1. Chemistry Chapter 20-21 Acids and Bases

2. (Self-Ionization of Water) H 2 O + H 2 O  H 3 O + + OH -  Two water molecules collide to form Hydronium and Hydroxide ions. 1 in 10,000,000 water molecules self-ionize. This is 10 -7 water molecules.

3. K w – Ionization Constant for Water In pure water at 25  C: [H + ] = [H 3 O + ] = 1 x 10 -7 mol/L [OH - ] = 1 x 10 -7 mol/L K w = Ion product constant ( for water at 25  C) Kw = [H + ] x [OH - ] Kw = (1 x 10 -7 )(1 x 10 -7 ) = 1 x 10 -14 1 10,000,000 (Dissociate) At 25  C, 1 in 10,000,000 water molecules will dissociate and form H + and OH -.

4. K w = [H + ] x [OH - ] Ion-product constant = Hydrogen-ion concentration x hydroxide-ion concentration

5. Ion Concentration in Water

6. Calculating pH “per Hydronium” pH = -log 10 [H 3 O + ] In other words: pH = -log[H + ] [H + ] = H + concentration in (mol/L) or (M) Example:[H + ] = 1 x 10 -7 M pH = -log(1 x 10 -7 ) pH = -(0-7) = 7.0 log(1) = 0log (10 -7 ) = -7

7. Acids Have a pH less than 7

8. Bases have a pH greater than 7

9. K w = [H + ] x [OH - ] Ion-product constant = Hydrogen-ion concentration x hydroxide-ion concentration Basic, neutral and acidic are all ways of describing pH concentrations in a solution.

10. Calculating pH Examples: [H + ] = 1 x 10 -1 pH = 1.0 [H + ] = 1 x 10 -4 pH = 4.0 [H + ] = 1 x 10 -7 pH = 7.0 [H + ] = 1 x 10 -9 pH = 9.0 [H + ] = 1 x 10 -13 pH = 13.0 [H + ] = 1 x 10 0 pH = 0 Acid Neutral Base Acid

11. Calculating pH Examples: [OH - ] = 1 x 10 -2 pOH = 2.0 Base pH = 12.0 pH + pOH = 14 14.0 [OH - ] = 1 x 10 -5 pOH = 5.0 Base pH = 9.0 14.0 [OH - ] = 1 x 10 -7 pOH = 7.0 Neutral pH = 7.0 14.0

12. Calculating pH [OH - ] = 1 x 10 -8 pOH = 8.0 Acid pH = 6.0 pH + pOH = 14 14.0 [OH - ] = 1 x 10 -10 pOH = 10.0 Acid pH = 4.0 14.0 [OH - ] = 1 x 10 -14 pOH = 14.0 Acid pH = 0 14.0 pH = -log [H + ]pOH = -log [OH - ]

13. Calculating pH [OH - ] = 1 x 10 0 pOH = 0 Base pH = 14.0 pH + pOH = 14 14.0 pH = -log [H + ]pOH = -log [OH - ]

14. Examples (Using Calculator): pH + pOH = 14 [H + ] = 2.5 x 10 -5 pH = -log(2.5 x 10 -5 ) Acid pH = -log [H + ]pOH = -log [OH - ] pH = 4.60 [H + ] = 3.0 x 10 -9 pH = -log(3.0 x 10 -9 ) Base pH = 8.52

15. Examples: pH + pOH = 14 [OH - ] = 6.0 x 10 -3 pOH = 2.22 Base pH = 11.78 14.0 pH = -log [H + ]pOH = -log [OH - ] pOH = -log(6.0 x 10 -3 ) [OH - ] = 9.0 x 10 -12 pOH = 11.05 Acid pH = 2.95 14.0 pOH = -log(9.0 x 10 -12 )

16. pH + pOH = 14 1 x 10 -5 M Acid 5 pH = -log [H + ]pOH = -log [OH - ] [H + ] = 10 -pH [OH - ] = 10 -pOH K w = [H + ] x [OH - ] 1 x 10 -14 = [H + ] x [OH - ] Equation Summary [H + ] = 10 -pH pH = ___[H + ] = __________ [OH - ] = __________ pOH = ___ [H + ] = 10 -5 pH + pOH = 14 pOH = 14 - pH pOH = 14 - 5 9 [OH - ] = 10 -pOH [OH - ] = 10 -9 1 x 10 -9 M pH = ___[H + ] = __________ [OH - ] = __________ pOH = ___ 6 [OH - ] = 10 -pOH [OH - ] = 10 -6 1 x 10 -6 M pH + pOH = 14 pH = 14 - pOH pH = 14 - 6 8 1 x 10 -8 M [H + ] = 10 -pH [H + ] = 10 -8 Base

17. [H + ] = __________ pH = ___ pH + pOH = 14 1.6 x 10 -3 M Acid 2.8 pH = -log [H + ]pOH = -log [OH - ] [H + ] = 10 -pH [OH - ] = 10 -pOH K w = [H + ] x [OH - ] 1 x 10 -14 = [H + ] x [OH - ] Equation Summary [H + ] = 10 -pH [OH - ] = __________ pOH = ___ [H + ] = 10 -2.8 pH + pOH = 14 pOH = 14 - pH pOH = 14 – 2.8 11.2 [OH - ] = 10 -pOH [OH - ] = 10 -11.2 6.3 x 10 -12 M [H + ] = 0.0016

18. Calculating pH, pOH pH = -log[H 3 O + ] or pH = -log[H + ] pOH = -log[OH - ] Relationship between pH and pOH pH + pOH = 14 Finding [H 3 O + ], [OH - ] from pH, pOH [H 3 O + ] = 10 -pH or [H + ] = 10 -pH [OH - ] = 10 -pOH SUMMARY