April 25, 2024, 10:27:16 PM
Forum Rules: Read This Before Posting


Topic: How to determine Ksp with titration?  (Read 21387 times)

0 Members and 1 Guest are viewing this topic.

Offline omgwtfitsp

  • Regular Member
  • ***
  • Posts: 13
  • Mole Snacks: +0/-0
How to determine Ksp with titration?
« on: December 25, 2010, 11:00:08 PM »
I was given this lab and my teacher absolutely did not explain anything to me, he just follows this other teacher and probably doesn't even know how to do this himself..

Anyways, I have to determine the Ksp of calcium hydroxide Ca(OH)2 after titrating it with HCl. * So please if you can check my work below and tell me if I did it right?*

Here's some equations I know I will have to use.

1) 2HCl + Ca(OH)2 -> CaCl2 + 2H2O  

2) Ca(OH)2 -> Ca 2+ + 2OH -  

So I started with 2mL of 0.05 M HCl and it took on average 4.6 mL of 0.07 M of the base to neutralize it and turn it pink.

So now I have to show my work for how to find the Ksp. But I really don't know how. My textbook doesn't have an example like I thought it would...

Firstly I'm confused as to why my teacher gave me a concentration of the base in the first place.

Anyways, I know that when it is neutralized, the moles of H+ would equal the moles of OH-. So, moles of H+ would be (0.05 M x 0.002L) 0.0001 moles. So moles of OH- will also be 0.0001 moles. So here's where I'm confused. The dissociation of the base brings out 2OH- so does that mean, the 0.0001 is for the 2OH- ions or just 1 of them?

So I went on and assumed it was for the 2OH- ions (if it is someone please explain why), and so then I used that to find the moles of Ca 2+, which is (0.0001/2) 0.00005 moles. So then I found the concentration of the Ca and OH ions using the volume of base that it took to titrate the solution. It turned out to be roughly: [OH-] = 0.0217 mol/L and for [Ca2+] = 0.0108 mol/L.

Assuming all the above I did was correct, I would just substitute the values into the Ksp formula for Ca(OH)2 right?

Ksp = [OH-]2x[Ca2+]

after substituting I got the answer: 5.08 x 10-6 which is really close to the actual Ksp of Ca(OH)2 which is 5.02 x 10-6.

So can someone look over my work and see if this is the correct way to determine the ksp? If so, I'd like to ask what the concentration of the base (0.07M) was used for? I didn't use it in my work so I keep thinking that I didn't do it correctly.

Also, when substituting the ion concentrations into the ksp formula, why don't we divide by [Ca(OH)2]? Isn't it an aqueous solution? Shouldn't it be divided as well making it: ksp = [OH-]2x[Ca2+] / [Ca(OH)2]??

Please, someone teach me the correct way to do this lab.

 

« Last Edit: December 25, 2010, 11:18:38 PM by omgwtfitsp »

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27663
  • Mole Snacks: +1801/-410
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: How to determine Ksp with titration?
« Reply #1 on: December 26, 2010, 06:01:14 AM »
It is not entirely clear to me what you were given and what you did, please be more specific.

In general you seem to be on the right track.

http://www.titrations.info/titration-calculation
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline igloo5080

  • Regular Member
  • ***
  • Posts: 16
  • Mole Snacks: +1/-1
Re: How to determine Ksp with titration?
« Reply #2 on: December 26, 2010, 10:45:26 AM »
Solubility products are concerned with the equilibrium between ions in solution and the solid. In this case the equilibrium is between the solid Ca(OH)2 and the Ca2+ and OH- ions in solution. It is assumed that no Ca(OH)2 'molecules' exist in solution and that the ions simply become dislodged from the surface of the solid during the dissolving process.

Since concentrations of solids are constant - linked to their densities - [Ca(OH)2] remains constant and therefore doesn't appear in the expression for KSP. So your calculations appear to have worked out well!

Sponsored Links