Multiple Choice
What is the expression for the equilibrium constant (Kb) for the dissociation of a weak base in water?
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17. Acid and Base Equilibrium
pH of Weak Bases
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What is the pH of a 0.225 M (CH3)2NH2Br solution, given that the Kb of (CH3)2NH is 5.4 × 10⁻⁴?
A
2.78
B
12.78
C
7.00
D
11.22
Verified step by step guidance1
Identify that (CH3)2NH2Br is a salt formed from the weak base (CH3)2NH and a strong acid HBr. In solution, (CH3)2NH2+ acts as a weak acid.
Write the hydrolysis reaction for (CH3)2NH2+: (CH3)2NH2+ + H2O ⇌ (CH3)2NH + H3O+. This shows the weak acid donating a proton to water.
Use the relationship between Ka and Kb for conjugate acid-base pairs: Ka × Kb = Kw, where Kw is the ion-product constant of water (1.0 × 10⁻¹⁴ at 25°C). Calculate Ka for (CH3)2NH2+ using Ka = Kw / Kb.
Set up the expression for the acid dissociation constant, Ka: Ka = [H3O+][(CH3)2NH] / [(CH3)2NH2+]. Assume initial concentration of (CH3)2NH2+ is 0.225 M and that the change in concentration is x, leading to [H3O+] = x.
Solve for x, which represents [H3O+], using the approximation method if x is small compared to the initial concentration. Finally, calculate pH using the formula pH = -log[H3O+].
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