Chemistry 1011

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Relation between K a and K b• Ammonia in aqueous solution is basic:NH 3(aq) + H 2 O (aq) NH+ 4 (aq) + OH − (aq)K b = [NH 4+ ] x[OH − ] = 1.8 x 10 −5[NH 3 ]• The ammonium ion is its conjugate acid• We can write an equation for NH 4+ acting as anacid as:NH+ + 4 (aq) H 2 O (aq) NH 3(aq) + H 3 O + (aq)K a = [NH 3 ] x[H 3 O + ] = 5.6 x 10 −10[NH 4+ ]Chemistry 1011 Slot 5 8
Relation between K a and K b• Add the two equations:H 2 O (aq) + H 2 O (aq) H 3 O + (aq) + OH − (aq)• The equilibrium constant of the resulting equationis the product of the two equilibrium constants:K w = [H 3 O + ] x [OH − ] = K a x K b= (5.6 x 10 −10 ) x (1.8 x 10 −5 )= 1.0 x 10 −14• This relationship can be used to determine K b for aweak acid from a measurement of K a for itsconjugate acid, and vice versaChemistry 1011 Slot 5 9
Page 1 and 2: Chemistry 1011TOPICAcids and BasesT
Page 3 and 4: Hydrolysis of Salts• All salts ar
Page 5 and 6: Hydrolysis of Salts• NH 4 Cl is t
Page 7: Anions as Weak BasesBase K b pK bPO
Page 11 and 12: Relative Strengths of Acids and Bas
Page 13 and 14: Predicting Acid-Base Propertiesof S

Chemistry 1011

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