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