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Lecture12_S08

Lecture12_S08 - Feb 15 2008 Lecture 12 Chapter 16 Acids and...

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Feb. 15, 2008 Lecture 12, Chapter 16: Acids and Bases H H H H H H H H O O O O + + 2 H 2 O (l) H 3 O + (aq) + OH - (aq) K w = [H 3 O + ] [OH - ] = [H + ] [OH - ] Equilibrium constant for ionization of water Use interchangeably with H 3 O + Water only ionizes to a very small extent: K w = 1x10 -14 Given K w , find [H+] and [OH-]: 2 H 2 O (l) H 3 O + (aq) + OH - (aq) Initially: 0 M 0M Change: +x +x Equilibrium: (x) (x) K w = [H 3 O + ][OH - ] = 1.0x10 -14 K w = (x)(x) 1x10 -14 = x 2 (1x10 -14 ) 1/2 = x 1x10 -7 = x = [H + ] = [OH - ] Often we want to know how much H + is in solution. For instance, we can measure pH of pure water to find K w . pH = -log [H+] or 10 -pH = [H+] for pure water [H+] = 1x10 -7 as above so, pH = -log (1x10-7) = - -7 pH = + 7 Reference : pH 10 -pH = [H+] -1 10+1 10 M 0 10 0 1 M 1 10 - 1 0.1 M 2 10 - 2 0.01 M 3 10 - 3 0.001 M 4 10 - 4 0.0001 M 5 10 - 5 0.00001 M 6 10 - 6 0.000001 M 7 10 - 7 0.0000001 M 8 10 - 8 0.00000001 M 9 10 - 9 0.000000001 M 10 10 - 10 0.0000000001 M 11 10 - 11 0.00000000001M 12 10 - 12 0.000000000001 M 13 10 - 13 0.0000000000001 M 14 10 - 14 0.00000000000001 M = Pretty dilute Small pH = large [H+] large pH = small [H+]

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Calculate the pH of a solution that is [OH-] = 0.010 M. K w = [H + ][OH - ] = 1.0x10 -14 [H + ] = K w /[OH - ] = (1.0x10 -14 )/(0.010) = 1x10 -12 pH = -log [H + ] = -log (1x10 -12 ) pH = 12 Alternatively, since [H + ][OH - ] = 1x10 -14 -log ( [H + ][OH - ] ) = -log (1x10 -14 ) -log [H + ] + log[OH - ] = -log (1x10
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