17) In a 0.450 M HONH2 solution, [OH] = 5.28 x 10-6 M. HONH2(aq) + H20(1) HONH3(aq) + OH(aq)
a. Find [HONH3).
b. Find Kb.
c. Find pOH.
d. Find H.
e. Find the percent ionization of HONH, in a 0.450 MHONH2 solution.
f. What concentration of NaOH would be required to make a solution with
the same pH that was calculated in part d.?
g. Find the percent ionization of NaOH in the above solution.

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%

each  column  of the  table  is called a group. elements in a group share similar properties. groups are  read  from  top  to  bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o.  2. 3. 4. 5. 6. 7. 8. 9.  2. beryllium. 9.012. 4.  sodium. 22.990.  11. 3.  magnesium. 24.305.  12.  potassium. 39.098.  19. 4.  calcium. 40.078.  20. i guess  i  

what does the table look like