Many analytical methods rely on equilibrium systems in aqueous solution. This unit will review General concepts of aqueous solutions Chemical equilibrium Equilibrium calculations Deviations from ideal behavior

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Many analytical methods rely on equilibrium systems in aqueous solution.

This unit will review

General concepts of aqueous solutions Chemical equilibrium

Equilibrium calculations Deviations from ideal behavior

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acid base conjugate

base of HF conjugate

acid of H₂O base acid conjugate

base conjugate

acid

acid strength base strength

For the general chemical reaction:

aA + bB cC + dD If A and B are brought together:

There is an initial reduction in the concentrations of A and B.

Both C and D will increase in concentration.

We reach a point where the concentrations no longer change.

A B

C D

time

kinetic equilibrium region region

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These are dynamic equilibria

At equilibrium, the forward and reverse rates of reaction are equal.

Any given species is constantly changing from one form to another.

Changes in the system will alter the rates and a new equilibrium will be achieved.

Equilibrium concentrations are based on:

The specific equilibrium The starting concentrations Other factors such as:

Temperature Pressure

Reaction specific conditions Altering conditions will stress a system,

resulting in an equilibrium shift.

K_eq = a_C^c a_D^d

a_A^a a_B^b a = activity

K_eq = [ C ]^c [ D ]^d [ A ]^a [ B ]^b

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We’ll be using molar concentrations when working with chemical equilibrium.

This introduces errors that you should be aware of.

While we will not work with activities, we need to know what they are.

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Except for dilute systems, the effective

concentration of ions is usually less than the actual concentration.

The term activity is used to denote this effective concentration.

activity a_i = f_i [ i ] where

f_i is the activity coefficient for i [ i ] is the molar concentration

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0.5 Z_i² µ ! 1 + 0.33 "iµ!

For very dilute solutions As f_i -> 1, a_i -> [ i ] For µ up to 0.1

f_i < 1 and a_i < [ i ] When u is > 0.1

Results in complicated behavior.

In general, if µ < 0.01, we can safely use molar concentrations.

K_eq = [ H₃O⁺ ] [ OH^- ] [ H₂O ]²

In dilute solutions, [ H₃O⁺] and [ OH^- ] is much smaller than [ H₂O ]. [ H₂O ] is essentially a constant of

~

55.5 M.

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K

_W

= 10

^-14

= [H

₃

O

⁺

] [OH

^-

^-8

) ( x)

Our equation can be rearranged as x² + 10^-8 x - 10^-14 = 0

This quadratic expression can be solved by:

x =

Only the positive root is meaningful in equilibrium problems.

-b + b² - 4ac 2a

x = -10^-8 + [ (10^-8)² + 4x10^-14]^1/2 / 2 = 1.9 x 10^-7 M

pH = 6.72

So adding a small amount of HCl to water DOES make it acidic.

While this approach is more time

consuming, you’ll find it very useful as our problems get more complex.

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K_SP expressions are used for ionic materials that are not completely soluble in water.

Their only means of dissolving is by dissociation.

AgCl_(s) Ag⁺_(aq) + Cl^-_(aq)

K_eq = [ Ag⁺_(aq) ] [ Cl^-_(aq)] [ AgCl_(s) ]

At equilibrium, our system is a saturated solution of silver and chloride ions.

The only way to know that it is saturated it to observe some AgCl at the bottom of the solution.

As such, AgCl is a constant and K_SP expressions do not include the solid form in the equilibrium expression

Determine the solubility of AgCl in water at 20^oC in grams / 100 ml.

K_SP = [Ag⁺] [Cl^-] = 1.0 x 10^-10 At equilibrium, [Ag⁺] = [Cl^-] so

1.0 x 10^-10 = [Ag⁺]² [Ag⁺] = 10^-5 M

g _AgCl = 10^-5 mol/l * 0.1 l * 143.32 g/mol Solubility = 1.43x10^-4 g / 100 ml

K_SP = 1.1 x 10^-12 = [ Ag⁺ ]² [ CrO₄^2- ] [CrO₄^2-] = [CrO₄^2-]Ag₂CrO₄ + [CrO₄^2-]Na₂CrO₄

With such a small value for KSP, we can assume that is [CrO₄^2-]Ag₂CrO₄ negligible.

If we’re wrong, our silver concentration will be significant (>1% of the chromate concentration.) Then you’d use the quadratic approach.

K_SP = 1.1 x 10^-12 = [ Ag⁺ ]² [ CrO₄^2- ] [CrO₄^2-] = 0.010 M

[ Ag⁺ ] = ( K_SP / [ CrO₄^2- ] ) ^1/2 = 1.1 x 10^-12 / 0.010 M = 1.1 x 10-5 M [ Ag⁺ ] << [ CrO₄^2- ]

so our assumption was valid.

