Calculate the pH of a 0.800 M CH₃NH₃Cl solution.K_b for methylamine,CH₃NH₂,is 3.7 × 10^-4. A) 1.76 B) 5.33 C) 8.67 D) 12.23

Question

Quizplus · Accepted Answer

First we need to write out the chemical equation for the reaction of methylamine with water:

CH3NH2 + H2O ⇌ CH3NH3+ + OH-

The equilibrium constant expression, K, can be written as:

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

We are given the value of Kb, the base dissociation constant for methylamine, which is related to K by the expression:

Kw/Kb = Ka = 1.0 × 10^-14

So we can calculate Ka:

Ka = 1.0 × 10^-14/3.7 × 10^-4 = 2.7 × 10^-11

Now we can use the Henderson-Hasselbalch equation to calculate the pH of the solution:

pH = pKa + log([A-]/[HA])

We know that [HA] = 0.800 M (the initial concentration of CH3NH2), and we can calculate [A-] as follows:

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

[OH-] = Kb[CH3NH2]/[CH3NH3+] = (3.7 × 10^-4)(0.800)/(0.800 + x)

where x is the concentration of CH3NH3+ at equilibrium. Since the initial concentration of CH3NH3+ is zero, we can approximate [CH3NH3+] as x.

Now we can substitute these values into the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

pH = 10.57 + log((3.7 × 10^-4)(0.800)/(0.800 + x))

At equilibrium, the amount of CH3NH2 that has reacted with water to form CH3NH3+ and OH- is negligible compared to the initial concentration, so we can assume that [CH3NH2] ≈ 0.800 M. This allows us to simplify the equation to:

pH = 10.57 + log(3.7 × 10^-4/0.800)

pH = 5.33

Calculate the PH of a 0