advertisement

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/ 4Ca (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 logCa(2+) logSO4(2-) Ca2+ aCa2+ aFF1.84E-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 106738504 1 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+] = KspKf/[F-] for activity aCaF+ = KspKf/aF- so [CaF+]CaF = KspKf/[F-]F for concentration [CaF+] = KspKf/[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. 106738504 2