Please help with chem homework!!??

Question

 A 155.0 mL solution of 2.996 M strontium nitrate is mixed with 220.0 mL of a 2.924 M sodium fluoride solution. Calculate the mass of the resulting strontium fluoride precipitate. Assuming complete precipitation, calculate the final concentration of each ion. If the ion is no longer in solution, enter a zero for the concentration. 

Na 

F 

Sr 

NO3

Answer 1

This is a rather long, involved, complex answer, so I hope you can follow the steps.

Sr(NO3)2(aq) + 2NaF(aq) ==> SrF2(s) + 2NaNO3(aq)

moles Sr^2+ = 0.155 L x 2.996 mole/L = 0.4644 moles

moles F^- = 0.220 L x 2.924 mole/L = 0.6433 moles

It takes 2 moles F^- to precipitate 1 mole Sr^2+, so F^- is limiting.

Since both Na^+ and NO3^- are spectator ions, and the final volume is 155 ml + 220 ml = 375 ml (0.375 L), the concentrations of Na^+ and NO3^- can be calculated as follows:

[Na^+]: (220 ml)(2.924 M) = (375 ml)(x M) and x = 1.715 M

[NO3^-]: (155 ml)(2.996 M)(2) = (375 ml)(x M) = 2.477 M

Now, for the [Sr^2+] and [F^-]

Initial moles of Sr^2+ = 0.4644 moles

Moles Sr^2+ precipitated by F^- = 0.6433/2 = 0.3217 (1 mole Sr^2+ precipitated by 2 moles F^-)

Moles Sr^2+ no precipitated (left over) = 0.4644 - 0.3217 = 0.1427 moles Sr^2+ unprecipitated

[Sr^2+] = 0.1427 moles/0.375 L = 0.3805 M

To find [F^-], one needs the Ksp for SrF2. There are several values listed in the literature. I am using a value of 2x10^-10.

SrF2(s) <==> Sr^2+(aq) + 2F^-(aq)

Ksp = [Sr^2+][F^-]^2

2x10^-10 = (0.3805)(x)^2

x^2 = 5.256x10^-10

x = [F^-] = 2.29x10^-5 M