the ionic product of water
DESCRIPTION
The Ionic Product of Water. K w. 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 - ] - PowerPoint PPT PresentationTRANSCRIPT
The Ionic Product of Water
Kw
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-]
Ionic Product of water, Kw
H2O(l) + H2O (l) H3O+
(aq) + OH-(aq)
H2O(l) H+
(aq) + OH-(aq)
Kc = [H+][OH-] [H2O]
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
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
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)
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+]
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-]
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
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
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
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