Asked by Hiba
75.0 mL 0.060 M NaF and 25 mL 0.150 M Sr (NO3) 2 are mixed. Calculate the Q.
For SrF2 Kc = 2.0 x 10-10
For SrF2 Kc = 2.0 x 10-10
Answers
Answered by
DrBob222
2NaF + Sr(NO3)2 ==>SrF2 + 2NaNO3
In the final solution before reaction:
(NaF) = 0.060 x (75/100) = ?
[Sr(NO3)2] = 0.150 x (25/100) = ?
Then Q = (Sr^2+)(F^-)^2
Plug in the final concentrations before mixing above and solve for Q.
You didn't ask but I expect it's part of the problem but
A ppt will occur IF Q > Ksp.
A ppt will not occur IF Q < Ksp
In the final solution before reaction:
(NaF) = 0.060 x (75/100) = ?
[Sr(NO3)2] = 0.150 x (25/100) = ?
Then Q = (Sr^2+)(F^-)^2
Plug in the final concentrations before mixing above and solve for Q.
You didn't ask but I expect it's part of the problem but
A ppt will occur IF Q > Ksp.
A ppt will not occur IF Q < Ksp
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