April 18, 2024, 11:35:48 PM
Forum Rules: Read This Before Posting


Topic: aqueous equilibria question  (Read 3266 times)

0 Members and 1 Guest are viewing this topic.

Offline hotvince

  • New Member
  • **
  • Posts: 5
  • Mole Snacks: +0/-0
aqueous equilibria question
« on: April 03, 2008, 09:48:33 PM »
Calculate the concentration of Cu+2 in 1.0 L of a solution that contains a total of 1 x 10-3 mol of copper(II) ion and that is 0.10 M in NH3.

Cu+2 + 4NH3 --> Cu(NH3)4+2 Kf= 5.0 x 1012

Offline Rabn

  • Chemist
  • Full Member
  • *
  • Posts: 284
  • Mole Snacks: +28/-13
Re: aqueous equilibria question
« Reply #1 on: April 03, 2008, 09:53:51 PM »
Let us know what your approach is and post any work you have done thus far.  Getting the answer isn't as important as understanding the reasoning behind it.

Offline hotvince

  • New Member
  • **
  • Posts: 5
  • Mole Snacks: +0/-0
Re: aqueous equilibria question
« Reply #2 on: April 03, 2008, 10:03:18 PM »
First, I switched the equation.

So my new equation is : Cu(NH3)4+2--> Cu+2+4NH3+

and my new Kf is 2 x 10 -13

Then I attempted to solve for [Cu+2] by using the following equation

Kf=[NH3]4[Cu+2]                                                   
     [Cu(NH3)4+2]

So I did (2 x 10-13) = (0.10)4(x)
                            (1 x 10-13 -x)

The answer I got was 2 x 10 -22, but the correct answer is 2 x 10-12

Offline Rabn

  • Chemist
  • Full Member
  • *
  • Posts: 284
  • Mole Snacks: +28/-13
Re: aqueous equilibria question
« Reply #3 on: April 03, 2008, 10:37:50 PM »
If you used the equation you posted you need to double check your numbers for correctness. Otherwise your good.

Sponsored Links