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Drever, Chapter 2

#1. Which is more stable at 25C, gypsum or anhydrite plus water?

Use the reaction: gypsum <==> anhydrite + water and calculate Delta G^o_R from:

Delta G^o_R = Delta G^o_{f anhydrite} + 2Delta G^o_{f water} - Delta G^o_{f gypsum}

Delta G^o_R = -1321.98 Kj +2(-237.14 Kj) - (-1797.36 Kj) = +1.1 Kj.

Because Delta G^o_R is a positive value, the reaction as written will not occur spontaneously, and thus gypsum is the stable form of calcium sulfate at 25 C.

#2 What is the solubility product of gibbsite (Al(OH)₃) at 25C?

Al(OH)₃ <==> Al³⁺ + 3(OH^-) and Delta G^o_R = Delta G^o_{f Al3+} + 3 Delta G^o_{f OH-} - Delta G^o_{f Al(OH)3}

Delta G^o_R = -487.65 Kj + 3(-157.2 Kj) - (-1154.86 Kj) = +195.61 Kj

Then use the relationship between Delta G^o_R and K:   Delta G^o_R = -2.303RT log K, or

log K = - Delta G^o_R/2.303RT = -(195.65)/(2.303)(8.3143x10^-3)(298.15) = -34.26

and K = 10 ^-34.26.

#3 Calculate the concentration of dissolved Al³⁺ in equilibrium with gibbsite at a pH of 4 and 5 C using the data from (a) Appendix II and (b) Appendix III. Does the solubility of gibbsite increase or decrease with decreasing temperature. How do the results from the two data sets compare?

From Appendix III: Al(OH)₃ + 3H⁺ <==> Al³⁺ + 3H₂O. Values of K and Delta H from two different references are: K = 10^7.74 and Delta H^o_R = -105.3 Kj/mole or K = 10^8.11 and Delta H^o_R = -95.4 Kj/mole. At 25 C: K = 10^7.74 = (Al³⁺)(H₂O)³/(Al(OH)₃)(H⁺)³.  So, (Al³⁺) = 10^7.74(Al(OH)₃)(H⁺)³/(H₂O)³. The concentration of the solid phase (Al(OH)₃) and the concentration of water are both = 1, and a pH of 4 means that (H⁺) = 10^-4. The results are: (Al³⁺) = (10^7.74)(10^-4)³ = 10^-4.26 = 5.5 x 10^-5 mole/kg or (Al³⁺) = (10^8.11)(10^-4)³ = 10^-3.89 = 1.3 x 10^-4 mole/kg. To obtain the solubility of gibbsite at 5 C instead of 25 C, use the van't Hoff equation to calculate K at the lower temperature, and then solve for (Al³⁺). So, the van't Hoff equation is: log K_T2 = log K_T1 + Delta H^o_R/2.303R(1/T₁ - 1/T₂). log K_T2 = 7.74 - 105.3 Kj/2.303(8.3143 x 10^-3)(1/298.15 - 1/278.15) = 9.065; K_T2 = 10^9.065 and the solubility is: (Al³⁺) = (10^9.065)(10^-4)³ = 10^-2.935 = 1.16 x 10^-3 mole/kg. The solubility of gibbsite increases with decreasing temperature.

From Appendix II, Delta G^o_R = Delta G^o_{f Al3+} + 3 Delta G^o_{f H2O} - [Delta G^o_{f Al(OH)3} + 3 Delta G^o_{f H+}] = -44.21 Kj.

Log K = -Delta G^o_R/2.303RT ₌ -(-44.21)/2.303(8.3143 x 10^-3)(298.15) = 7.74, and K = 10^7.74. This value compares very well with the previous answer.

#4. Al³⁺ forms a complex AlSO⁺₄ in the presence of sulfate:

Al³⁺ + SO₄⁼ <==> AlSO₄⁺; K_stab = 10^3.01. How many ppm of Al (total) would be in equilibrium with gibbsite at pH 4 and 25 C in the presence of 10^-3 moles of SO₄⁼?

