Question

200 mL of 0.0010 M Sr(NO₃)₂ (aq) are mixed with 800 mL of 0.0050M NaF (aq) to make 1.0 L of solution. The Ksp value of SrF₂ is 4.3 x 10^-9. Which is correct? I know the answer, I just don't know how they got to it.

a. SrF₂ will not precipitate.

b. SrF₂ will precipitate. NaF is the limiting reactant.

c. SrF₂ will precipitate. NaNO₃ is the limiting reactant.

d. SrF₂ will precipitate. Sr(NO₃)₂ is the limiting reactant.

e. SrF₂ will precipitate. SrF₂ is the limiting reactant.

Answer 1

200 mL of 0.0010 M Sr(NO₃)₂ (aq) = 0.200 L * 0.0010 mole / L = 2.0 * 10^-4 mole.

800 mL of 0.0050 M NaF (aq) = 0.800 L * 0.0050 mole / L = 4.0 * 10^-3 mole,

Sr^2+ + 2 F- ...........> SrF2

total volume = (200 +800) = 1000 ml = 1.00 L

thus

[Sr^2+] = 2.0 * 10^-4 mole / 1.0 L = 2.0 * 10^-4 M

and

[F-] = 4.0 * 10^-3 mole / 1.00 L = 4.0 * 10^-3 L

Q = [Sr^2+][F-]^2 = 2.0 * 10^-4 * (4.0 * 10^-3)^2 = 3.2 * 10^-9

we have

Ksp value of SrF₂ is 4.3 x 10^-9

as Ksp > Q

and

hence SrF₂ will not precipitate.

a. SrF₂ will not precipitate. is the answer.