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Topic: Calculating Ksp from Flame Photometry?  (Read 1624 times)

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Offline Halogen876

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Calculating Ksp from Flame Photometry?
« on: July 16, 2014, 12:12:54 PM »

I am struggling with this question:

A saturated solution of CaF2 (in equilibrium with solid CaF2) produces an intensity reading of 116.1 on the flame photometer. A 5.00ppm standard solution of Ca 2+ ion reads 66.8 and distilled water reads 8.7. From these data, calculate the soulbility product of CaF2.

First, I subtracted 8.7 from the 2 readings to get 107.4 for thr saturated solution and 58.1 for the Ca standard.

Then I set up I=kc and filled in 58.1 for I and 5.00ppm for c and solved for k; which I found to be 11.6ppm -1. I then filled this value of k and the reading of 107.4 back into I=Kc and solved the concentration of Ca 2+ to be 9.24ppm so 9.24mg/L or 1.18E-4 mol/L.

The concentration of F- would be twice this so 2.36E-4mol/L.

Ksp=[Ca 2+ ][F - ]2 and when I fill the concentrations into that equation, I get Ksp=6.57E-12. The answer in the book is 4.91E-11 though so I must be missing something but I'm not sure what it is...any help would be greatly appreciated :)

Offline mjc123

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Re: Calculating Ksp from Flame Photometry?
« Reply #1 on: July 17, 2014, 04:31:40 AM »
9.24 ppm Ca is 0.00942/40 = 2.36 x10-4 M. That (Ca concentration) is what the flame photometer tells you. You worked out the concentration of CaF2, on the assumption that you had 9.24 ppm of CaF2.

That is why I dislike talking of salt concentrations in ppm, and wish people would always use molar. Why not say the standard solution was 0.125mM Ca?

Offline Halogen876

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Re: Calculating Ksp from Flame Photometry?
« Reply #2 on: July 17, 2014, 08:52:06 AM »
I see what you mean. That all makes perfect sense now! Thank you so much for your help  :)

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