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The Ionic Product of Water K w

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Page 1: 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

The Ionic Product of Water

Kw

Page 2: 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

Ionic Product of water, Kw

• 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-]

Page 3: 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

Ionic Product of water, Kw

H2O(l) + H2O (l) H3O+

(aq) + OH-(aq)

H2O(l) H+

(aq) + OH-(aq)

Kc = [H+][OH-] [H2O]

Page 4: 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

Ionic Product of water, Kw

H2O(l) H+

(aq) + OH-(aq)

Kc = [H+][OH-] [H2O]

Since [H2O] is effectively constant and in large excess

Kw = [H+][OH-] mol2dm-6

Page 5: 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

Ionic Product of water, Kw

At 298K the value of Kw is 1 x 10-14 mol2dm-6

Kw = [H+][OH-] mol2dm-6

In pure water [H+] = [OH-]

So Kw = [H+]2 Hence [H+] = √Kw

At 298K [H+] = √(1 x 10-14) = 1 x 10-7

So at 289K the pH of pure water is 7

Page 6: 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

Effect of temperature on Kw

Dissociation of water is endothermic

Increasing temp increases [H+]

Increasing temp increases Kw

At higher temps pH of water will decrease

H2O(l) H+

(aq) + OH-(aq)

Page 7: 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

At 321K the value of Kw is 4 x 10-14 mol2dm-6

calculate the pH of water at this tempKw = [H+][OH-] mol2dm-6

In pure water [H+] = [OH-]

So Kw = [H+]2 Hence [H+] = √Kw

At 321K [H+] = √(4 x 10-14) = 2 x 10-7

So at 321K the pH of pure water is 6.70

pH = -log10[H+]

Page 8: 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

The pH of Strong Bases

Kw gives us a method of calculating the pH of strong bases

Kw = [H+][OH-] mol2dm-6

So [H+] = Kw [OH-]

Page 9: 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

Calculate the pH of 0.1M KOH at 298K

Kw = [H+][OH-] = 1 x 10-14 mol2dm-6

Since KOH is a strong base [OH-] = 0.1

[H+] = Kw

[OH-]

pH = -log10[H+]

[H+] = 1 x 10-14

0.1= 1 x 10-13

pH = 13

Page 10: 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

At 338K the pH of pure water is 6.5. Calculate Kw at this temp. Hence calculate the pH of 0.01M NaOH at this temp

Kw = [H+][OH-]

[H+] = 3.162 x 10-7

pH = -log10[H+]

Kw = [H+]2 for pure water

[H+] = [OH-] for pure water

Page 11: 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

At 338K the pH of pure water is 6.5. Calculate Kw at this temp. Hence calculate the pH of 0.01M NaOH at this temp

Kw = (3.162 x 10-7)2

Kw = [H+]2 for pure water

Kw = 1x 10-13

Kw = [H+][OH-]

[H+] = Kw

[OH-]= 1 x 10 -13

0.01= 1 x 10 -11

Page 12: 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

At 338K the pH of pure water is 6.5. Calculate Kw at this temp. Hence calculate the pH of 0.01M NaOH at this temp

pH = -log10 [H+]

pH = 11

[H+] = 1 x 10 -11