We can approach this in a couple of steps: 1. we know that Al³⁺ reaches equilibrium with gibbsite (K = 10^7.74 at 25 C), so we can calculate the amount of Al³⁺ that will be in equilibrium. 2. some of that Al³⁺ will react with SO₄⁼ to form the complex AlSO₄⁺, thereby reducing the Al³⁺ concentration , so more gibbsite will have to dissolve, until eventually equilibrium in both sequences is attained. The TOTAL aluminum in solution is (Al³⁺) + (AlSO₄⁺).

a. How much Al³⁺ will form? Al(OH)₃ + 3H⁺ <==> Al³⁺ + 3H₂O, K = 10^7.74, and (Al³⁺) = 10^-4.26 mole/kg.

b. How much AlSO₄⁺ will form? Al³⁺ + SO₄⁼ <==> AlSO⁺₄; K_stab = 10^3.01, so (AlSO₄⁺) = K (Al³⁺)(SO₄⁼) = 10^3.01(10^-4.26)(10^-3) = 10^-4.25 mole/kg.

Total aluminum = (Al³⁺) + (AlSO₄⁺) = 10^-4.26 + 10^-4.25 = 1.112 x 10^-4 mole Al/kg of solution = 0.11 mmole x 26.98 g/mole = 2.97 ppm.

7. a. What are the concentrations of each ion in mmol/kg?

b. Do the charges balance (total positive charges = total negative charges)?

c. What are the activity coefficients for Sr²⁺ and SO₄^2- in the water?

d. By how much is the water supersaturated or undersaturated with respect to SrSO₄?

a. To convert ppm to mmol/kg, divide each ppm value by the appropriate atomic or molecular weight:

b. Do the charges balance? Mmol of positive charges = 5.22 + 0.38 + 2(9.48) + 2(0.9) + 2(0.009) = 26.38. Mmol of negative charges = 2(11.61) + 0.42 + 2.46 = 26.10. Charges balance very closely; charge imbalance is (positive - negative)/(positive + negative) x 100 = 0.6%.

c. To calculate activity coefficients, we must first calculate the ionic strength (I) of the water: I = ½ Sum (c_iZ_i²) = ½(5.22 + 0.38 +(9.48)2² + (0.9)2² + (0.2)2² + (11.61)2² + 0.42 + 2.46) = 48.26 mmol/kg = 0.0483 mol/kg. Then use I to calculate the activity coefficients:  log Gamma = -AZ_i²I/(1 + Ba_oI)

log Gamma _Sr2+ = -0.5085(4)(0.0483)/(1 + (0.3281 x 10⁸)(5 x 10^-8)(0.0483)) = -0.3285 and so Gamma _Sr2+ = 10^-0.3285 = 0.469.

log Gamma _SO4⁼ = -0.5085(4)(0.0483)/(1 + (0.3281 x 10⁸)(5 x 10^-8)(0.0483)) = -0.3285 and so Gamma _SO4= = 10^-0.3285 = 0.469.

d. To determine the relative amount a solution is supersaturated or undersaturated with respect to a particular mineral, calculate the Saturation Ratio by: SR = (IAP/K_sp)^1/nu, where nu is the number of ions involved, in this case, 2. From Appendix III, K_sp = 10^-6.63 for SrSO₄. IAP = (concentration of Sr²⁺)(Gamma_Sr2+)(concentration of SO₄^2-)(Gamma_SO4=) = 10^-7.639. SR = (10^-7.639/10^-6.63)^½ = (10^-1.006)^½ = 10^-0.503 = 0.314. The solution is about 31% saturated with respect to SrSO₄.

8. Basically perform the calculation in #7 again, but this time use WATEQF to obtain the answer.

The answers should be pretty close. Differences may result because of:  1. WATEQF accounts for complexes in the solution, whereas we didn't in the manual calculation; complexes change the actual concentrations of the free ions used to calculate the ionic strength.  2. the log of the solubility product we used was -6.63, whereas WATEQF used a value of -6.58.

updated 15 Sept 1999