2. Vocabulary Hydronium = H 3 O + Proton = H+ or H 3 O + Hydroxide = OH- Amphoteric = a compound that can act as an acid or base. Alkaline = When a base completely disassociates in water to yield OH- ions.

3. H 2 O HCl H + + Cl - Acid is a proton donor (H+), which creates hydronium ions. H 2 O NaOH Na + + OH - A base is a proton acceptor

4. Acid Bronsted-Lowry – H+ (proton) donor (doesn’t have to happen in an aqueous solution) –Ex., HCl + NH 3  NH 4 + + Cl - Arrhenius – produce H+ in aqueous solution [H+] > [OH-]

5. Base Bronsted-Lowry – H+ (proton) acceptor Arrhenius – produce hydroxide (OH-) in an aqueous solution [H+] < [OH-]

6. http://www.chem.utk.edu/yang/chap8.ppt

7. Strong Acids Chloric Acid HClO 3 Hydrobromic Acid HBr Hydrochloric Acid HCl Hydriodic Acid HI Nitric Acid HNO 3 Perchloric AcidHClO 4 Sulfuric AcidH 2 SO 4 Hydronium Ion H 3 O +

8. Strong Bases Carbonate IonCO 2/3 - Phosphate IonPO 3/4- Hydroxide IonOH - Amide Ion NH 2- Hydride Ion H -

9. http://www.chem.utk.edu/yang/chap8.ppt

10. Conjugate acid – base pair –Conjugate acid-base pair: –Conjugate acid-base pair: any pair of molecules or ions that can be interconverted by transfer of a proton http://www.chem.utk.edu/yang/chap8.ppt

11. In the reaction of a Bronsted-Lowry acid, the molecule or ion that remains is the conjugate base.

12. Neutralization Reactions Occurs in an aqueous solution between a strong acid and a strong base. When neutralization occurs, a salt and water are formed. Ex., HCl + NaOH  H 2 O + NaCl

13. Lewis Acid- an atom, ion or molecule that accepts an electron pair to form a covalent bond. Applies to any ion that can accept an electron pair. H + +2:NH 3  [H-NH 3 ] + or [NH 4 ] + A Lewis base is an atom or molecule that donates an electron pair to form a covalent bond.

14. Net Ionic Equations A net ionic equation includes only those compounds and ions that undergo a chemical change. –Write an overall ionic equation –All soluble compounds are shown as dissociated ions in solution, ppt are shown as solids. –Ex., ppt of cadmium sulfate Cd 2+ (aq) + 2NO 3 - + 2NH 4 + (aq) SO 4 2-  CdSO 4 (s) + 2NO 3 - (aq) +2NH 4 + (aq)

15. Net Ionic Equation Cd 2+ (aq) + SO 4 2- (aq)  CdSO 4 (s)

16. HCl + H 2 O H 3 O + + Cl - (acid) (base) (conjugate acid) (conjugate base)

17. Water can be both a acid and a base H 2 O + H 2 O H 3 O + + OH - The animation here shows the formation of H 3 + O ions and OH - ions in an aqueous system

18. K w = ion-product constant K w = in any aqueous solution @ 25 degrees Celcius the product of [H 3 O+][OH-] must always equal 1.0 X 10 -14 [ H + ] = [OH - ] = 1.0 * 10 -7 K w = [H + ] [OH - ] = (1.0 *10 -7 )(1 *10 -7 ) K w = 1.0 * 10 -14

19. Calculating pH and pOH Next let's look at values for pH and pOH. pOH is simply the power of hydroxide ion concentration and is figured the same way as pH but using the concentration of hydroxide ion instead.) Source: http://dl.clackamas.cc.or.us/ch105-05/calculat1.htmhttp://dl.clackamas.cc.or.us/ch105-05/calculat1.htm

20. Let's start by working with the concentrations that are 1.0 x 10 (raised to some power). These pH and pOH values can be figured very simply. When [H 3 O + ] is 10 -7 M, the pH is 7. Also the [OH - ] is 10 -7 M and the pOH is 7. Note that the pH and pOH add up to 14. pH[H 3 O + ][OH - ]pOH 71.0 x 10 -7 M 7

21. Now look at the acidic solutions. When the [H 3 O + ] is 10 -6 M, the pH is 6. Also, the [OH - ] is 10 -8 M and the pOH is 8. Again, the pH and the pOH add up to 14. When the [H 3 O + ] is 10 -5 M, pH is 5, [OH - ] is 10 -9 M, and pOH is 9. pH + pOH = 14. 6565 1.0 x 10 -6 M 1.0 x 10 -5 M 1.0 x 10 -8 M 1.0 x 10 -9 M 8989 pH[H 3 O + ][OH - ]pOH

22. Next, the basic solutions. When [OH - ] is 10 -6 M, the pOH is 6. Since [H 3 O + ] 10 -8 M, the pH is 8. When [OH - ] is 10 -5 M, and [H 3 O + ] is 10 -9 M, the pOH is 5 and pH is 9. Again in both cases the sum of pH and pOH is 14. 8989 1.0 x 10 -8 M 1.0 x 10 -9 M 1.0 x 10 -6 M 1.0 x 10 -5 M 6565 pH[H 3 O + ][OH - ]pOH

23. However, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. For example, we skipped over the value of 2.0 x 10 -7. This is more complicated. However, if you use a calculator that will handle logarithms, it is a very simple calculation. First you enter the hydronium ion concentration. You can use decimal format or scientific notation. Next push the log button. Then change the sign by pushing the +/- button. In this case we get 6.70 for the pH. The other values can be obtained in the same way. 6.70 7.30 2.0 x 10 -7 M 1.0 x 10 -7 M 1.0 x 10 -7 M 2.0 x 10 -7 M 7.30 6.70 pH[H 3 O + ][OH - ]pOH

24. PH Scale PH = -log [H + ] Step 1 : Enter [Molarity of H + ] Step 2 : Press “log” and hit +/- key Reverse for graphing calculator

25. More vocabulary Diprotic Acid H 2 CO 3 = H + + HCO 3 - HCO 3 - = H + + CO 3 -2 Triprotic acid –H 3 PO 4 +H 2 O-->H 3 O+H 2 PO 4 –It's second ionization: H 2 PO 4 +H 2 O-->H 3 O+HPO 4 –It's third ionization: HPO 4 +H 2 O-->H 3 O+PO 4 Oxyacid – compound of H, O, and a third element usually a nonmetal. Ex HNO 3 Organic acid – has a carbon