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
To calculate the concentrations of the ions in the final solution, we need to consider the reaction that occurs when NaF and Sr(NO3)2 are mixed. The reaction can be represented as:
2 NaF + Sr(NO3)2 -> SrF2 + 2 NaNO3
From the balanced equation, we can see that for every 2 moles of NaF, 1 mole of SrF2 is formed. Therefore, the concentration of Sr2+ ion will be equal to the concentration of NO3- ion.
Let's calculate the concentration of Sr2+ ion (and consequently the concentration of NO3- ion) first:
1. Calculate the moles of Sr(NO3)2:
moles = volume (in L) x concentration (in M)
moles = 19 mL x (1 L / 1000 mL) x 0.10 M
moles = 19 x 0.001 x 0.10 = 0.0019 mol
2. Since 2 moles of NaF react with 1 mole of Sr(NO3)2 to form SrF2, we have:
moles of SrF2 formed = 0.0019 mol / 2 = 0.00095 mol
3. Calculate the concentration of Sr2+ ion:
concentration (in M) = moles / volume (in L)
concentration = 0.00095 mol / (57 mL + 19 mL) x (1 L / 1000 mL)
concentration = 0.00095 / 0.076 L
concentration = 0.0125 M
So, the concentration of Sr2+ ion in the final solution is 0.0125 M, and the concentration of NO3- ion is also 0.0125 M.
Now let's calculate the concentration of Na+ ion and F- ion:
- The concentration of Na+ ion will be the same as the initial concentration, because Na+ ion does not react with any of the ions in the solution.
Therefore, the concentration of Na+ ion in the final solution is 0.040 M.
- Since F- ion reacts with Sr2+ ion to form SrF2, all of the F- ions will be consumed in the reaction. Therefore, the concentration of F- ion will be zero in the final solution.
So, the concentrations of the ions in the final solution are:
[NO3-] = 0.0125 M
[Na+] = 0.040 M
[Sr2+] = 0.0125 M
[F-] = 0 M
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.