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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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Presentation on theme: "Ionization constant, K a, for a weak acid HA  H + + A - HA  H + + A - K a = [H + ][A - ] K a = [H + ][A - ] [HA] [HA]"— Presentation transcript:

1 Ionization constant, K a, for a weak acid HA  H + + A - HA  H + + A - K a = [H + ][A - ] K a = [H + ][A - ] [HA] [HA]

2 Weak AcidEquationKaKa acetic acidHC 2 H 3 O 2  H + + C 2 H 3 O x10 -5 benzoic acidC 6 H 5 CO 2 H  H + + C 6 H 5 CO x10 -5 chlorous acidHClO 2  H + + ClO x10 -2 formic acidHCHO 2  H + + CHO x10 -4 hydrocyanic acidHCN  H + + CN 6.2 x 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 x lactic acidCH 3 CH(OH)CO 2 H  H + + CH 3 CH(OH)CO x10 -4 nitrous acidHNO 2  H + + NO x10 -4 phenolHOC 6 H 5  H + + OC 6 H x10 -10

3 What is the [H + ] in 0.100M formic acid? K a for formic acid is 1.77 x HCOOH  H + + COOH - Since this is a weak acid, [HCOOH] ͌ 0.100M K a = [H + ][COOH - ] = 1.77 x [HCOOH] [HCOOH] Let x = [H + ] = [COOH - ] x 2 = 1.77 x x 2 = 1.77 x X = 4.21 x M

4 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 M, [HCOOH] = 0.100M 4.21 x = 4.21% 0.100

5 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.

6 Base Ionization Constants SubstanceFormulaKbKb AmmoniaNH x AnilineC 6 H 5 NH x Dimethylamine(CH 3 ) 2 NH5.1 x EthylamineC 2 H 5 NH x HydrazineN2H4N2H4 1.7 x Urea NH 2 CONH x


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