1.The pH of a 0.10 mol/L aqueous solution of Fe(NO3)3 is not 7.00. The equation that best accounts for this observation is:

Question

a.
Fe3+(aq) + 3H2O(l)<-->Fe(OH)3(aq) + 3H+(aq)
b.
NO3-(aq) + H2O(l)<-----> HNO3(aq) + OH-(aq)
c.
Fe(H2O)63+(aq) + H2O(l)<---->Fe(H2O)5(OH)2+(aq) + H3O+(aq)
d.
Fe(H2O)63+(aq) + H2O(l)<--->Fe(OH2)5(H3O)4+(aq) + OH-(aq)
e.
HNO3(aq) + H2O(l)<---->H3O+(aq) + NO3-(aq)

My correct answer is supposed to be c but I don't know why. An explanation would be greatly appreciated

2.The percentage ionization in a 0.05 mol/L NH3(aq) solution whose pH is 11.00 is
a. 11%.
b. 0.02%.
c. 3%.
d. 5%.
e. 2%.
The correct answer is supposed to be c
I did 10^(-11)=[H+]
(1 x 10^(-11)/0.05) x 100%
Then I get 2 x 10^(-8)

Answer 1

Although we commonly write a solution of iron(III) as Fe^3+ actually it is [Fe(H2O)6]^3+. Then itg reacts with a H2O molecule from the solute. You know about the auto ionization of H2O. Right? That is
H2O + H2O ==> [H3O]^+ + OH^- and that is happening here except one of the 6 H2O molecules in the hexahydrate is used. It will have an equilibrium constant etc etc. Actually, that is just the first step. The second one is as follows:
[Fe(H2O)5(OH)]^2+ + H2O ==> [Fe(H2O)4(OH)2]^+ + H3O and it goes through the same process a third time, too but I'll leave that to you.

2. Your answer of 2E-8 is not right but c is not the right answer either.
If pH = 11 then pOH is 3 and (OH^-) -0.001
.......NH3 + H2O ==> NH4^+ + OH^-
I.....0.05............0.......0
C......-x.............x.......x
E....0.05-x...........x.......x

So x is 0.001 which is the amount ionized. The starting amount is 0.05 so percent is (0.001/0.05) x 100 = ? but not 3%

Answer 2

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