CH302_021711 - If you know [H+] you know [OH-] Kw =...

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Principles of Chemistry II © Vanden Bout If you know [H + ] you know [OH - ] K w = [H + ][OH - ] log(K w ) = log([H + ][OH - ]) log(K w ) = log[H + ]+ log[OH - ] log(10 -14 ) = log[H + ]+ log[OH - ] -14 = -pH - pOH 14 = pH + pOH
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Principles of Chemistry II © Vanden Bout Weak Acid HA(aq) H + (aq) + A - (aq) HA H + A - I C E C -x C-x O +x +x K a = [H + ][A - ] [HA] O +x +x (x)(x) C-x = really 10 -7 x ~ K a C assuming x << C
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Principles of Chemistry II © Vanden Bout This is a great simple result [H + ] K a C a C a is the concentration of the acid K a is the equilibrium constant for the acid This assumes the concentration is large and that K a is small
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Principles of Chemistry II © Vanden Bout What is the pH of a 1M solution of weak acid with a K a = 10 -6 ? A. 1 B. 3 C . 7 D . 8 E. 9 [H + ] =sqrt(1x10 -6 ) = 10 -3
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Principles of Chemistry II © Vanden Bout B(aq) +H 2 O(l) BH + (aq) + OH - (aq) Weak Base K b = [BH + ][OH - ] [B] identical result as before (same assumptions) [OH - ] = K b C b
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Principles of Chemistry II © Vanden Bout Weak Acids HA(aq) H + (aq) + A - (aq) with the proton acid without the proton base HA weak acid A - weak base
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Principles of Chemistry II © Vanden Bout A - (aq) +H 2 O(l) HA(aq) + OH - (aq) Weak Base K b = [HA][OH - ] [A - ] identical result as before (same assumptions) [OH - ] = K b C
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Principles of Chemistry II © Vanden Bout Same with the base B(aq) +H 2 O(l) BH + (aq) + OH - (aq) with the proton acid without the proton base BH + (aq) H + (aq) + B(aq)
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This note was uploaded on 03/27/2011 for the course CHEM 302 taught by Professor Mccord during the Spring '10 term at University of Texas at Austin.

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CH302_021711 - If you know [H+] you know [OH-] Kw =...

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