Among others, this is a limiting reagent (LR) problem.
millimols NaF = mL x M = 57 x 0.04 = 2.28 initially = I
millimols Sr(NO3)2 = 19 x 0.10 = 1.9 initially
C = change
E = equilibrium
..............2NaF + Sr(NO3)2 ==> 2NaNO3 + SrF2
I.............2.28........1.90....................0................0
C..........-2.28.......-1.14.................2.28.............1.14(s)
E.............0.............0.76................2.28............1.14(s)
(NO3^-) never entered into the reaction so you have 1.90*2 = 3.8 millimols in 57 + 19 = 76 mL. (NO3^-) = millimols/mL = ?
Na^+ never entered into the reaction. (Na^+) = mmols/mL = ?
(Sr^2+). You started with 1.90 mmols Sr^2+ from Sr(NO3)2. You used 1.14 to ppt the SrF2 leaving you with 0.76 mmols excess Sr(NO3)2 so mmols/mL = (Sr^2+). Technically, this should be added to the minuscule amount from the SrF2 ppt and you can do that if you wish. It will so small, however, that the difference is not worth the trouble.
(F^-) All of the F^- has ppted as SrF2 so all of the F^- in solution comes from the SrF2 with the laws of Ksp for SrF2 and the common ion of Sr^2+.
Ksp = (Sr^2+)(F^-)^2 = 2E-10
Plug in (Sr ^2+) from above to the Ksp expression and solve for F^-. It will be a small number.
Post your work with any questions you may have.
A volume of 57 mL of 0.040 M NaF is mixed with 19 mL of 0.10 M Sr(NO3)2. Calculate the concentrations of the following ions in the final solution.(Ksp for SrF2 2.0 × 10−10)
[NO3−]
M
[Na+]
M
[Sr2+]
M
[F−]
M
1 answer
1 answer