Weak Bases and Kb

Methylamine is a weak base. The hydrolysis equation is:

CH3NH2 (aq) + H2O (l) ⇄ CH3NH3+ (aq) + OH (aq)

The base, CH3NH2, abstracts a proton from water. The conjugate acid is CH3NH3+. This is a basic solution because hydroxide ion is produced. The base hydrolysis constant for this reaction, Kb is:

\(\displaystyle K_b\;=\;\frac{[\mathrm{CH_3NH_3^+}][\mathrm{OH^-}]}{[\mathrm{CH_3NH_2}]}\)

 
The base hydrolyis constant can either be looked up or determined from Ka. The value of Ka for CH3NH3+ is 2.3 x 10-11.

Here, we will relate Ka, Kb, and Kw. Consider the following equations:

HA (aq) + H2O (l) ⇄ A (aq) + H3O+ (aq)    \(\displaystyle K_a\;=\;\frac{[A^-][H_3O^+]}{[HA]}\)
 
A (aq) + H2O (l) ⇄ HA (aq) + OH (aq)    \(\displaystyle K_b\;=\;\frac{[HA][OH^-]}{[A^-]}\)

 
Sum the equations:

HA (aq) + H2O (l) ⇄ A (aq) + H3O+ (aq)    \(\displaystyle K_a\;=\;\frac{[A^-][H_3O^+]}{[HA]}\)
 
A (aq) + H2O (l) ⇄ HA (aq) + OH (aq)    \(\displaystyle K_b\;=\;\frac{[HA][OH^-]}{[A^-]}\)
___________________________________________________________________________
2 H2O (l) ⇄ H3O+ (aq) + OH (aq)    Ka x Kb = Kw

 
We see that Ka x Kb = Kw = 1.0 x 10-14

If Ka is known, we can find Kb. If Kb is known, we can find Ka.

Ka x Kb = 1.0 x 10-14

For example acetic acid has Ka equal to 1.8 x 10-5. We can calculate Kb

\(\displaystyle K_b\;=\;\frac{1.0\times 10^{-14}}{1.8\times 10^{-5}}\;=\;5.6\times 10^{-10}\)

 

Weak bases will undergo hydrolysis with water to produce the conjugate acid and hydroxide ion. For example, fluoride ion will react with water to produce hydrofluoric acid and hydroxide ion.

F (aq) + H2O (l) ⇄ HF (aq) + OH (aq)

\(\displaystyle K_b\;=\;\frac{[HF][OH^-]}{[F^-]}\;=\;1.5\times 10^{-11}\)

 

Let’s calculate the pH of a solution that is 0.25 M CH3NH2. Kb = 4.4 x 10-4.

First write the hydrolysis reaction.

CH3NH2 (aq) + H2O (l) ⇄ CH3NH3+ (aq) + OH (aq)   \( K_b=\frac{[\mathrm{CH_3NH_3^+}][\mathrm{OH^-}]}{[\mathrm{CH_3NH_2}]}=4.4\times 10^{-4}\)
 
Set up an ICE table.

ICE table for 0.25 M methylamine

Assume x is neglible and 0.25 M – x ≅ 0.25 M

\(\displaystyle \frac{x^2}{0.25}\;=\;1.5\times 10^{-11}\)
 
Solve for x.

\(\displaystyle x\;=\;\sqrt{0.25\times\;(1.5\times 10^{-11})}\;=\;1.9\times\;10^{-6}\;M\)
 
Check assumption.

\(\displaystyle \frac{1.9\times 10^{-6}\;M}{0.25\;M}\times\;100\;=\;0.00076\%\)
 
The assumption is valid.

The [OH] = 1.9 x 10-6 M. Use pOH to calculate the pH.

pH = 14.00 – (-log(1.9 x 10-6) = 8.28

The solution is basic as we expected.

Watch the following video.

 

Exercises

Exercise 1. Calculate Kb for NH3 if Ka for NH4+ is equal to 5.6 x 10-10.

Check Solution to Exercise 1

Exercise 2. Write a balanced net ionic equation and the equilibrium constant expression for the weak base ethlyamine, C2H5NH2, in water. Ka (C2H5NH3+) = 8.6 x 10-4.

Check Solution to Exercise 2

Exercise 3. Pyridine is a base with Kb = 2.0 x 10-9. What is the pH of a 0.75 M pyridine solution?

Check Solution to Exercise 3

Exercise 4. Ammonia, NH3 has a Kb = 1.8 x 10-5. What is the pH of an aqueous solution that is 0.35 M NH3?

Check Solution to Exercise 4

Exercise 5. A 0.75 M aqueous solution of methylamine, CH3NH2, has a pH of 12.26. Calculate Kb and the equilibrium concentrations of [CH3NH2], [CH3NH3+], and [OH].

Check Solution to Exercise 5

Back to Acids and Bases: Aqueous Equilibria
Back to Study Guides for General Chemistry 2

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