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)
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.
a. Your answer is incorrect. To determine the minimum concentration of the precipitating agent to cause precipitation, you need to calculate the concentration at which the value of Q (reaction quotient) equals or exceeds the value of Ksp (solubility product constant).
In this case, the concentration of BaF2 is unknown, so let's assume it is x M. The reaction will be:
BaF2 <--> Ba2+ + 2F-
The equilibrium expression is:
Ksp = [Ba2+][F-]^2
2.45 x 10^-6 = (x)(2x)^2
2.45 x 10^-6 = 4x^3
x^3 = 6.125 x 10^-7
x = (6.125 x 10^-7)^(1/3)
x = 0.0086 M
Therefore, the minimum concentration of BaNO3 required to cause precipitation of the cation is 0.0086 M.
b. Your answer is correct.
c. Your answer is correct.
Your answers for parts a, b, and c are incorrect. Let's go through each one step-by-step:
a. BaNO3 + NaF → BaF2 + NaNO3
NaNO3 is soluble, so it will not cause precipitation.
For BaF2, the Ksp is given as 2.45 x 10^-6.
The equation for Ksp is: Ksp = [Ba2+][F-]^2.
Since the concentration of F- is unknown, let's represent it as x.
Therefore, we have (0.035)(x)^2 = 2.45 x 10^-6.
Solving for x, we get x = 9.31 x 10^-3 M.
Therefore, the minimum concentration of the precipitating agent (NaF) is 9.31 x 10^-3 M.
b. CaI2 + K2SO4 → CaSO4 + 2KI
KI is soluble, so it will not cause precipitation.
For CaSO4, the Ksp is given as 7.10 x 10^-5.
The equation for Ksp is: Ksp = [Ca2+][SO4^2-].
Since the concentration of SO4^2- is unknown, let's represent it as x.
Therefore, we have (0.085)(x) = 7.10 x 10^-5.
Solving for x, we get x = 8.35 x 10^-4 M.
Therefore, the minimum concentration of the precipitating agent (K2SO4) is 8.35 x 10^-4 M.
c. AgNO3 + RbCl → AgCl + RbNO3
RbNO3 is soluble, so it will not cause precipitation.
For AgCl, the Ksp is given as 1.77 x 10^-10.
The equation for Ksp is: Ksp = [Ag+][Cl-].
Since the concentration of Cl- is unknown, let's represent it as x.
Therefore, we have (0.0018)(x) = 1.77 x 10^-10.
Solving for x, we get x = 9.83 x 10^-8 M.
Therefore, the minimum concentration of the precipitating agent (RbCl) is 9.83 x 10^-8 M.
Your answers seem to be correct. You have correctly set up the equilibrium expressions for the precipitation reactions and solved for the concentration of the precipitating ions.
For part a, you correctly calculated the concentration of fluoride ions ([F-]) needed to reach the Ksp value of BaF2 and found it to be 0.0265 M.
For part b, you correctly calculated the concentration of sulfate ions ([SO4^2-]) needed to reach the Ksp value of CaSO4 and found it to be 8.35 x 10^-4 M.
For part c, you correctly calculated the concentration of chloride ions ([Cl-]) needed to reach the Ksp value of AgCl and found it to be 9.83 x 10^-8 M.
Overall, your approach and calculations are correct. Well done!