[docnet]

Determine the molar solubility of Chromium (III) hydroxide ion in a buffered solution with pH = 11.45.

The reactions:

Dissolution of chromium hydroxide:   Cr(OH)3(s) <=> Cr3+(aq) + 3OH-(aq)
Formation of the complex ion:          Cr3+(aq) + 4OH-(aq) <=> Cr(OH)4-(aq)

Given:
Ksp of Cr(OH)3(s) = 3.0E-29
Kf of Cr(OH)4-(aq) = 8.0E29


What I've tried so far:

-I added the reactions: Cr(OH)3(s) + OH-(aq) <=> Cr(OH)4-(aq)
-I multiplied the K's and got 24 as the new equilibrium constant

constructed an equilibrium table and ended up with the equilibrium expression  24 = x/(.0028-x)
I didn't include the Cr(OH)3 because it's a solid.

(.0028 is the molarity of the OH-ion calculated from the given pH of 11.45):
10^-(14-11.45) = [OH-] = .0028

x = .002688M


The Answer from the book is 6.8E-2M but no work is shown. Thanks for any input!

[mjc123]

What does "buffered" mean?

[docnet]

I assume the book is throwing that word a bit loosely to mean that there is added OH- ions in the solution that Chromium (III) hydroxide will be dissolved in. In this case I'm guessing that we are only concerned with the concentration of the Cr3+ ions dissolved. The traditional sense of the word buffer (the ability to resist pH change) wouldn't make sense because no other information on the word buffer is given, but the formation of the complex ions would result in a lower pH than 11.45 because most the OH- ions would be consumed (indicated by the high Kf value). In this situation the result of the "buffer" means the increased solubility of the Chromium (III) hydroxide by a formation of a complex ions..

This concept is still new to me because I first started learning it last night. I'm going to tackle more reading and retry this problem when I get back home tonight. I don't think I have a very solid conceptual understanding on this topic

Any input from the more salt chemists would be appreciated (no pun intended) .

[Borek]

I assume the book is throwing that word a bit loosely

No, it has a very precise meaning.

The traditional sense of the word buffer (the ability to resist pH change) wouldn't make sense because no other information on the word buffer is given

There is all information given that you need - you are told solution is buffered at pH 11.45. Doesn't matter how.

the formation of the complex ions would result in a lower pH than 11.45 because most the OH- ions would be consumed

Nope. Buffered means pH remains constant, and the concentration of OH^- doesn't change (despite ions being consumed).

In reality pH is never exactly constant and it changes a bit when H⁺/OH^- are consumed, but these changes are usually negligible. You are not asked here to calculate full equilibrium, just to assume pH is constant.

[docnet]

Borek,

Thank you for the correction. That approach to the problem (the pH staying constant) makes more sense. I'll re-try the problem as soon as I get home tonight (after some more reading).

[docnet]

Well, I just got back from a day long bike ride and I'm absolutely wrecked.. Cycled around 70 miles for the first time today. I'm going to try again tomorrow. Planning to do lots of practice problems as well as spend time in the textbook to nail this into my head. The reason is that Gen Chem II starts next week and I'm taking a condensed course-load this year.

My only experience with chemistry is gen chem I so far. I'm also taking ochem and biochem this year and want to be as prepared as I can be.

[docnet]

Tonight I practiced some basic molar solubility problems and "common ion effect" type solubility problems and it was much easier this time around.


[OH-] = 10^-2.55 = .002818M

Ksp * Kf = K overall  = 24


24 = [Cr(OH)4-] / [OH-]
24 = x/.002818

Molar solubility of Cr(OH)3(s) in the buffered solution of pH 11.45 = x = 24(.002818) = 6.8E-2M

It was basically the value of K times the concentration of the Hydroxide ion. it made things so much easier. thank you Borek.