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Solubility of Salts. A saturated solution of a salt is in a state of equilibrium The solubility of a salt is the amount that dissolves at equilibrium Usually reported as weight/volume (g/L or mg/L)

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solubility of salts
Solubility of Salts
  • A saturated solution of a salt is in a state of equilibrium
  • The solubility of a salt is the amount that dissolves at equilibrium
    • Usually reported as weight/volume (g/L or mg/L)
    • Can determine from equilibrium expression and amount of ions produced (use stoichiometric relationships)
  • Use saturation index to determine whether a solution is saturated with respect to a mineral
  • Can predict how much salt must dissolve or precipitate to reach equilibrium
    • Point at which IAP = Ksp
changing solution composition due to precipitation
Changing solution composition due to precipitation
  • As salts precipitates, ratios of ions changes
    • The precipitation of a salt reduces the concentrations of ions and changes the chemical composition of remaining solution
  • The initial ratio of species can affect which minerals precipitate
    • GEOCHEMICAL DIVIDE
    • End up with a different final solution
    • May lead to precipitation of different minerals
precipitation of salts in natural waters
Precipitation of Salts in Natural Waters
  • If 2 minerals have a common ion, we can determine the ratio of the couterions at equilibrium
    • The more insoluble mineral contributes a relatively small amount of the common ion
  • Replacement of 1 mineral by another is common in geology
    • Introduction of a common ion causes solution to become supersaturated with respect to the less soluble compound
    • Thus the more soluble compound is always replaced by less soluble
    • Makes sense: less soluble happier as solid, more soluble happier dissolved (relatively)
supersaturation
Supersaturation
  • Solutions in nature become supersaturatedwith respect to a mineral by:
    • Introduction of a common ion
    • Change in pH
    • Evaporative concentration
    • Temperature variations
      • In general solubilities increase with increasing T, but not always (e.g., CaCO3)
calcite solubility
Calcite Solubility
  • Dissolution of calcite done primarily by acid
    • In natural systems, primary acid is CO2
    • CO2(g) CO2(aq)
    • CO2(aq) + H2O  H2CO3
    • CaCO3 + H2CO3 Ca2+ + 2HCO3-
    • Increasing PCO2 increases H2CO3, which increases amount of CaCO3 dissolved
    • Calcite cannot persist in even mildly acidic waters
    • Solubility of calcite decreases with increasing temperature
incongruent dissolution
Incongruent Dissolution
  • KAlSi3O8 + 9H2O + 2H+ Al2Si2O5(OH)4 + 2K+ + 4H4SiO4
    • If products such as K+and/or H4SiO4are removed by flowing groundwater, achieving equilibrium (saturation) may not be possible
  • Water/rock ratio is a key variable whether a chemical reaction achieve equilibrium in nature
    • The higher the water/rock ratio, the more likely the reaction goes to completion, not equilibrium
      • Products removed
    • If the ratio is small, the reactions can control the environment and equilibrium is possible
geochemical cycles
Geochemical Cycles
  • Material is being cycled continuously in the Earth’s surface system
    • All chemical elements are cycled
  • We can think of the Earth’s surface as consisting of several reservoirs connected by “pipes” through which matter moves
transfers between reservoirs
Transfers between reservoirs

d4,1n

dt

Reservoir 1

d1,2n

dt

Reservoir 4

Reservoir 2

d3,4n

dt

d2,3n

dt

Reservoir 3

n = component concentration

t = time

dn = rate of transfer of a component

dt from one reservoir to another

= Flux

steady state
Steady State
  • Steady state means that the composition of reservoirs in a cycle does not change over time
    • No accumulation or loss of the material of interest
    • Input + any production in the reservoir = outflow + any consumption in the reservoir
    • The mass balance = 0; no creation or loss of material
  • In the Earth surface environment, especially the oceans and atmosphere, cycling of materials has been occurring at near steady-state
steady state1
Steady State

Reservoir 1

d1,2n

dt

d2,3n

dt

d1,2n

dt

=

Reservoir 2

d2n

dt

= 0

d2,3n

dt

Reservoir 3

residence time
Residence Time
  • Residence time is the average time a molecule spends in a reservoir between the time it arrives and the time it leaves
  • Determined by dividing the amount in the reservoir by the flux in (or out)

