Question

17) In a 0.450 M HONH2 solution, [OH] = 5.28 x 10-6 M.

Answer 1

Answer:

Given

[HONH2] = 0.45M

[OH-] = 5.26 x 10-6 M

                                           HONH2 + H2O -------------> HONH3+ + OH-

Initial                                     0.45           55                         0             0

at equilibrium                        0.45-x        55-x                      x             x

Given

[OH-] = x =  5.26 x 10-6 M

Therefore [HONH3+] = x =  5.26 x 10-6 M

pOH = -log[OH-] = -log(5.26 x 10-6) = 5.279

=> pH = 14- pOH = 8.72

From hendersen-hasselbach equaiton

8.72 = pKa + log(0.45/5.26 x 10-6)

=> pKa = 3.788

=> pKb = 14-3.788 = 10.21

percent of ionization = 5.26 x 10-6 * 100/0.45 = 1.17 x 10-3 %

concentration of NaOH required to make the same pH= [OH-] = 5.26 x 10-6 M

Percent of ionization of NaOH = [OH-]*100/NaOH = 5.26 x 10-6 *100/5.26 x 10-6 = 100%