[Halogen876]

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

[mjc123]

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?

[Halogen876]

I see what you mean. That all makes perfect sense now! Thank you so much for your help