Final answer:
The pH of a 0.25 M solution of a weak base with a Kb of 1.3 x 10^-6 is approximately 12.08.
Step-by-step explanation:
To find the pH of a 0.25 M solution of a weak base, we first need to find the concentration of hydroxide ions ([OH-]). We can use the Kb (base ionization constant) to calculate [OH-].
First, let's set up the equation using the given Kb:
(CH3)2NH + H2O ⇌ (CH3)2NH2+ + OH-
The Kb expression for this reaction is:
Kb = [CH3)2NH2+][OH-] / [CH3)2NH]
Using the values given, we can substitute and rearrange the equation to solve for [OH-].
[OH-] = √(Kb × [CH3)2NH])
Plugging in the values:
[OH-] = √((5.9 × 10^-4)(0.25))
[OH-] = √(1.475 × 10^-4)
[OH-] ≈ 1.215 × 10^-2 M
Now, we can find the pOH:
pOH = -log([OH-])
pOH ≈ -log(1.215 × 10^-2)
pOH ≈ 1.92
Finally, we can find the pH using the equation:
pH = 14 - pOH
pH ≈ 14 - 1.92
pH ≈ 12.08