What is the pH of a 0.33 M solution of methylamine (CH₃NH₂,K_b = 4.4 × 10^-4)at 25^oC? (K_w = 1.01 × 10^-14) A) 5.56 B) 0.48 C) 12.07 D) 13.52 E) 1.93

Question

Quizplus · Accepted Answer

Methylamine (CH3NH2) is a weak base, and when dissolved in water it undergoes the following reaction:

CH3NH2 + H2O ↔ CH3NH3+ + OH-

The equilibrium constant for this reaction is given by the expression:

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

The value of Kb for methylamine is related to the value of Ka for its conjugate acid, ammonium ion (NH4+), by the expression:

Kb × Ka = Kw

where Kw is the ion product constant for water (1.0×10^-14 at 25°C).

We can use this relationship to find the value of Kb for methylamine:

Kb × 6.31×10^-10 = 1.0×10^-14

Kb = 1.58×10^-5

Now, we can use the expression for Kb to find the concentration of hydroxide ion (OH-) in a 0.33 M solution of methylamine:

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

[OH-] = Kb × [CH3NH2] / [CH3NH3+]

[OH-] = 1.58×10^-5 × 0.33 / 0.33

[OH-] = 1.58×10^-5

Finally, we can use the relationship between pH and [OH-] to calculate the pH of the solution:

pH = 14 - pOH = 14 - (-log[OH-]) = 12.07

Therefore, the answer is C.

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