Answer the following questions about an aqueous 00353 M solu

Answer the following questions about an aqueous 0.0353 M solution of CsH5NH2. The Kb is 3.9 x 10-10 What is [OH]: 49D What is the pH: What is [H3o: 9 What is the percent ionization: 4

Solution

Balanced equation with ICE table

C6H5NH2 + H2O <----> C6H5NH3+ + OH-

I 0.0353

C - x +x +x

E 0.0353-x x x

Equilibrium constant expression of the reaction

Kb = [C6H5NH3+] [OH-]/[C6H5NH2]

3.9 x 10^-10 = x^2 / 0.0353-x

Since 0.0353 >> x

3.9 x 10^-10 = x^2 / 0.0353

x^2 = 1.3767 x 10^11

x = 3.71 x 10^-6

[OH-] = 3.71 x 10^-6 M

pOH = - log [OH-]

= - log (3.71 x 10^-6) = 5.43

pH = 14 - 5.43 = 8.57

[H3O+] = 10^-pH = 10^-8.57 = 2.69 x 10^-9 M

% ionization = 3.71 x 10^-6 *100/0.0353

= 0.0105%