2n

d1,2n

dt

T =

cycles and reaction rates
Cycles and reaction rates
  • As materials cycle through the Earth, they are moved and transformed at various rates (the “pipes” connecting reservoirs)
  • Transport processes and chemical reactions take time
  • Kinetics is the study of reaction rates
kinetics vs thermodynamics
Kinetics vs. Thermodynamics
  • Thermodynamics tells us what should happen
    • All reactions tend to move towards equilibrium
    • Not concerned with time or steps involved in reaction
    • Only starting and ending points are of interest
    • Reactions cannot occur in contradiction to thermodynamics
  • Kinetics tells us whether a reaction actually does occur in a reasonable interval of time
    • Most geochemical reactions take place slowly
    • The reaction pathway is important
    • Kinetics is the study of what goes on in between the start and end points
reaction pathway
Reaction Pathway
  • CaCO3 + H2CO3 Ca2+ + 2HCO3-
  • This is the overall reaction for calcite dissolution; there are actually many steps:
    • Dissolution and hydration of CO2(g) (→ H2CO3)
    • Transport of H2CO3 to mineral surface
    • Adsorption of H2CO3 to mineral surface
    • Detachment of products (Ca2+, HCO3-) from mineral surface
    • Transport of products to the bulk solution
  • There is a rate associated with each of these steps, and the rates may not be constant
reaction rates
Reaction Rates
  • Controlled by a variety of factors
    • Temperature
      • Generally rates increase as T increases
    • Concentrations of reactants and products
    • Mass transfer rates
    • Etc.
  • An overall (complete) reaction is the end product of several elementary reactions
    • Elementary reactions = reaction that actually occurs as written at molecular level
    • Each elementary reaction proceeds at its own rate
    • The rate of an overall reaction is controlled by the slowest step in the process
overall vs elementary reactions
Overall vs. Elementary Reactions
  • Consider dissolution of nephaline to produce gibbsite
  • Elementary reactions:
    • NaAlSiO4(s) + 4H+ Al3+ + Na+ + H4SiO4
    • Al3+ + 3H2O  Al(OH)3(s) + 3H+
  • Overall reaction:
    • NaAlSiO4(s) + H+ 3H2O  Al(OH)3(s) + Na+ + H4SiO4
    • Al3+ is a reactive intermediate that doesn’t persist
    • Because it is being both produced and consumed, Al3+ never reaches equilibrium concentration with respect to either nephaline or gibbsite
  • Add elementary reactions to get overall reaction
groundwater and partial equilibrium
Groundwater and Partial Equilibrium
  • Low temperature systems (groundwater) can usually be described by partial equilibrium
    • For faster reactions, there is equilibrium between the water and rocks
    • For slow reactions, equilibrium is not achieved
  • Residence time: the longer water is in contact with rock, the higher the probability of equilibrium being reached.
  • In general (for near surface groundwater):
    • High temperature primary minerals unstable at near surface temperatures, are rarely in equilibrium with groundwater
      • Remember Bowen’s reaction series
    • Low temperature secondary minerals are generally at or near equilibrium
reaction pathway1
Reaction Pathway
  • Each step in an overall reaction has a rate
  • Oftentimes there is one step significantly slower than the others
    • The complete reaction can proceed only as fast as the slowest individual step
    • This step controls the overall rate; called the rate-determining step
rate determining step
Rate-Determining Step
  • Climb rungs
  • Sit
  • Slide
  • Go back to ladder

“Reaction” Steps

reaction rates and equilibrium
Reaction Rates and Equilibrium
  • Rates decline with time as a reaction reaches equilibrium or goes to completion
  • In other words, the further away from equilibrium a reaction is, the more rapidly it occurs (in relative terms)
slide24

Slow

Fast

reaction order
Reaction Order
  • The rate of change of the concentration of a reactant generally varies as some power of its concentration
  • X  Y + Z
  • The consumption of reactant X
    • -d(X) = k(X)n

dt

      • (X) = concentration of reactant X
      • t = time
      • n = reaction order (determined experimentally)
      • k = specific rate constant (determined experimentally)
      • Negative sign indicates the rate decreases with time as X reacts; as X is used up there is less of it to react so the rate slows down
reaction order1
Reaction Order
  • Reaction order determined by measuring how the initial reaction rate varies with the initial concentration of X
    • -d(X) = k(X)1: first order

dt

    • -d(X) = k(X)2: second order

dt

    • -d(X) = k(X)0: zeroth order

dt

      • (rate doesn’t depend on reactant concentration)
    • Could be a fraction as well (e.g., 1/2)
reaction order examples
Reaction Order Examples
  • Zero:
    • Biodegradation when organic compound concentration is high
  • 1st:
    • Radioactive decay
    • Biodegradation at lower concentrations
reaction rates1
Reaction Rates
  • As reactants react, products are produced
  • Rates can be expressed in terms of an increase in concentration of a product as well as the decrease in a reactant
rate constants
Rate Constants
  • X  Y + Z
  • Rate constants are determined experimentally by measuring concentration of reactant at known intervals of time
rate constants1
Rate Constants
  • X  Y + Z
  • Rate constants are determined experimentally by measuring concentration of reactant at known intervals of time
      • k = rate constant; the larger it is, the faster the reaction
        • For 1st order rate, units = time-1
      • n = reaction order
rate constant1
Rate Constant
  • Plot ln(X) vs. t, get a straight line
  • Negative slope, concentration of reaction X decreases with time
  • Absolute value of slope = k
slide37

