At 25 degree C, the initial concentration of 0.20M solution of ammonia, NH3, in

Question

At 25 degree C, the initial concentration of 0.20M solution of ammonia, NH3, in water, H2O, is left to ionize (see the equilibrium reaction below). The equilibrium constant, Kb, of the solution is 1.8 x 10^-5 at 25 degrees C.

NH3 (aq) + H2O (l) ---------> <--------- NH4^+ (aq) + OH^ - (aq)

a. Determine the concentrations of all species in solution at equilibrium using the expression of Kb and its value as well as the low ionization approximation of a weak base (summarize your findings in a table)

b. Calculate the concentration in hydronium ions, [H3O+], knowing the value of the concetration in hydroxide ions, [OH-] from (a.) Then calculate the pH of the soultion of ammonia 0.20 M at 25 degrees C. Note that Kw= [H3O+] x [OH^ -] = 1.00 x 10^-14 at 25 degrees C.

Explanation / Answer

a) NH3 + H2O -------------> NH4+ + OH-

0.20M-X X X

we know that

Kb = [NH4+] [OH-] / [NH3]

1.8 x 10^-5 = X^2 / ( 0.20 -X)

the low ionization approximation of a weak base [NH3] = 0.20 -X =0.20

therefore

1.8 x 10^-5 = X^2 / 0.20

x^2 = 3.6 x 10^-6

X = 1.897 x 10^-3 M

therefore

[nh3] = 0.20 -0.001897 = 0.198103M

[NH4+] = 1.897 x 10^-3 M

[OH-] = 1.897 x 10^-3 M

b)

from Kw = [H3O+][OH-]

we have

[H3O+]= 5.27 X 10^-12 M

therefore

pH = -log[H3O+]= -log ( 5.27 X 10^-12 ) = 11.278

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