Aqueous geochemistry of acid mine drainage

N.V. Sidenko & B.L Sherriff

Objective:

Understand major factors controlling metal mobility concentrations and pH.

Predict changes in pH and metal concentration using equilibrium calculations

Outline:

Equilibrium

pH, neutralization/acidification

Complexation in the solution

Solubility of metal solids

Ksp, IAP and saturation index,

System, closed, open, components, phases, species.

Modeling:

Active treatment (lime use)

Equilibrium

R_b

H₂O <--> H⁺ + OH^-

_reactant R_{f products}

EQ when R_f =R_b

EQ constant , K = activity of products/activity of reactant

In diluted solutions, activity is equal to concentration measured in mole per liter (mole/L or M)

Molar concentrations is input for most software

Law of mass action

K = K_{W (water dissociation)} = [H⁺][OH^-]/[H₂O]

a _H2O = ?

K_W = [H⁺][OH^-]/[H₂O]= 10^-14, in neutral solution [H⁺]=[OH^-]= 10^-? M

pH = -Log[H⁺], in neutral solution pH = -Log10^-? = ? M

What is [H⁺] in stream with pH=3?

Answer:

Lime slurry has pH=10, what is [OH-] concentration

Answer:

Acidification -when H⁺ is produced or OH^- is consumed by reaction:

In AMD, acid is produced

1) by sulfide oxidation,

FeCuS_{2(chalcopyrite)} + 7/2O2 +H₂O -> Fe²⁺ + Cu²⁺ +2SO₄^2- + 2H⁺

then one H⁺ is consumed (left side of the reaction) by Fe oxidation

Fe²⁺ + 1/4O₂ + H⁺ -> Fe³⁺ +1/2H₂O

2) by complexation and precipitation, when some OH^- is taken out of water molecule

Fe³⁺ + 2H₂O -> Fe(OH)₂⁺_{(aqueous complex with OH-)} +2H⁺

Fe³⁺ + 3H₂O -> Fe(OH)_{3(Fe-hydroxide precipitate)} +3H⁺

Neutralization – when H⁺ is consumed or OH^- is produced by reaction

When:

Ca²⁺ + 2OH^- _{(slaked lime)} + 2H⁺ -> Ca²⁺ + 2H₂O (hydroxide addition)

H⁺+ CaCO_3(calcite) -> Ca + HCO₃^- (_{aqueous complex with H+) ,} or reaction with salt of strong base and weak acid)

Q; Find examples of minerals which are salts of strong base and weak acid

How much grams of calcite do we need to neutralize 1 liter of solution from ph 3 to pH 7.

a)Moles acid to be neutralized = 10^-3 - 10^-7 (_{very small copared to 10-3}) ~ 10^-3 M (moles/L)

b) 1 mole of calcite neutralizes 1 mole of acid (see reaction above) means ~ 10^-3 moles of calcite (M CaCO₃ = 100 g/mole)

c) convert moles to grams

Metal complexes in the solution:

in solution, ions of opposite charge tend to associate and form complexes:

Fe(OH)₂⁺, Al(OH)₃^o_aq, PbSO₄ ^o_aq, HSO₄^-, FeHSO₄⁺²

Dissolution/precipitation of metal solids. pH dependence:

Dissolution of metal hydroxides is key factor controlling metal migration and concentration during the treatment.

We assume that all metals (consider Zn as example) during lime treatment will precipitate in form of hydroxide. Metal hydroxides are considered in equilibrium with solution.

At what pH solubility of c is lowest, meaning lowest Zn concentrations.

