1. Acid-Base Chemistry Tentative Test Date Monday, April 19 th Todays Objectives Distinguish among the three definitions of acids and bases. Calculate [H + ], [OH - ], and pH.

2. Definitions of Acids and Bases ACIDSBASES Arrhenius Proton Producer in Aqueous Solutions Hydroxide Producer in Aqueous Solutions Brønsted- Lowry Proton DonorProton Acceptor LewisElectron AcceptorElectron Donor More General Proton = H +

3. Identifying Acids and Bases For the following neutralization reactions, identify the acid and the base. 1) HF + H 2 O F - + H 3 O + 2) NH 3 + H 2 O NH 4 + + OH – Since water can act as both an acid and a base, it is called amphoteric. Acids = Lose H + Bases = Gain H + AcidBase Acid Base

4. Auto-Ionization of Water H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH - (aq) hydronium ion hydroxide ion water molecule

5. Auto-Ionization of Water H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH - (aq) K w = [H 3 O + ][OH - ] K w = Water Dissociation Constant At 25 o C, K w equals 1.0 x 10 -14. [H 3 O + ] = Concentration of H 3 O + [OH - ] = Concentration of OH -

6. Example Problem At 25 o C, the concentration of hydroxide ions in an aqueous solution is 4.5 x 10 -6 M. Determine the concentration of hydronium ions or [H 3 O + ]. K w = [H 3 O + ][OH - ] [H 3 O + ] == 2.2x10 -9 M [OH - ]

7. Simplified Equation H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH - (aq) H 2 O(l) H + (aq) + OH - (aq) K w = 1.0 x10 -14 = [H + ][OH - ] H + and H 3 O + are interchangeable in the equation. Simplified

8. Now You Try A solutions hydrogen ion concentration is 9.01 x 10 -10 M. What is the concentration of hydroxide ions in this solution?

9. pH pH = - log [H + ] More Acidic More Basic

10. Simple pH Calculations Ex. #1:What is the pH of a solution with a hydronium ion concentration of 1.0 x10 -4 M? pH = -log[H 3 O + ] = - log (1.0x10 -4 M) = 4 Ex. #2: Calculate the pH of a solution if [H + ] equals 1.0x10 -9 M. pH = -log[H + ] = - log (1.0x10 -9 M) = 9 Which is more acidic: Example #1 or 2? Example #1 is more acidic since it has a lower pH.

11. pH with Log Tables Ex. #3:Calculate pH of 4.58 x 10 -3 M H +. pH = - log [H + ] pH = - log (4.58 x 10 -3 M) pH = - [log (4.58) + log (10 -3 )] pH = -[0.661 + -3] = - [ -2.339] pH = 2.339

12. pH with Log Tables Ex. #3:Calculate pH of a solution if the hydronium ion concentration is 1.79 x 10 -8 M. pH = - log [H 3 O + ] pH = - log (1.79 x 10 -8 M) pH = - [log (1.79) + log (10 -8 )] pH = -[0.253 + -8] = - [ -7.647] = 7.647

13. Homework Lab notebooks due tomorrow!! Worksheet on Naming and Writing Formulas for Acids and Bases Book Problems pg. 582 # 6 & 7 pg. 586 # 8 & 9

14. Acid-Base Chemistry Tentative Test Date Monday, April 19 th Todays Objective Calculate [H + ], [OH - ], pH and pOH.

15. Review from Yesterday Calculate the pH of a solution whose hydronium ion concentration is 2.71 x 10 -7 M. pH = - log (2.71 x 10 -7 M) pH = -[log (2.71) + log (10 -7 )] pH = - [0.433 + -7] pH = 6.567

16. Combining the Problems Using your calculator, calculate the pH of a solution whose hydroxide ion concentration is 1.3 x 10 -3 M. Step 1: Find [H + ]. Step 2: Calculate pH. pH = -log (7.69 x 10 -12 M) pH = - [log (7.69) + log (10 -12 )] pH = -[0.886 + -12] pH = 11.114

17. Relationships between pH, [H 3 O + ], and [OH - ] As pH increases… The solution becomes more (acidic or basic). The [H 3 O + ] (increases or decreases). The [OH - ] (increases or decreases).

18. [H + ] from pH pH = -log[H + ] -pH = log[H + ] 10 You will either need to be able to derive this formula or memorize this formula.

19. [H + ] from pH Ex #1: Calculate the hydronium ion concentration for a solution with a pH of 6.719. [H + ] = 10 –pH = 10 –6.719 Using the Log Tables, 10 -6.72 = 10 -7 x 10 0.281 1.91 x 10 -7 M From Log Tables

20. [H + ] from pH Ex #2: Calculate the hydronium ion concentration for a solution with a pH of 10.5. [H + ] = 10 –pH = 10 –10.5 Using Your Calculator, press the following buttons for most calculators. 2 nd Log – 10.5 3.16 x 10 -11 M

21. pOH pOH = -log [OH - ] Using your log tables, calculate the pOH of a solution that contains 0.0807 M OH -. 0.0807 M = 8.07 x 10 -2 M pOH = - log (8.07x10 -2 M) = -[log (8.07) + log(10 -2 )] pOH = -[0.907 + -2] = 1.093

22. Relating pH and pOH K w = [H + ][OH - ] 1.0x10 -14 = [H + ][OH - ] -log(1.0x10 -14 ) = -log([H + ][OH - ]) -log(1.0x10 -14 ) = -log[H + ] + -log[OH - ] 14 = pH + pOH

23. A Summary of Calculations Important Formulas K w = [H + ][OH - ]pH + pOH = 14 pH = -log[H + ][H + ] = 10 -pH pOH = -log[OH - ] [OH - ] = 10 -pOH Types of SolutionpH[H + ] vs [OH - ] AcidicLess than 7[H + ] > [OH - ] Neutral7.000…[H + ] = [OH - ] BasicGreater than 7[H + ] < [OH - ]