The Ionic Product of Water KwKw. Ionic Product of water, K w Just because a solution contains [H + ] it doesn’t necessarily mean it’s acidic. All aqueous.

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Presentation transcript:

The Ionic Product of Water KwKw

Ionic Product of water, K w Just because a solution contains [H + ] it doesn’t necessarily mean it’s acidic. All aqueous solutions will contain H + and OH - ions For an acid solution [H + ] > [OH - ] For an alkaline solution [H + ] < [OH - ] For an neutral solution [H + ] = [OH - ]

Ionic Product of water, K w H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH - (aq) H 2 O (l) H + (aq) + OH - (aq) K c = [H + ][OH - ] [H 2 O]

Ionic Product of water, K w H 2 O (l) H + (aq) + OH - (aq) K c = [H + ][OH - ] [H 2 O] Since [H 2 O] is effectively constant and in large excess K w = [H + ][OH - ] mol 2 dm -6

Ionic Product of water, K w At 298K the value of K w is 1 x mol 2 dm -6 K w = [H + ][OH - ] mol 2 dm -6 In pure water [H + ] = [OH - ] So K w = [H + ] 2 Hence [H + ] = √K w At 298K [H + ] = √(1 x ) = 1 x So at 289K the pH of pure water is 7

Effect of temperature on K w Dissociation of water is endothermic Increasing temp increases [H + ] Increasing temp increases K w At higher temps pH of water will decrease H 2 O (l) H + (aq) + OH - (aq)

At 321K the value of K w is 4 x mol 2 dm -6 calculate the pH of water at this temp K w = [H + ][OH - ] mol 2 dm -6 In pure water [H + ] = [OH - ] So K w = [H + ] 2 Hence [H + ] = √K w At 321K [H + ] = √(4 x ) = 2 x So at 321K the pH of pure water is 6.70 pH = -log 10 [H + ]

The pH of Strong Bases K w gives us a method of calculating the pH of strong bases K w = [H + ][OH - ] mol 2 dm -6 So [H + ] = K w [OH - ]

Calculate the pH of 0.1M KOH at 298K K w = [H + ][OH - ] = 1 x mol 2 dm -6 Since KOH is a strong base [OH - ] = 0.1 [H + ] = K w [OH - ] pH = -log 10 [H + ] [H + ] = 1 x = 1 x pH = 13

At 338K the pH of pure water is 6.5. Calculate K w at this temp. Hence calculate the pH of 0.01M NaOH at this temp K w = [H + ][OH - ] [H + ] = x pH = -log 10 [H + ] K w = [H + ] 2 for pure water [H + ] = [OH - ] for pure water

At 338K the pH of pure water is 6.5. Calculate K w at this temp. Hence calculate the pH of 0.01M NaOH at this temp K w = (3.162 x ) 2 K w = [H + ] 2 for pure water K w = 1x K w = [H + ][OH - ] [H + ] = K w [OH - ] = 1 x = 1 x

At 338K the pH of pure water is 6.5. Calculate K w at this temp. Hence calculate the pH of 0.01M NaOH at this temp pH = -log 10 [H + ] pH = 11 [H + ] = 1 x