Write Zn(OH)_2(solid) solubility equation

Zn(OH)2(solid) + 2H+ -> Zn + 2H2O;

K = [Zn^2+]/[H^+]^2 = 10^11.5;

LgK=11.5

Take log of the law of mass action: LgK=Lg[Zn²⁺] – 2log[H⁺]= Lg[Zn²⁺] + 2pH

Rearrange for Lg[Zn²⁺]:

Lg[Zn^2+] = LgK- 2pH = 11.5 - 2pH

(a)

Plot this line in coordinates pH- Log[Zn²⁺]

At pH 4 [Zn²⁺] = 10^3.5M; at pH 8 [Zn²⁺] = 10^-4.5M

However it will be concentration of free Zn ion or Zn²⁺, which is not the same as total Zn, which is measured (and regulated in most cases). Reason is formation of complexes. Hydroxide complexes can be suspected at first because reaction occurs in the water.

Zn^+2 + H2O = ZnOH^+ + H^+;

K = [H^+][ZnOH^+]/[Zn^2+] = 10^-9;

LgK=-9.0

Take log of the law of mass action:

LgK= Lg[H⁺] + Lg[ZnOH⁺] - Lg[Zn²⁺]

= - pH + Lg[ZnOH⁺] - Lg[Zn²⁺]

Solve for Lg[ZnOH⁺]; Lg[ZnOH⁺] = pH + Lg[Zn²⁺] + LgK

Lg[ZnOH^+] = Lg[Zn^2+] + pH -9.0

(b0)

Zn⁺² concentrations are controlled by equilibrium with Zn(OH)_2(solid). Therefore, substitute for Lg[Zn²⁺] right side of equation (a)

Lg[ZnOH^+] = 11.5 - 2pH + pH -9.0	(b1)
Lg[ZnOH^+] = 2.5 - pH	(b2)

Plot concentration of complex vs pH,

At pH 4 [ZnOH⁺] = 10^-1.5M; at pH 8 [ZnOH⁺] = 10^-5.5M, in both cases [ZnOH⁺] lower than [Zn²⁺]

Another complex can have one more hydroxide:

Zn^+2 + 2H2O = Zn(OH)2^oaq + 2H^+;

K = [H^+]^2[Zn(OH)2^o]/[Zn^2+] = 10^-17.0;

Take log law of mass action LgK= 2Lg[H⁺] + Lg[Zn(OH)₂^o] - Lg[Zn²⁺]

Solve for Lg[Zn(OH)₂^o]; Lg[Zn(OH)₂^o] = Lg[Zn²⁺] - 2Lg[H⁺] + lgK

Lg[Zn(OH)2^o] = Lg[Zn^2+] + 2pH -17.0

(c0)

substitute for Lg[Zn²⁺] right side of equation (a)

Lg[Zn(OH)2^o] = -5.5

(c1)

Concentration of Zn(OH)₂^o_aq = 10^-5.5M and does not depend on pH and this line will be parallel to pH axis

at pH > 8.5, [Zn(OH)₂^o_aq] > [Zn²⁺] (10^-5.5M). Thus, pH just above 8.5 will require minimum use of lime for neutralization and achieve MINIMUM total concentration of Zn 10^-5.5M.

Another complex can have one more hydroxide in alkaline conditions:

Zn^+2 + 3H2O = Zn(OH)3^- + 3H^+;

K = [H^+]^3[Zn(OH)3^-]/[Zn^2+] = 10^-28.4;

Take log law of mass action LgK= 3Lg[H⁺] + Lg[Zn(OH)₃^-] - Lg[Zn²⁺]

Solve for Lg[Zn(OH)₃^-]; Lg[Zn(OH)₃^-] = Lg[Zn²⁺] - 3Lg[H⁺] + lgK

Lg[Zn(OH)2^o] = Lg[Zn^2+] + 3pH - 28.4=11.5 - 2pH3pH - 28.4

(c0)

substitute for Lg[Zn²⁺] right side of equation (a)

Lg[Zn(OH) 3^-] = pH - 17.9

(c1)

complex concentration increases with pH,

over pH 12.4, [Zn(OH) ₃^-] > Zn(OH)₂^o (10^-5.5M), this means that if we increase pH further solubility of Zn(OH)_{2 solid} will increase.