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[H₃O⁺] [A^-] [HA]

[OH^-] [BH⁺] [B]

Water is not included in the

expressions because it is a

constant

[ H₃O⁺ ] [ A^- ] [ HA ]

Since both a H₃O⁺ and a A^- is produced for each HA that dissociates:

[ H₃O⁺ ] = [ A^- ]

[ HA ] = 0.1 M - [ H₃O⁺ ] Lets set X = [ H₃O⁺ ]

K_A = = 2.24 x 10^-5 K_A = 2.24 x 10^-5 = X² / (0.1 - X) Rearranging give us:

X²+ 2.24x10^-5X - 2.24x10^-6 = 0 We can solve this quadratic or possibly

assume that the amount of acid that dissociates is insignificant (compared to the undissociated form)

K_A = 2.24 x 10^-5 [ H₃O⁺ ] = [ A^- ] [ HA ] = 0.1 M - [ H₃O⁺ ]

K_A / 0.1 < 10^-3 - can assume that [HA] = 0.1 M 2.24x10^-5 = X² / 0.1M

X = (2.24x10^-6)^1/2 = 0.00150 pH = 2.82

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Now, lets go for the exact solution Earlier, we found that

X²+ 2.24x10^-5X - 2.24x10^-6 = 0

X = -b + b² - 4ac 2a

X = 0.00149 pH = 2.82

No significant difference between our two answers.

X = -2.24x10^-5 + [(2.24x10^-5)² +4x2.24x10^-6]^1/2 2

If you are starting with an acid, acidic conditions or the conjugate acid of a base

Do your calculations using K_A

If starting with a base, basic conditions or the conjugate base of an acid

Do your calculations using K_B

You can readily convert pH to pOH and K_A to K_B values.

With complex formation, two or more species will join, forming a single, new species.

Mⁿ⁺ + xL M(L)_xⁿ⁺

K_F = [ M(L)_xⁿ⁺ ] [ Mⁿ⁺ ] [ L ]^m

If we are evaluating the decomposition of a complex, we can use a K decomposition.

M(L)_xn+ Mⁿ⁺ + xL K_D =

Since water is not a portion of these equilibria, K_D = 1 / K_F

[ Mⁿ⁺ ] [ L ]^m [ M(L)_xⁿ⁺ ]

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Equilibrium expressions for REDOX systems are derived from standard electrode potentials.

6Fe²⁺ + Cr₂O₇^2- + 14 H₃O⁺ 6Fe³⁺ + 2Cr³⁺ + 21H₂O

K_REDOX =

We’ll review how to determine and work with K_REDOX when we cover the units on

electrochemistry.

[Fe³⁺]⁶ [Cr³⁺]² [Fe²⁺]⁶ [Cr₂O₇^2-][H₃O⁺]¹⁴

K_A₃ K_A₂

K_A₁

[H₃O⁺] [PO₄^3-] [HPO₄^2-] K_A₃ =

[H₃O⁺] [HPO₄^2-] [H₂PO₄^-] K_A₂ =

[H₃O⁺] [H₂PO₄^-] [H₃PO₄]

K_A₁ = ^Note:

[H₃O⁺] is the same for each expression.

The relative amounts of each species can be found if the pH is known.

The actual amounts can be found if pH and total H₃PO₄ is known.

One possible ‘first step’ would be to determine the relative amounts of each species.

[H₂PO₄^-] [H₃PO₄] K_A₁

[H₃O⁺]

K_A₃ [H₃O⁺]

[PO₄^3-] [HPO₄^2-] K_A₂

[H₃O⁺]

[HPO₄^2-] [H₂PO₄^-]

=

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[H₂PO₄^-] [H₃PO₄] 7.5 x 10^-2

10^-7

4.8 x 10^-13 10^-7

[PO₄^3-] [HPO₄^2-] 6.2 x 10^-8

10^-7

[HPO₄^2-] [H₂PO₄^-]

=

= 750000 0.62

4.8x10^-6

These ratios show that only H₂PO₄^- and HPO₄^2- are present at significant levels at pH 7.

Total phosphate = 1.0 M so

1.0 = [H₃PO₄] + [H₂PO₄^-] + [HPO₄^2-] + [PO₄^3-] This is referred to as a mass balance.

Based on the ratio, we know that [H₃PO₄] and [PO₄^3-] are not present at significant levels for this mass balance so:

1.0 = [H₂PO₄^-] + [HPO₄^2-]

[HPO₄^2-] = 0.62 [H₂PO₄^-]

1.0 = [H₂PO₄^-] + 0.62 [H₂PO₄^-] = 1.62 [H₂PO₄^-]

[H₂PO₄^-] = 1.0 / 1.62 = 0.617 M [HPO₄^2-]

[H₂PO₄^-] = 0.62 Since you now know the concentration of H₂PO₄^-, you can now sequentially solve the other equilibrium expressions.

This approach is very useful when dealing with complex equilibria.

The mass balance is on means of eliminating

‘insignificant’ species based on addition or subtraction