calculate the equilibrium concentration for the nonionized bases and all ions in a solution that is 0.25M

The equilibrium concentration of [CH₃NH₂] = 0.23965 M.

The equilibrium concentration of CH₃NH₃⁺ and  OH⁻ =  0.01035 M respectively.

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.