Thus, reaction Zn(OH)₂^o_aq <=> Zn(OH)_2(solid) will control MINIMUM total concentration of Zn 10^-5.5M = 3.1x10^-6M What is that in mg/l? This reaction will dominate between pH 8.5 and 12.4.

Ksp, IAP and saturation index

If we know concentration of ions in the solution we can predict possible metal solids which can be in equilibrium with solution (not always 100%). Consider solution is in equilibrium with anglesite

PbSO_{4 (solid, anglesite)} <->Pb²⁺ + SO₄^2-

Activity of PbSO_{4 solid}?

K=K_{sp (solubility product)} = [Pb²⁺][SO₄^2-]=10^-7.79 = tabulated value; LogK_sp =-7.79

Your solution has 10^-1M of [SO₄^2-] and 10^-6.8 M of [Pb²⁺],

Ion Activity Product (IAP)= [Pb²⁺][SO₄^2-]=10^-1 x 10^-8 = 10^-7.8; LogIAP =-7.8

Saturation Index (SI) is comparison of tabulated solubility to ion concentrations in the solution.

SI _anglesite = Log (IAP/K_sp); if IAP= K_sp then SI= 0

Log (IAP/Ksp) = Log(IAP) – LogK_sp=-9 – (-7.79) = -1.21.

Solution is undersaturated with respect to anglesite (it cannot precipitate). Normally, if saturation index is 0±0.5, solid is considered in equilibrium in the solution.

System: volume surrounded by boundaries:

Closed system – maintain initial mass (confined groundwater)

Open system – open to mass transfer trough boundary.(surface water)

AMD normally open to atmosphere PO₂ = 0.2 PCO₂ = 10^-2.5

Phases :solid, gases and solution.

Species: forms of the element (in soulution in our case) ions or complexes.

Fe²⁺,Fe³⁺, FeSO₄^-, Fe(OH)₂⁰

Figure: concentration of aqueous Zn species at different pH in equilibrium with solid Zn(OH)₂.

Assignment, class &home:

1) Open file Sat_ind.phrq. Solution represents seepage at toe of tailings.

Change database to wateq4f.dat.

a) Run the program, identify, and list the most significant aqueous complexes for Fe, Al, Cu, Zn, Ni and Pb.

b) List minerals of Fe, Al, Cu, Zn, Ni and Pb, which might be in equilibrium with solution.

2) Open file Lime_tr.phrq. We need to estimate how much lime (Ca(OH)2 or Portlandite) should be added to reach saturation of a) Al(OH)₃(a), b) Cu(OH)₂, c) Zn(OH)₂-g, c) Ni(OH)₂, d) Pb(OH)₂ (consider 0<SI<1)? Vary amount of lime (Ca(OH)2 or Portlandite) and see how SI of the metal hydroxides will change. System open to atmosphere during the treatment, therefore pressures of gases are fixed, Fe(OH)₃ and gypsum are minerals which expected to precipitate. Therefore, all these phases are in equilibrium with solution.

Indicate pH at saturation of every hydroxide (a to d). What pH should be reached to treat all these metals?

e) List the most significant aqueous complexes for Fe, Al, Cu, Zn, Ni and Pb when the mental hydroxides precpitate.

Home assignment.

3) Construct diagram solubility of Al(OH)₃ solid in coordinates pH – lg[Al form in solution] using following data:

Al(OH)3(solid) + 3H^+ = Al^+3 + 3H2O	log_k 10.8
Al^+3 + H2O = AlOH^+2 + H^+	log_k -5.0
Al^+3 + 2H2O = Al(OH)2^+ + 2H^+	log_k -10.1
Al^+3 + 3H2O = Al(OH)3^0aq + 3H^+	log_k -16.9
Al^+3 + 4H2O = Al(OH)4^- + 4H^+	log_k -22.7

Define lowest pH at lowest concentration of Al in the solution in equilibrium with Al(OH)_3.