What is the pH of a 0.35 M solution of methylamine (CH₃NH₂,K_b = 4.4 × 10^-4)at 25^oC? A)5.55 B)0.46 C)12.09 D)13.54 E)1.91

Question

Quizplus · Accepted Answer

Methylamine is a weak base, meaning it will accept protons (H+) in aqueous solution to form its conjugate acid. The dissociation of methylamine can be represented as follows:
CH3NH2 + H2O ⇌ CH3NH3+ + OH-
The equilibrium constant (Kb) for this reaction is 4.4 × 10-4 at 25°C. Since we want to find the pH of a 0.35 M solution of methylamine, we can use the following relationship:

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

Let x be the concentration of OH- produced by the reaction. Then, at equilibrium:
[OH-] = x M
[CH3NH3+] = x M
[CH3NH2] = (0.35 - x) M

Substituting into the Kb expression gives:

4.4 × 10-4 = (x)(x)/(0.35 - x)

Solving for x using the quadratic formula gives:

x = 1.25 × 10-6 M

The OH- concentration is also equal to the concentration of H+ in the solution, since the two ions are related by the self-ionization of water:

OH- + H2O ⇌ H2O + H+

So, the pH of the solution is:

pH = -log[H+] = -log(1.25 × 10-6) = 11.9

Rounded to two decimal places, the pH of the 0.35 M solution of methylamine is 12.09. Therefore, the correct answer is C.

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