Look up Ksp for CaF2.
Calculate Qsp = (Ca^2+)(F^-)^2 and compare with ksp.
You will need to change 1.0 mg/L to mols/L and 200 mg/L to mols/L.
Will a fluoride concentration of 1.0 mg/L be soluble in a water containing 200 mg/L of calcium?
3 answers
This is what i found for ksp
CaF2<--> Ca(2+) +2F(-) Ksp=(x)(4x^2) where x=solubility Therefore, Ksp=3.7 x 10^-11
and for Qsp I am not sure how to go on about calculating that.
CaF2<--> Ca(2+) +2F(-) Ksp=(x)(4x^2) where x=solubility Therefore, Ksp=3.7 x 10^-11
and for Qsp I am not sure how to go on about calculating that.
Change 1 mg/L F^- to mols/L.
Change 200 mg/L Ca^2+ to mols/L.
Then substitute those values into
Qsp = (Ca^2+)(F^-)^2 = ?
(By the way, DON'T multiply F^- by 2; i.e., substitute mols/L for F^-.)
Then compare Qsp with Ksp. If Qsp is larger than Ksp, a ppt occurs. If Qsp is less than Ksp, no ppt will occur.
Change 200 mg/L Ca^2+ to mols/L.
Then substitute those values into
Qsp = (Ca^2+)(F^-)^2 = ?
(By the way, DON'T multiply F^- by 2; i.e., substitute mols/L for F^-.)
Then compare Qsp with Ksp. If Qsp is larger than Ksp, a ppt occurs. If Qsp is less than Ksp, no ppt will occur.
3 answers