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Published byAustin Beckles Modified over 10 years ago
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Ionization constant, K a, for a weak acid HA H + + A - HA H + + A - K a = [H + ][A - ] K a = [H + ][A - ] [HA] [HA]
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Weak AcidEquationKaKa acetic acidHC 2 H 3 O 2 H + + C 2 H 3 O 2 1.8 x10 -5 benzoic acidC 6 H 5 CO 2 H H + + C 6 H 5 CO 2 6.4 x10 -5 chlorous acidHClO 2 H + + ClO 2 1.2 x10 -2 formic acidHCHO 2 H + + CHO 2 1.8 x10 -4 hydrocyanic acidHCN H + + CN 6.2 x10 -10 hydrofluoric acidHF H + + F 7.2 x10 -4 hypobromous acidHOBr H + + OBr 2 x10 -9 hypochlorous acidHOCl H + + OCl 3.5 x10 -8 hypoiodous acidHOI H + + OI 2 x10 -11 lactic acidCH 3 CH(OH)CO 2 H H + + CH 3 CH(OH)CO 2 1.38 x10 -4 nitrous acidHNO 2 H + + NO 2 4.0 x10 -4 phenolHOC 6 H 5 H + + OC 6 H 5 1.6 x10 -10
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What is the [H + ] in 0.100M formic acid? K a for formic acid is 1.77 x 10 -4 HCOOH H + + COOH - Since this is a weak acid, [HCOOH] ͌ 0.100M K a = [H + ][COOH - ] = 1.77 x 10 -4 [HCOOH] [HCOOH] Let x = [H + ] = [COOH - ] x 2 = 1.77 x 10 -4 x 2 = 1.77 x 10 -4 0.100 0.100 X = 4.21 x 10 -3 M
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Percent ionization [amount ionized] [amount ionized] [original acid] [original acid] What is the percent ionization of [H + ] from the previous problem? [H + ] [HCOOH]0.100M [H + ] = 4.21 x 10 -3 M, [HCOOH] = 0.100M 4.21 x 10 -3 = 4.21% 0.100
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KbKbKbKb base-dissociation constant base-dissociation constant NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH - (aq) K b = [NH 4 + ][OH - ] K b = [NH 4 + ][OH - ] [NH 3 ] [NH 3 ] if K b is large, the products of the dissociation reaction are favored if K b is large, the products of the dissociation reaction are favored if K b is small, undissociated base is favored. if K b is small, undissociated base is favored.
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Base Ionization Constants SubstanceFormulaKbKb AmmoniaNH 3 1.8 x 10 -5 AnilineC 6 H 5 NH 2 4.2 x 10 -10 Dimethylamine(CH 3 ) 2 NH5.1 x 10 -4 EthylamineC 2 H 5 NH 2 4.7 x 10 -4 HydrazineN2H4N2H4 1.7 x 10 -6 Urea NH 2 CONH 2 1.5 x 10 -14
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