3

26. Making solutions 1)Pour in a small amount of the solvent, maybe about one-half 2)Then add the pre-massed solute (and mix by swirling to dissolve it) 3)Carefully fill to final volume. –Fig. 16.8, page 481 Can also solve: moles = M x L Sample Problem 16.3, page 482

27. Dilution Adding water to a solution will reduce the number of moles of solute per unit volume but the overall number of moles remains the same! Think of taking an aspirin with a small glass of water vs. a large glass of water You still have one aspirin in your body, regardless of the amount of water you drank, but a larger amount of water makes it more diluted.

28. Dilution water (solvent)solute concentrated, M initial diluted, M final adding water lowers the solute concentration moles of solute remain constant V initial V final moles initial = moles final M final x V final = M initial x V initial

29. Dilution The number of moles of solute in solution doesn’t change if you add more solvent! The # moles before = the # moles after Formula for dilution: M 1 x V 1 = M 2 x V 2 M 1 and V 1 are the starting concentration and volume; M 2 and V 2 are the final concentration and volume. Stock solutions are pre-made solutions of known Molarity. Sample 16.4, p.484

30. Percent solutions can be expressed by a) volume or b) mass Percent means parts per 100, so Percent by volume: = Volume of solute x 100% Volume of solution indicated %(v/v) Sample Problem 16.5, page 485

31. Percent solutions Percent by mass: = Mass of solute(g) x 100% Volume of solution (mL) Indicated %(m/v) More commonly used 4.8 g of NaCl are dissolved in 82 mL of solution. What is the percent of the solution? How many grams of salt are there in 52 mL of a 6.3 % solution?

32. Titration Calculation HCl + NaOH  NaCl + HOH at equivalence point: mole HCl = mole NaOH moles = M x V L M acid x V initial acid = M base x V buret A way to analyze solutions! indicator PGCC CHM 101 Sinex If the mole to mole ratio = 1:1

33. Titration Calculation HCl + NaOH  NaCl + HOH at equivalence point: mole HCl = mole NaOH moles = M x V L A way to analyze solutions! indicator PGCC CHM 101 Sinex

34. M acid x V initial acid = M base x V buret If the mole to mole ratio = 1:1 If the mole to mole ratio is not 1:1 you need to add a mole factor to the equation above based on the coefficients in the balanced equation. Example: 2 HCl + Sr(OH) 2  SrCl 2 + 2H 2 O Mole to mole ratio = 2 acid : 1 base

35. The number of moles of acid is twice the number of moles of base, therefore it will take twice as much base to neutralize the acid. moles acid = 2(moles base ) The '2' in this formula is called the mole factor. Substituting into equation (1) we have: M A V A = 2 M B V B

36. Sample Problem: If 25.0 mL of a standard 0.05 M HCl solution is required to neutralize 20.0 mL of a solution of Sr(OH) 2, what is the concentration of the base? (Use the equation in the example above.) M A V A = 2 M B V B (0.05 M)(25.0 mL) = 2(X)(20.0 mL) X = 0.0313

37. 35.62 mL of NaOH is neutralized with 25.2 mL of 0.0998 M HCl by titration to an equivalence point. What is the concentration of the NaOH? LAB PROBLEM #1: Standardize a solution of NaOH — i.e., accurately determine its concentration.

38. PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Add water to the 3.0 M solution to lower its concentration to 0.50 M Dilute the solution!