Answer:
The equilibrium concentration of [CH₃NH₂] = 0.23965 M.
The equilibrium concentration of CH₃NH₃⁺ and OH⁻ = 0.01035 M respectively.
Step-by-step explanation:
The first step is to write out the dissociation reaction. Therefore, the equation showing the dissociation is given as below;
CH₃NH₂ + H₂0 <--------------------------------------------------------> CH₃NH₃⁺ + OH⁻.
Kindly note that ''<----------->'' arrow shows that the reaction is an equilibrium reaction.
Therefore, at the start of the reaction [that is time, t =0], we have that the concentration of CH₃NH₂ = 0.25M, thus, the concentration of CH₃NH₃⁺ and OH⁻ is zero respectively at this time, t =0.
At equilibrium, the concentration of CH₃NH₂ = 0.25M - x, thus, the concentration of CH₃NH₃⁺ and OH⁻ is x respectively.
Therefore, kb = 4.47 × 10-4 = [CH₃NH₃⁺ ][OH⁻]/[CH₃NH₂]. Hence, slotting in the values into this equilibrium equation showing the relationship between kb and concentration of the species involved, we have that;
kb = 4.47 × 10⁻⁴ = x² /0.25 - x.
x² + 4.47 × 10⁻⁴x - 1.1175 × 10⁻⁴ = 0.
Solving this quadratic equation gives us the value of x as 0.01035 M.
Thus, the concentration of [CH₃NH₂] = 0.25 M - 0.01035 M = 0.23965 M
The equilibrium concentration of [CH₃NH₂] = 0.23965 M.
The equilibrium concentration of CH₃NH₃⁺ and OH⁻ = 0.01035 M respectively.