Slope = -0.079

k = 0.079 sec-1

temperature dependence of rates
Temperature dependence of rates
  • Temperature has a strong influence on reaction rates
  • Consider X + Y  Z
    • X and Y must collide with sufficient energy to form a bond
    • Minimum energy is activation energy (Ea)
    • Ea acts as a barrier that must be overcome
  • Reaction rates ultimately depend on frequency with which atoms/molecules of reactants achieve necessary energy to form bonds
activation energy e a1
Activation energy (Ea)
  • Ea explains why quartz does not usually control aqueous SiO2 concentrations
    • Equilibrium thermodynamics say it should
    • However, there is a high Ea for quartz dissolution that results in a long rate
    • Also helps explain why quartz is very resistant to chemical weathering
e a and arrhenius equation
Ea and Arrhenius Equation
  • Ea can be determined experimentally
  • The Arrhenius Equation relates reaction rate to Ea and T:
      • k = rate
      • A = experimental constant for a specific reaction
      • R = gas constant
      • T in Kelvin
  • k ~doubles for every 10°C increase
e a and arrhenius equation1
Ea and Arrhenius Equation
  • Can re-write as
    • Get a straight line by plotting 1/T vs. log k
      • y = mx + b
      • Slope =
      • y intercept = log A
radioactive decay
Radioactive decay
  • Radioactive decay follows a first-order rate law
    • A = rate of disintegration
    • k = rate constant
    • t = time
radioactive decay1
Radioactive decay
  • A = A0e-kt
    • Ao = initial radioactivity
    • Sometimes λ is used instead of k
      • Decay constant
half lives
Half Lives
  • ln(1) – ln(2) = -kt½
    • ln(1) = 0
    • t1/2 = half-life; the length of time required to remove half the initial concentration
      • these are known with precision
half lives1
Half Lives
  • ln(1) – ln(2) = -kt½
    • ln(1) = 0
    • t1/2 = half-life; the length of time required to remove half the initial concentration
      • these are known with precision
tritium
Tritium
  • Half life = 12.3 yr
  • Decay constant?
    • k = 0.693/12.3 = 0.0563 yr-1
  • How long before 75% gone?
    • t = -ln(0.25) / (0.0563) = 24.6 yr
mass transfer
Mass Transfer
  • There are 2 main processes transferring mass in the subsurface
    • Advection
    • Diffusion
advection
Advection
  • Advection involves the displacement of matter in response to action of a force; it is a Flux (mass per time)
  • Groundwater flow = advection
    • Movement may provide reactants and remove products for reactions
    • Darcy’s Law is the governing law for groundwater flow
darcy s law
Darcy’s Law
  • F = K(h/L)
    • h/L = the hydraulic gradient
    • h = change in head; difference in the water elevation (measured in boreholes) between 2 points of an aquifer
    • L = the distance between those 2 points measured along the flow path
    • K = hydraulic conductivity (units = distance/time)
      • K is a measure of how easily water can flow through the rock or sediments; it is related to permeability
hydraulic conductivity
Hydraulic Conductivity
  • Hydraulic conductivity (K) is related to permeability
  • K covers 12 orders of magnitude for geologic materials
diffusion
Diffusion
  • Molecules/ions of a solute in a solution tend to disperse through a stationary solvent by diffusion
  • In groundwater, the slower the rate of advection, the more important diffusion becomes
  • Most important in aquitards (very low flow geologic units), such as clays
redox introduction
Redox: Introduction
  • Redox reactions occur when an element changes its valence state
    • Some elements have 2 or more valence states
    • e.g., Iron can be +3, +2, or 0
    • Nitrogen can be +5, +4, +3, +2, +1, 0, -3
  • Electrons (e-) are transferred from one compound to another
    • e--donors (lose electrons)
    • e--acceptors (gain electrons)
oxidation vs reduction
Oxidation vs. Reduction
  • Loss of e- by e--donor = oxidation
  • Gain of e- by e--acceptor = reduction
  • Oxidation and reduction always accompany one another
    • Electrons cannot exist freely in solution
electron donors and acceptors
Electron donors and acceptors
  • Electron-donors vs. e--acceptors
    • Some elements always give up or accept same # of e- = fixed valences
      • Alkali metals and alkaline earths give up (donate) 1 or 2 e-
      • Halogens accept 1 e-
    • Some elements can lose varying numbers of electrons depending on e--acceptor strength
    • Others can take up varying numbers depending on e- availability in environment
slide58

