The Equilibrium Constant

We have the general reaction

dDcCbBaA

Equilibrium Constant• From this we can write an expression for

our equilibrium

Where we have the products over the reactants and each compound is represented by its concentration

ba

dc

BA

DCK

A Word on Units

• What are the units on K?

• From the formula we might guess that they would depend on the specific equilibrium involved.

• This is not the case however.

• Let us visit the concept of standard state.

Standard States• The terms and unit that we will use are

dictated by convention

• Solutions M

• Gas bar

• The values used in our equilibrium expression will be based on some reference value

Standard States• For solutions the standard state is 1.00M

• Gas 1.00 bar

• Pure solids or solvents the standard state is the substance.

• So if a solution is 2.31 x 10-3 M then the value we use in the equilibrium expression is....

Standard State

• 2.31 x 10-3 M / 1.00 M = 2.31 x 10-3 with the units factoring out.

• Since gas values are usually expressed as pressure instead of seeing [X], you will see a gas expressed as PX

• Different standard states could be picked but convention is well established on these terms in Analytical Chemistry.

Equilibrium Constants

• The equilibrium expression will depend on how the expression is written.

• As a dissociation HA = H+ + A-

[HA]

]][A[HK1

Equilibrium• But written as an association

H+ + A- = HA

• Which we can see that K1 = 1/ K2

• So if we write an equation in reverse we must take the reciprocal of the constant

]][A[H

[HA]K 2

Equilibrium

What about added equilibria.

H2S = H+ + HS-

then HS- = H+ + S2-

][

]][[

21 SH

HSHKa

][

]][[ 2

2

HS

SHKa

Equilibrium

• If we add the two reactions we get.

H2S = 2 H+ + A2-

Giving us the product of the equilibria.

][

][][

2

22

SH

SHK

Ta

Thermodynamics

• How does this all relate to “thermo”

• Two contributions– Enthalpy H, a measure of the heat of a

reaction. + endothermic, - exothermic– Entropy S, a measure of disorder in a system.

+ for more disorder, - for less disorder

ThermodynamicsWhen HCl (which is a gas) is bubbled into

water the following reaction occurs

H2O + HCl(g) = H+(aq) + Cl-(aq)

Which has a Ho of -75.15 kJ/mole @ 25 C

Heat can be viewed as ending up in bonds. What bonds are in this reaction?

The naught (o) means that the product and reactants are in the standard state.

Thermodynamics

• When we dissolve KCl, a salt similar to NaCl then we get an entropy effect. Since the ions in solution are more disordered than the ions in the solid.

• H2O + KCl(s) = K+ (aq) + Cl-(aq)

So is + 76 J/(K.mole) @ 25C and entropy favors this reaction

Thermodynamics

• But for the reaction of HCl with water the S value is negative. Why would that be the case?

• What happens if the terms work against each other.

Thermodynamics• This is where Gibbs free energy comes in.

G = H - TS• For our reaction with HCl in water what do we

get? G = -75.15 kJ/mole - (298K)(-131.5J/K.mole)

and calculating we get G = -35.94 kJ/mole• If G is negative then the reaction is favored.

Exergonic• If G is positive then the reaction is disfavored

Endergonic

Thermodynamics• If G is zero then the reaction is at

equilibrium.

• Remember that the values we have been looking at are in their standard state and these values will vary with concentration.

• What is the relationship between Go and K

RT

Go

eK

La Chatelier’s Principle

• If a change disturbs a equilibrium system that system will proceed back to equilibrium to partially offset the change.

• Use Q, same form as the equilibrium expression. If Q is less than K then the system must proceed to the right.

• If Q is more than K then the system must go to the left.

La Chatelier’s Principle

• Acetic acid in waterHAc = H+ + Ac-

• If you double the hydrogen ion concentration by adding a strong acid you will cause the amount of HAc to increase

51075.1][

]][[

xHAc

AcHK a

Effect of heat on K

• The equilibrium constant of an endothermic reaction increases if temperature is raised.

• The equilibrium constant of an exothermic reaction decreases if temperature is raised.

• Look at it this way

reactants = products + heat

Solubility • A very important parameter in many areas

of chemistry.– Likes dissolve likes– Polar compounds will dissolve in polar solvents

• sugar in water• alcohols in water

– Non polar compounds will dissolve in non polar solvents.

• oils in carbon tetrachloride• flavor in fats

Salts in water

• Some salts are ‘soluble’ in water and some are not.– Soluble

• NaCl KCl Na3PO4

– Not Soluble• AgCl LaF3 Cu(OH)2

• Yet all lead to varying amounts of ions in solution that are very polar.

Solubility Products• Let us look at an example

AgCl(s) = Ag+ + Cl-

• Ksp = [Ag+][Cl-]

• If we look up the value for this Ksp in appendix F we find that the value listed is 1.8 x 10-10, what is [Ag+] for a saturated solution?

• The Ksp for NaCl would be 49. What does that mean?

Ksp Problem

• What will be the concentration of La3+ if 1.00 grams of the solid LaF3 is added to 10 mL of water and allowed to come to equilibrium?

• Ksp = 2 x 10-19

• Best way to solve. Set up a table!

Ksp solutions

LaF3 = La3+ 3F-

Initial Conc. solid 0 0

Final Conc. solid x 3x

Where x is the concentration of La3+ at equilibrium

Ksp = [La3+][F-]3

Ksp Solutions

Ksp = [La3+][F-]3

From last slide [La3+] = x and [F-] = 3x

so 2 x 10-19 = x (3x)3 = 27x4

so x = 9 x 10-6 M La3+ and

what value for F- ? ____________________

Common Ion Effect

• We can figure out how much AgCl would dissolve in pure water. How much would dissolve if the solution already had some of either ion already in it?

