Question

Find the [OH−] of a 0.33 M ethylamine (C₂H₅NH₂) solution. (the value of kb for ethylamine (C₂H₅NH₂) is 5.6×10−4.)

a. 0.014 M
b. 0.022 M
c. 0.028 M
d. 0.035 M

Answer 1

Using the provided value of K₋ for ethylamine, we find the hydroxide ion concentration, [OH−], of a 0.33 M ethylamine solution to be 0.014 M, which corresponds to option a.

Step-by-step explanation:

To find the concentration of hydroxide ions, [OH−], in a 0.33 M ethylamine (C2H5NH2) solution, we use the base dissociation constant for ethylamine (Kb). The base dissociation reaction is as follows:

C2H5NH2 + H2O → C2H5NH3+ + OH−

The equilibrium expression for this reaction is:

Kb = [C2H5NH3+][OH−]/[C2H5NH2]

Assuming that the degree of dissociation is small and the concentration of ethylamine does not change significantly, we can simplify the expression to:

Kb ≈ x2/0.33

where x is equal to the concentration of hydroxide ions. Thus:

x2 = Kb × 0.33 = (5.6 × 10−5) × 0.33

x ≈ √((5.6 × 10−5) × 0.33)

After calculating x, we find that x ≈ 0.014 M, which is the concentration of hydroxide ions in the solution.

Therefore, the correct answer is option a. 0.014 M for the hydroxide ion concentration in a 0.33 M solution of ethylamine.