What must be the concentration of F- to just start precipitation of SrF2 from a solution which is 1.202*10-5M i Sr(NO3)2? Ksp = 2.900*10-9
2.9*10-9/1.202*10-5 and find the square root of the number = 0.0155
I agree.
To determine the concentration of F- required to just start the precipitation of SrF2, we need to use the concept of the solubility product constant (Ksp) and set up an equilibrium expression.
The given Ksp value for SrF2 is 2.900*10^-9. The chemical equation representing the dissolution of SrF2 in water is:
SrF2(s) ⇌ Sr2+(aq) + 2F-(aq)
According to the stoichiometry of the equation, for every Sr2+ ion that dissolves, two fluoride ions (F-) are produced. Therefore, the concentration of F- in the solution will be twice the concentration of Sr2+ in the solution.
Given that the concentration of Sr(NO3)2 is 1.202*10^-5 M, we can assume that all of the Sr(NO3)2 dissociates into Sr2+ and 2NO3- ions. This means that the concentration of Sr2+ is also 1.202*10^-5 M.
Since the concentration of F- is twice the concentration of Sr2+, we have:
[F-] = 2 * [Sr2+]
[F-] = 2 * (1.202*10^-5 M)
[F-] = 2.404*10^-5 M
Therefore, the concentration of F- needed to just start the precipitation of SrF2 is 2.404*10^-5 M.