Barium nitrate is added to a solution of .025 M sodium fluoride.
a) At what concentration of Ba+2 does a precipitate start to form?
b) Enough barium nitrate is added to make [Ba+2]= .0045 M. What percentage of the original fluoride ion has precipitated?
I'm having a lot of trouble writing an equilibrium equation for some reason.
Would it be something along the lines of:
Ba(NO3)2 + 2NaF(aq) <--> 2NaNO3 + Ba+2 + 2F-
?
BaF2 is a precipitate. So a molecular equation would be
Ba(NO3)2 + 2NaF ==>2Na^+ + 2NO3^- + BaF2
For the solubility of BaF2,
BaF2(s) ==> Ba^+2 + 2F^-
Ksp = (Ba^+2)(F^-)^2
(a) You know F ion. It is 0.025M. Plug into Ksp and solve for Ba ion.
(b) Knowing Ba ion is 0.0045M, solve for F ion. That subtracted from the inbitial amount will be the about pptd. Then take the percent of that. Post your work if you get stuck.
The equilibrium equation you wrote is incorrect. The correct equation for the reaction between barium nitrate and sodium fluoride is:
Ba(NO3)2 + 2NaF -> 2NaNO3 + BaF2
BaF2 is the precipitate in this reaction.
To answer the questions:
(a) To find the concentration at which a precipitate starts to form, you need to determine the solubility product constant (Ksp) for BaF2. The solubility product constant expression for BaF2 is:
Ksp = [Ba2+][F-]^2
Given that the concentration of sodium fluoride (NaF) is 0.025 M, you can assume that the concentration of fluoride ions ([F-]) is also 0.025 M.
Let's assume the concentration of barium ions ([Ba2+]) is x (M). At the point where a precipitate starts to form, [Ba2+] and [F-] will be in equilibrium, so you can set up the equation:
Ksp = x * (0.025)^2
Solve for x:
x = Ksp / (0.025)^2
Plug in the value of the solubility product constant for BaF2 to find the concentration of barium ions at which a precipitate starts to form.
(b) Now, let's say enough barium nitrate is added to make [Ba2+] = 0.0045 M. You need to determine what percentage of the initial fluoride ion (F-) has precipitated.
Substitute the [Ba2+] = 0.0045 M into the Ksp expression:
Ksp = (0.0045) * ([F-])^2
Solve for [F-]:
([F-])^2 = Ksp / 0.0045
[F-] = √(Ksp / 0.0045)
Subtract this value from the initial concentration of fluoride ions (0.025 M):
Percentage of precipitation = ([F-] initial - [F-] final) / [F-] initial * 100
Plug in the values to calculate the percentage of the initial fluoride ion that has precipitated.
Remember to use appropriate significant figures in your calculations and final answer.
Correct, the equilibrium equation for the reaction between barium nitrate and sodium fluoride is:
Ba(NO3)2 + 2NaF(aq) ⇌ 2NaNO3 + BaF2
(a) To determine the concentration of barium ions at which a precipitate begins to form, you will need to compare the values of the ion product (Q) and the solubility product constant (Ksp) for barium fluoride (BaF2). The solubility product constant (Ksp) expression for BaF2 is:
Ksp = [Ba+2][F-]^2
Assuming that the initial concentration of sodium fluoride (NaF) remains constant and equal to 0.025 M, the initial concentration of fluoride ions (F-) is also 0.025 M.
Since barium nitrate is being added, the concentration of barium ions (Ba+2) will increase as the reaction progresses until a precipitate begins to form. The concentration of barium ions at which a precipitate starts to form can be determined by comparing the Q value to the Ksp value.
Q = [Ba+2][F-]^2
If Q < Ksp, no precipitate is formed. If Q = Ksp, the solution is at equilibrium and a precipitate may just start to form. If Q > Ksp, a precipitate will definitely form.
(b) To determine the percentage of the original fluoride ion that has precipitated when [Ba+2] = 0.0045 M, you will need to calculate the change in concentration of fluoride ions (F-) and then determine the percentage of that change relative to the initial concentration.
Let's solve part (a) first.