Ka and Kb

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NH 4 + (aq) + H 2 O (l) NH 3 (aq) + H 3 O + (aq) NH 3 (aq) + H 3 O + (aq) NH 4 + (aq) + H 2 O (l) We could write this reaction two different ways (A and B). These give us the same info, but their equilibrium constants might have different names and might be calculated differently. If we write an equilibrium constant for the top reaction, where NH 4 + donates H + to water, we are describing the dissocation of a weak acid (the equilibrium of an acid with its conjugate base), and we would call the equilibrium constant K a (A) (B) If we write an equilibrium constant for the bottom reaction, where NH 3 accepts H + from water, we are describing the equilibrim of a base with its conjugate acid, and we would call the equilibrium constant K b K a = [NH 3 ][H 3 O + ] [NH 4 + ] K b = [NH 3 ][H 3 O + ] [NH 4 + ] = 1.8 × 10 5 = K a 1 = 1.8 × 10 5 = K b 1 So, if ALEKS ever tells you the value of K b , remember that K b = 1/K a

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Ka and Kb

Transcript of Ka and Kb

Page 1: Ka and Kb

NH4+ (aq) + H2O (l) NH3 (aq) + H3O+ (aq)

NH3 (aq) + H3O+ (aq) NH4+ (aq) + H2O (l)

We could write this reaction two different ways (A and B). These give us the same info, but their equilibrium constants might have different names and might be calculated differently.

If we write an equilibrium constant for the top reaction, where NH4+ donates H+ to water, we are describing the dissocation of a weak acid (the equilibrium of an acid with its conjugate base), and we would call the equilibrium constant Ka

(A)

(B)

If we write an equilibrium constant for the bottom reaction, where NH3 accepts H+ from water, we are describing the equilibrim of a base with its conjugate acid, and we would call the equilibrium constant Kb

Ka = [NH3][H3O+][NH4+]

Kb =[NH3][H3O+]

[NH4+]= 1.8 × 10−5 =

Ka

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= 1.8 × 105 =Kb

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So, if ALEKS ever tells you the value of Kb, remember that Kb = 1/Ka