Question

The degree to which a weak base dissociates is given by the base-ionization constant, Kb. For the generic weak base, B B(aq)+H2O(l)⇌BH+(aq)+OH−(aq) this constant is given by Kb=[BH+][OH−][B] Strong bases will have a higher Kb value. Similarly, strong bases will have a higher percent ionization value. Percent ionization=[OH−] equilibrium[B] initial×100% Strong bases, for which Kb is very large, ionize completely (100%). For weak bases, the percent ionization changes with concentration. The more dilute the solution, the greater the percent ionization. Ammonia, NH3, is a weak base with a Kb value of 1.8×10−5. Part A What is the pH of a 0.300 M ammonia solution? Express your answer numerically to two decimal places. View Available Hint(s) pH = nothing Part B What is the percent ionization of ammonia at this concentration? Express your answer with the appropriate units. View Available Hint(s) % ionization = nothing nothing

Answer 1

a) pH = 11.37

b) %ionization = 0.774%

Step-by-step explanation:

a) the equilibrium reaction is as follows:

NH3 + H2O = NH4+ + OH-

equilibrium (moles) 0.3 x x x

The equilibrium equation is equal to:

Kb = ([NH4+]*[OH-])/[NH3]

Since the value of x is very small compared to 0.3, we can simplify the expression to:

Kb = x^2/0.3

Clearing x, we have:

x = (Kb * 0.3)^1/2 = (0.3 * 1.8x10^-5)^1/2 = 0.00232 M

Due [OH-] = x = 0.00232 M, we can calculate the pOH

pOH = -log[OH-] = -log(0.00232) = 2.63

the pH is equal to:

pH = 14 - pOH = 14 - 2.63 = 11.37

b) Since NH3 is a weak base, we can use the Ostwald equation to calculate the degree dissociation

α = (Kb/C)^1/2, where C is the molar concentration

α = (1.8x10^-5/0.3)^1/2 = 0.00774

%α = 0.00774 * 100 = 0.774%