Download Fluorite exam ans EA02

EA2002
Geochemistry equilibrium exam 2.
Name
Do your own work, without help, open notes and book. Show all work. Excel permitted for
iterative calculations but only after you show the formulas involved. Note, most problems have
multiple parts, circle your answers.
Fluorite is the most common fluoride bearing mineral. Locally it occurs where ground water
contamination can occur. The EPA has a limit of 4 mg/L for fluoride ion in drinking water.
Two. Determine the value for fluoride in equilibrium with fluorite in pure water in moles per liter
and milligrams per liter. Assume activity coefficients equal one (straight concentration
calculation). Does this exceed EPA guidelines?
Data
CaF2 = Ca+2 + 2 F- log_k = -10.600 Temperature = 25 degrees C and pressure = 1 bar.
Ans. Ksp = [(Ca2+)(F-)2)]/CaF2 (solid)
log Ksp = -10.600
Two fluoride anions are produced for every calcium cation, so the concentration table is:
Before
After
CaF2
solid
solid
Ca2+
0
x
2F0
2x
Substituting x for Ca2+ and 2x for F- (the solid does not change and has a concentration of 1)
gives:
Ksp = (x)(2x)2 and 4x3 = 10-10.600 so x = (2.512E-11/4)1/3 = 1.84E-04
Ca2+ = 1.8E-04 molal; and F- = 3.7E-04 molal (molal = molar in this dilute solution)
Ca2+ = 1.84E-04 mol/L x 40.078 g/mol x 1000 mg/g = 7.9 mg/L
F- = 3.69E-04 mol/L x 18.9984 g/mol x 1000 mg/g = 7.0 mg/L
F Exceeds EPA limits
Three. Write an activity equilibrium expression for fluorite, Ca2+, and F- ions. Find the total
concentration of fluoride ion in pure water in equilibrium with fluorite using activities. You
will need an iterative approach. Give fluoride in moles per liter and mg/L. How much
higher is this (%)?
Same as above but with activity
aCaF2 = aCa+2 + a2 FKsp = ((aCa2+)(aF-)2)/aCaF2 (solid)
Ksp = (x)Ca(2x)2 (F)2
Ksp/ Ca (F)2 = 4(x)3
x = (Ksp/ 4Ca (F)2)1/3
note (F) must be squared, just like the F concentration term
Table of iterations using log  = -0.5085 z2 I equation for log activity coefficient. m = molal.
Ca2+ m
F- m
I logCa(2+) logSO4(2-) Ca2+
aCa2+
aFF1.84E-04 3.69E-04 5.53E-04 -0.0479 -0.0116 0.896 0.974 1.65E-04 3.59E-04
1.95E-04 3.90E-04 5.85E-04 -0.0492 -0.0119 0.893 0.973 1.74E-04 3.79E-04
1.95E-04 3.90E-04 5.85E-04 -0.0492 -0.0119 0.893 0.973 1.74E-04 3.80E-04
Ca2+ = 2.0E-04 molal and F- = 3.9E-04 molal F = (3.90E-04/3.69E-04)100= 5.6% higher
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Three. Ca2+ and F- form a CaF+ species in solution. Note this is a charged species. Will the ionic
strength of the water affect its concentration? CaF+ has an association constant of log kf = 0.940.
Based on the association constant do you think CaF+ will be a major or minor component of the
species in solution?
Minor component, the formation constant is quite small so not much will form.
Find its activity and concentration in equilibrium with fluorite in pure water.
CaF2 = Ca+2 + 2 FKsp = [Ca+2][F-]2 or to see the relationship better Ksp = [Ca+2][F-][F-]
Ca+2 + F- = CaF+
Kf = [CaF+]/ ([Ca+2][F-]) and rearranging
[Ca+2][F-] = [CaF+]/Kf
Rearranging Ksp = [Ca+2][F-][F-] to [Ca+2][F-] = Ksp/[F-]
substituting above in [Ca+2][F-] = [CaF+]/Kf gives
[CaF+] = KspKf/[F-]
for activity aCaF+ = KspKf/aF- so [CaF+]CaF = KspKf/[F-]F
for concentration [CaF+] = KspKf/[F-]
Results: Simple Debye-Huckel should do as solution very dilute. Same formula as used in
iterations.
log CaF+ = -0.5085 12 I. Use I from problem 1. I = 5.85E-04
log CaF+ = -1.19E-02 and CaF+ = 0.973. (As expected, the log activity coefficient is the same
as for F-.)
activity CaF+ = (10-10.60 x 100.940)/ (3.90E-04 x 0.972) = 5.77E-07 in molal or M units, with F+
value from problem 1.
activity CaF+ = 5.8E-07 mol/L
Concentration, same as above without CaF so CaF+ = (10-10.600 x 100.940)/ 3.90E-04 = 5.9E-07
Concentration CaF+ =5.9E-07 mol/L
Four. Based on problem one, the simplest case, how much Ca2+ would you have to add to the
water supply, in moles per liter, to get the fluoride content down to 1.0 mg/L?
In this problem you know the value of F- in the solution. The added Ca2+ acts as a common
ion. Problem states just use concentrations not activities. The solubility relations are
(again):
From problem 1.
CaF2 = Ca+2 + 2 FKsp = [Ca+2][F-]2
with Ksp and F- known rearrange to solve for Ca+2.
[Ca+2] = Ksp /[F-]2
F- is known and = 1mg/L so we need to convert 1 mg/L to moles per liter.
(1 mg F+ /L x 1 g/1000 mg) / 18.9984 g F+ /mole = 5.26E-05 moles F+ /L
From [Ca+2] = Ksp /[F-]2
[Ca+2] = 10-10.60 /[5.26E-05]2
Solving for total [Ca+2] gives 9.07 E-03 moles/L
Amount already present from CaF2; Ca2+ = 1.84 E-04 molal
so that added = Difference moles Ca2+ /L = (9.07 E-03 - 1.84 E-04) moles Ca2+ /L = 8.9 E-03
moles Ca2+ /L added. Notice Ca from fluorite small, as we could have guessed.
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