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Chemistry Chapter 20-21 Acids and Bases. (Self-Ionization of Water) H 2 O + H 2 O  H 3 O + + OH -  Two water molecules collide to form Hydronium and.

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Presentation on theme: "Chemistry Chapter 20-21 Acids and Bases. (Self-Ionization of Water) H 2 O + H 2 O  H 3 O + + OH -  Two water molecules collide to form Hydronium and."— Presentation transcript:

1 Chemistry Chapter 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 water molecules.

3 K w – Ionization Constant for Water In pure water at 25  C: [H + ] = [H 3 O + ] = 1 x mol/L [OH - ] = 1 x mol/L K w = Ion product constant ( for water at 25  C) Kw = [H + ] x [OH - ] Kw = (1 x )(1 x ) = 1 x ,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 M pH = -log(1 x ) 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 pH = 1.0 [H + ] = 1 x pH = 4.0 [H + ] = 1 x pH = 7.0 [H + ] = 1 x pH = 9.0 [H + ] = 1 x pH = 13.0 [H + ] = 1 x 10 0 pH = 0 Acid Neutral Base Acid

11 Calculating pH Examples: [OH - ] = 1 x pOH = 2.0 Base pH = 12.0 pH + pOH = [OH - ] = 1 x pOH = 5.0 Base pH = [OH - ] = 1 x pOH = 7.0 Neutral pH =

12 Calculating pH [OH - ] = 1 x pOH = 8.0 Acid pH = 6.0 pH + pOH = [OH - ] = 1 x pOH = 10.0 Acid pH = [OH - ] = 1 x pOH = 14.0 Acid pH = pH = -log [H + ]pOH = -log [OH - ]

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

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

15 Examples: pH + pOH = 14 [OH - ] = 6.0 x pOH = 2.22 Base pH = pH = -log [H + ]pOH = -log [OH - ] pOH = -log(6.0 x ) [OH - ] = 9.0 x pOH = Acid pH = pOH = -log(9.0 x )

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

17 [H + ] = __________ pH = ___ pH + pOH = x M Acid 2.8 pH = -log [H + ]pOH = -log [OH - ] [H + ] = 10 -pH [OH - ] = 10 -pOH K w = [H + ] x [OH - ] 1 x = [H + ] x [OH - ] Equation Summary [H + ] = 10 -pH [OH - ] = __________ pOH = ___ [H + ] = pH + pOH = 14 pOH = 14 - pH pOH = 14 – [OH - ] = 10 -pOH [OH - ] = x M [H + ] =

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


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