Calculate the percent ionization in the following solution. 0.10 M CH3NH2

Answer 1

To calculate the percent ionization, we need to know the initial concentration of the weak base (CH3NH2) and the concentration of the conjugate acid formed when it ionizes (CH3NH3+).

In this case, CH3NH2 can ionize to form CH3NH3+ and OH-. Therefore, the equilibrium equation is:

CH3NH2 + H2O ⇌ CH3NH3+ + OH-

Since the initial concentration of CH3NH2 is 0.10 M, we can assume that the concentration of CH3NH3+ and OH- at equilibrium is x M.

Using the equilibrium expression and the given concentrations, we can write:

Ka = ([CH3NH3+][OH-])/[CH3NH2]

Since OH- is formed in a 1:1 ratio with CH3NH2, we can simplify the equation as:

Ka = x*x/(0.10-x)
x^2 = Ka(0.10-x)

Assuming x << 0.10, we can approximate the expression as:
x^2 = Ka(0.10)

Solving for x, we find:
x = √(Ka(0.10))

Using the known value of Ka for CH3NH2 (5.6 x 10^-4), we can substitute it into the equation to calculate x:

x = √((5.6 x 10^-4)(0.10))

x ≈ 1.67 x 10^-3 M

The percent ionization can be calculated as the ratio of x to the initial concentration of CH3NH2, multiplied by 100:

Percent ionization = (x/[CH3NH2]) * 100
Percent ionization = (1.67 x 10^-3 / 0.10) * 100
Percent ionization ≈ 1.67%