Question

Determine the minimum concentration of the precipitating agent on the right to cause precipitation of the cation from the solutions to the left.

a. 0.035 M BaNO3; NaF
b. 0.085 M CaI2 ; K2SO4
c. 0.0018 M AgNO3 ; RbCl

My work:

Are my answers correct?

a. BaNO3 + NaF --> BaF2 + NaNO3
NaNO3: soluble
BaF2: Ksp= 2.45 x 10^-6
BaF2 <--> Ba2+ + 2F-
Q= [Ba2+][F-]^2
Q= (0.035)(F-)^2 = Ksp= 2.45 x 10^-5
[F-]= 0.0265 M (final answer for part a)

b. CaI2 + K2SO4 --> CaSO4 + KI
KI: soluble
CaSO4: Ksp= 7.10 x 10^-5
CaSO4 <--> Ca2+ + SO4^2-
Q= [Ca2+][SO4^2-]
Q= (0.085) (SO4^2-)=Ksp=7.10 x 10^-5
[SO4^2-]= 8.35 x 10^-4 M (final answer for part b)

c. AgNO3 + RbCl --> AgCl + RbNO3
RbNO3: soluble
AgCl: Ksp= 1.77 x 10^-10
Q= [Ag+][Cl-]
Q= (0.0018)(Cl-) = 1.77 x 10^-10
[Cl-]= 9.83 x 10^-8 M (final answer for part c)

Answer 1

That looks ok to me but I object somewhat to you saying Q = Ksp.
For example, Ksp for c is
Ksp = [Ag^+][Cl^-]. In my book, (just an expression) Q is used to multiply Ag^+ and Cl^- and use the answer to compare with Ksp to see if a ppt will take place. That is, for c, it's Ksp that is [Ag^+][Cl^-] and not Qsp. Ksp = Qsp ONLY at one condition and that is when the first molecule of a ppt occurs.