Halogens

Alkali

Metals

Alkaline

Earths

valence numbers
Valence numbers
  • Use valence number = electrical charge an atom acquires if ions are in aqueous solution
    • e.g., H+, Cl-, Fe2+, Fe3+
  • Conventions for valences numbers:
    • All elements in pure form = 0 (e.g., O2)
    • H = +1 except in metal hydrides (e.g. LiH) = -1
    • O = -2 except in peroxides (H2O2) = -1
    • Valence numbers assigned to elements in molecules or complex ions
      • Algebraic sum of atoms’ valence states
      • Total of neutral molecule = 0 or equals charge of ion
valence numbers examples
Valence numbers (examples)
  • Sulfate (SO42-)
    • O is -2
    • 4 x (-2) = -8
    • So S must be +6 [-2 - (-8)]
  • Methane (CH4)
    • H is +1
    • 4 x (+1) = +4
    • So C is -4
  • Nitrate (NO3-): what is valence of N?
  • Ammonium (NH4+): what is valence of N?
balancing redox reactions
Balancing Redox Reactions
  • e.g., Fe + Cl2 Fe3+ + 2Cl-
    • Fe loses 3 e- (oxidized; charge increases)
    • Cl gains 2 e- (reduced; charge decreases)
    • Reaction is mass balanced, but not charge balanced
    • To balance charge, cross multiply reactants
      • Fe x 2 and Cl2 x 3
    • 2Fe + 3Cl2 Fe3+ + 2Cl-
    • Thenbalance mass on right side of reaction
    • 2Fe + 3Cl2 2Fe3+ + 6Cl-
      • Both mass and charge balanced
half reactions
Half Reactions
  • Redox reactions can also be written as half reactions, with e- included
    • Fe  Fe3+ + 3e- [oxidation half-reaction]
    • 2e- + Cl2 2Cl- [reduction half-reaction]
  • Multiply so that e- are equal (LCM)
    • 2(Fe  Fe3+ + 3e-) = 2Fe  2Fe3+ + 6e-
    • 3(2e- + Cl2 2Cl-) = 6e- + 3Cl2 6Cl-
  • Combine:
    • 2Fe + 3Cl2 2Fe3+ + 6Cl-
balancing redox reactions1
Balancing Redox Reactions
  • Pyrite oxidation to Fe(OH)3 in the presence of oxygen:
    • FeS2 + O2 Fe(OH)3 + SO42-
  • Determine valence states
    • FeS2:
      • Fe = +2, S = -1
    • O2:
      • O = 0
    • Fe(OH)3:
      • Fe = +3, O = -2
    • SO42-:
      • S = +6, O = -2
balancing redox reactions2
Balancing Redox Reactions
  • FeS2 + O2 Fe(OH)3 + SO42-
  • Determine the number of electrons transferred
    • Fe:
      • +1 (from Fe2+ to Fe3+)
    • S:
      • -7 (from S- to S6+); multiply by 2 = 14
    • So for every mole of FeS2, 15 electrons are transferred
    • O:
      • +2 (from O to O2-); multiply by 2 = 4
  • So 15 electrons donated per mole of FeS2, 4 electrons accepted per mole of O2
    • Note: Fe and S are e--donors, O2 e--acceptor
balancing redox reactions3
Balancing Redox Reactions
  • Next, cross multiply reactants to account for electrons
    • 4FeS2 + 15O2 Fe(OH)3 + SO42-
  • Then balance the non-O and H elements on the right side (Fe and S)
    • 4FeS2 + 15O2 4Fe(OH)3 + 8SO42-
  • Then balance O using H2O
    • 4FeS2 + 15O2 + 14H2O  4Fe(OH)3 + 8SO42-
  • Finally balance H and charge using protons (H+)
    • 4FeS2 + 15O2 + 14H2O  4Fe(OH)3 + 8SO42- + 16H+
balancing redox reactions4
Balancing Redox Reactions
  • Can also do using Half Reactions
    • FeS2 + 11H2O  Fe(OH)3 + 2SO42- + 19H+ + 15e-
    • O2 + 4H+ + 4e- 2H2O
  • Multiply so that e- drop out
    • 4(FeS2 + 11H2O  Fe(OH)3 + 2SO42- + 19H+ + 15e-)
    • 15(O2 + 4H+ + 4e- 2H2O)
  • When half reactions are added
    • 4FeS2 + 15O2 + 14H2O  4Fe(OH)3 + 8SO42- + 16H+