pH of methylamine solution.

Question

CH₃NH₂ (aq) + H₂O (l) ßà CH₃NH₃⁺ (aq) + OH^- (aq)          K_b = 4.00 x 10^-4

a.) Determine the [OH^-] for 0.25 M methylamine, CH₃NH₂.

b.) Determine the pH of the methylamine solution.

Answer 1

Methylamine in aqueous (water) solution behaves as a weak base, removing a proton from water and becoming CH₃NH₃⁺. It behaves this way because of the lone pair of electrons on the nitrogen atom. This lone pair will form a new bond with one of H₂O's hydrogens.

Methylamine is a weak base, meaning that not 100% of CH3NH2 converts to CH3NH3+; the base and its conjugate acid both exist in solution. You can find their relative amounts using the equation

[CH3NH3+][OH-]

Kb == --------------

[CH3NH2]

To find [OH-], rearrange the above equation to

[OH-] == Kb * [CH3NH2] / [CH3NH3+]

From the ICE table (not shown here),

[OH-] == x

[CH3NH2] == 0.25 - x

[CH3NH3+] == x

Kb = 4.00 x 10^-4

Plugging into our above equation, we get

x == 4.00x10^4 * (0.25-x)/x

From here, it's just algebra to get A. :)

For B, start with the definition of pH and equations

pH == -log[H+]

pOH == -log[OH-]

pH + pOH == 14

Combining these gives you,

pH == 14 - pOH == 14 - (-log[OH-])

and you get [OH-] from part A.

Does this help? Need a hand with the algebra.