[Captain Proton]

Here's the problem:

Consider the following reaction:
Ba(NO3)2(aq) + 2KOH (aq) -->
My answer: Ba(NO3)2(aq) + 2 KOH (aq) <--> 2 KNO3(aq) + BaOH2
a. Write the net ionic equation for this reaction:
My answer: Ba+2  +  2 NO3-1 +  2 K+1  +  2 OH-1 --> 2 K+1  + 2 NO3-1 + Ba+2 + 2 OH-1
I think that they are all part of the equation, even though Barium Hydroxide is only slightly soluble, but I am probably wrong on that one.
b.  Using just the net ionic equation, write the equilibrium expression for this reaction.
 I think that the equation would be either...
             [K+1]^2 * [NO3-1]^2 * [Ba+2] + [OH-1]^2
 Keq  =  -----------------------------------------------------
             [Ba+2] * [NO3-1]^2 * [K+1]^2 * [OH-1]^2

or

            [KNO3]^3 * [Ba(OH)2]
 Keq =  ---------------------------
            [Ba(NO3)2][KOH]2

Alas, these are probably wrong too... sigh...

c. A student mixed 50 ml of 0.1M barium nitrate with 75 ml of 0.1 M potassium hydroxide. The student waited several hours to be sure equilibrium was reached and then measured the concentration of barium ion and the concentration of hydroxide ion present in the solution. The following data were obtained:

Now I don't know what I am supposed to do.  I haven't the faintest idea how to incorporate the data into an equation or anything.

   [Ba+2 ion] = 0.021 M

   [OH-] = 0.021 M

   What is the value of the equilibrium constant that corresponds to your equilibrium expression in part b?

I am sorry that this post is so very long, but I am really stuck.  I've IM'ed people in my class, and they all agree that it isn't like anything we've ever seen in class.  Any hints, answers, or mathwork that you can provide would be much appreciated.  Thanks.

[Borek]

Hint: barium hydroxide is weakly soluble.

If so, what will be the net ionic equation?

[Captain Proton]

            [K+1]^2 * [NO3-1]^2 * [Ba+2] + [OH-1]^2
Keq  =  -----------------------------------------------------
            [Ba+2] * [NO3-1]^2 * [K+1]^2 * [OH-1]^2

Hmmm... I think this was right...

Ba(NO3)2(aq) + 2 KOH (aq) <--> 2 KNO3(aq) + BaOH2

Ba+2  +  2 NO3-1 +  2 K+1  +  2 OH-1 --> 2 K+1  + 2 NO3-1 + Ba+2 + 2 OH-1

Oh.. and then my net equation would be the above big one with the reactants on the top and the products on the bottom (without spectator ions).

Ba+2 (aq) +  2 OH-1 (aq) --> Ba(OH)2 (aq)


            [Ba+2][OH-1]^2
Keq = -----------------------
               [Ba(OH)2]

Ok, so then I think I can use the I.C.E. method...

     Ba+2 (aq) +  2 OH-1 (aq) --> Ba(OH)2 (aq)
 ------------------------------------------------------
| I  |   0.1 M   |    0.1 M    |        0.0 M             |
-------------------------------------------------------
| C |     -x      |     -2x      |          +x               |
===================================
| E  | (0.1-x) M  *  (0.1-2x) M

 But what does the third column equal, and where do I go from here?  Could I fill in the [Ba+2][OH-1]^2 with the 0.021 M from part C, but what about the other information?

[Borek]

[Ba+2] + [OH-1]^2

Why sum?

Ba+2 (aq) +  2 OH-1 (aq) --> Ba(OH)2 (aq)


            [Ba+2][OH-1]^2
Keq = -----------------------
               [Ba(OH)2]

Seems OK, although solids activities are considered to equal 1.

Ok, so then I think I can use the I.C.E. method...

ICE is of no use here as you don't know Initial concentraions. However, Equilibrium concentrations are given.