lec16_whc

# lec16_whc - Last Time Weak acids and bases Ka, pKa and Kb,...

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Last Time • Weak acids and bases •K a , pK a and K b , pK b • Relative strengths • Factors that determine acidity • pH calculations

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Weak Acids/Bases • Total charge in the solution must be zero!! • For acids : [H 3 O + ] = [A ] + [OH ] K w = [H 3 O + ][OH ] = [H 3 O + ] ( [H 3 O + ] – [HA] ) • For bases : [OH ] = [H 3 O + ] + [M + ] K w = [H 3 O + ][OH ] = [H 3 O + ] ( [H 3 O + ] + [B ] )
Dilute Solutions – Weak Acids • Problem: Acid with K a and [H 3 O + ] ~ 10 –7 • Wrong answer: use “x” and find pH directly • Correct answer – solve for [H 3 O + ] using K w = [H 3 O + ][OH ] K a = [H 3 O + ] = [OH ] + [A ] A = [HA] + [A ] [H 3 O + ][A ] [HA]

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Last Time: pH Strong acid or base: Weak acid or base: Very dilute (e.g. 10 -8 M) strong acid or base: pH = -log 10 [H 3 O + ] = -log 10 [HCl] initial HCl(aq) + H 2 O(l) Æ H 3 O + (aq) + Cl (aq) HC 2 H 3 O 2 (aq) + H 2 O(l) H 3 O + (aq) + C 2 H 3 O 2 (aq) K a = [CH 3 CO 2 H] [CH 3 CO 2 ][H 3 O + ] Æ equilibrium/ICE table! 2H 2 O(l) H 3 O + (aq) + OH (aq) K w =[H 3 O + ][OH ] Self ionization of water very important
An aqueous solution of acetic acid has pH of 2.87. From pK a =4.74 calculate the percent ionization! Example 16D – Percent Ionization HC 2 H 3 O 2 + H 2 OH 3 O + + C 2 H 3 O 2 ¯ Percent ionization = [H 3 O + ] [HC 2 H 3 O 2 ] originally × 100% Before we start: pH= -log[H 3 O + ] = 2.87 Æ [H 3 O + ] = 10 -2.87 = 1.35 × 10 -3 mol/L pK a = -log K a = 4.74 Æ K a = 10 -4.74 = 1.8 × 10 -5 Similar example text p 681

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HC 2 H 3 O 2 + H 2 OH 3 O + + C 2 H 3 O 2 ¯ But we know at equilibrium [H 3 O + ] = 1.35x10 -3 M = x Solution 16D – Percent Ionization (#1) -x +x +x Change: y0 0 Init: y-x x x Equil: Amount ionized = x = [C 2 H 3 O 2 - ] = [H 3 O + ] = 1.35x10 -3 M y- 1.35x10 -3 1.35x10 -3 1.35x10 -3 Æ So, at equil: This number came from the pH measurement.
32 3 2 232 [] [ C H O ] [C H O ] a HO K H +− == -3 2 5 -3 -3 2 -3 5 (1.35x10 ) 1.8 10 1.35x10 (1.35x10 ) [ C H O ] +1.35x10 135 1.8 10 init y Hy →= = × =0 . 1 0 m o l / L Solution 16D – Percent Ionization (#2) To find original amount of acid (y) need K a : % ionization = 0.00135 0.10135 × 100% = 1.33% Ionized Original × 100% = Amount ionized = x = [C 2 H 3 O 2 - ] = [H 3 O + ] = 1.35x10 -3 M

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Sulfuric Acid - A diprotic acid. H 2 SO 4 + H 2 O H 3 O + + HSO 4 - HSO 4 - + H 2 O H 3 O + + SO 4 2- K a1 = very large K a2 = 1.1x10 -2 • No need for a double equilibrium calculation because can simplify when K a1 is very large and K a2 small!
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## This note was uploaded on 06/13/2010 for the course CHEM 995940767 taught by Professor Topadakis during the Spring '10 term at UC Davis.

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lec16_whc - Last Time Weak acids and bases Ka, pKa and Kb,...

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