• Let’s look at the more complex example however.

Common Ion SolutionsLet the initial Fluoride concentration be 0.001 M

LaF3 = La3+ 3F-

Initial Conc solid 0 0.001

Final Conc solid x 3x + 0.001

Where x is the concentration of La3+ at equilibrium

Common Ion

Ksp = [La3+][F-]3

From last slide[La3+] = x and [F-] = 3x + 0.001

so 2 x 10-19 = x (3x + 0.001)3 = quadric equation to solve. Fun!!!!!

What do we do??????

Common Ion

We know that the amount of F- we are going to get will be small. We will make an approximation. We will assume that x is small in the F- term. Giving2 x 10-19 = x (0.001)3

Which gives us x = 2 x 10-10 M for La3+

Now what is F-? Does this check out?

Approximations

• Knowing when to make approximations will be very helpful to us.

• When approximations are made then we must check our answer to make sure that this approximation make some sense.

• Is the final F- concentration close to the 0.001 M as we assumed? If not then we will need to solve without making the approximation which can be difficult.

Separation by Pcpt

• In theory this will work great.

• Co-precipitation will lead to significant error.

Complex Formation

• Many different equilibria can be going on at the same time. The second type is the formation of complexes. The example in the book is the formation of iodides of lead(II).

Complexes

• With lead and iodide we can have solid PbI2 formed. Ksp = [Pb2+][I-]2

• With extra added iodide we have complex formed.

• As iodide is added we form the mono, di, tri and tetra iodide species. Each step has and equilibrium constant. These are formation constants.

Complexes• Pb2+ + I- = PbI- K1

• Pb2+ +2I- = PbI2 2

• Pb2+ +3I- = PbI32- 3

• Pb2+ + 4I- = PbI42- 4

• Each step also has a stepwise function also.• PbI3 + I- = PbI4

2- K4

Complexes

• All equilibria must be satisfied at the same time. In this case here we can control the solubility of the PbI2 by added iodide ion in increasing amounts giving a mixture of the species that were mentioned.

Protic Acid and Bases

• Bronsted and Lowry – Acids are proton donors– Bases are proton acceptors

• We increase or decrease the amount of hydronium ion (H3O+) in solution with these acids and bases.

• We also have the Lewis acid base definition. Deals with electron pairs.

Salts

• When acids and bases are brought together they will neutralize each other. This produces a salt. Most salts added to water might not dissolve completely but will dissociate completely.

• NaCl Na2SO4 Na3PO4 NH4Cl

Conjugates

• Products of acid/base reactions are also classified as acid/base.

• Acetic acid + Ammonia = Acetate Ion + Ammonium ion

Acetic acid acid

Acetate ion conjugate base

Ammonia base

Ammonium ion conjugate acid

Water

• Water is our most common solvent. Its behavior and properties are important to world as we know it.

• Water can either gain a proton or lose a proton.

• We can use the hydronium ion H3O+ in our reactions or we often get lazy and just use H+

Water

• Water can react with itself. One water could donate a proton and another water could lose that proton. This is called autoprotolysis.

• H2O + H2O = H+3O + OH-

• Other protic solvents can do this ie. Acetic acid• 2CH3COOH = CH3COOH2

+ + CH3COO-

• There are solvents that are called aprotic. These solvents do not give up or accept the proton

• Diethyether, hexane, acetonitrile, toluene.

Water

• Autoprotolysis Constant• H2O = H+ + OH- has a special equilibrium

constant. Kw

• Kw = [H+][OH-] Kw = 1.01X 10-14 @25C• Kw varies with temp. We will often see this

expressed on the log scale.• pKw = -log Kw

Water

• What is pH? • pH = -log[H+]• What is the pH of pure water.

For each H2O that dissociates we get one H+ and one OH-,

so [H+] = [OH-]

Kw = [H+]2 = 1 x 10-14

H+ = 1.00 x 10-7

pH = 7.0

Water will usually have traces of CO2 and other ions that change the pH

Acid and Base Strength

• Some acids and base are considered strong! This means that they fully dissociate in water.– Acids HCl, H2SO4, HNO3, HClO4

– Bases NaOH, KOH

• Many other acid are weak. This only means that they do not fully dissociate in water. Do not infer that they are less hazardous.– Acids Acetic, H3PO4, Benzoic etc.

– Bases Ammonia, other amines.

Weak Acids

• The dissociation of weak acid have and equilibrium constant called Ka. If more that one proton can dissociate, i.e. H3PO4 each added loss will have its own equilibrium constant.

Acid/Conjugate Relationship

• Acetic acid HAHA = H+ + A-

Ka = [H+][A-]/[HA]

• The conjugate base. NaA Sodium AcetateNaA + H2O = HA + OH- + Na+

Kb = [HA][OH-]/[A-]

Acid/Conjugate Relationship• What is Ka*Kb

Ka*Kb = [H+][A-]/[HA]*[HA][OH-]/[A-] = [H+][OH-] = Kw

So once you know the Ka for the acid form you know the Kb for the conjugate base. It must be the same conjugate pair.

Goodbye bases, You will note that there are no Kb values listed in the Appendix in the text. This the the way that biochemists look at the world.