05 aqueous solutions

11
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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Page 1: 05 aqueous solutions

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

Page 2: 05 aqueous solutions

acid base conjugate base of HF

conjugate acid of H2O

acid base conjugate base

conjugate acid

ac

id s

tre

ng

th

ba

se stre

ng

th

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

BC

D

time

kinetic equilibrium region region

Page 3: 05 aqueous solutions

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.

Keq = aC

c aDd

aAa aB

b a = activity

Keq = [ C ]c [ D ]d

[ A ]a [ B ]b

Page 4: 05 aqueous solutions

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.

( )

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 ai = fi [ i ] where fi is the activity coefficient for i [ i ] is the molar concentration

Page 5: 05 aqueous solutions

0.5 Zi2 µ !

1 + 0.33 "i µ!

For very dilute solutions As fi -> 1, ai -> [ i ]

For µ up to 0.1 fi < 1 and ai < [ i ]

When u is > 0.1 Results in complicated behavior.

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

Keq = [ H3O+ ] [ OH- ]

[ H2O ]2

In dilute solutions, [ H3O+] and [ OH- ] is much smaller than [ H2O ]. [ H2O ] is essentially a constant of ~55.5 M.

Page 6: 05 aqueous solutions

KW = 10-14 = [H3O+] [OH-]

[H3O+] = [H3O+]water + [H3O+]HCl

[OH-] = [H3O+]water

Lets set x = [H3O+]water then

[H3O+] = x + 1.0 x 10-8

[OH-] = x

10-14 = (x + 1.0 x 10-8) ( 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.

Page 7: 05 aqueous solutions

KSP 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)

Keq = [ 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 KSP expressions do not include the solid form in the equilibrium expression

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

KSP = [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

KSP = 1.1 x 10-12 = [ Ag+ ]2 [ CrO42- ]

[CrO42-] = [CrO4

2-]Ag2CrO4 + [CrO42-]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.

KSP = 1.1 x 10-12 = [ Ag+ ]2 [ CrO42- ]

[CrO42-] = 0.010 M

[ Ag+ ] = ( KSP / [ CrO42- ] ) 1/2

= 1.1 x 10-12 / 0.010 M

= 1.1 x 10-5 M

[ Ag+ ] << [ CrO42- ]

so our assumption was valid.

Page 8: 05 aqueous solutions

[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+ ]

KA = = 2.24 x 10-5 KA = 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)

KA = 2.24 x 10-5

[ H3O+ ] = [ A- ]

[ HA ] = 0.1 M - [ H3O+ ]

KA / 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

Page 9: 05 aqueous solutions

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 KA

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

Do your calculations using KB

You can readily convert pH to pOH and KA to KB values.

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

Mn+ + xL M(L)xn+

KF = [ 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

KD =

Since water is not a portion of these equilibria, KD = 1 / KF

[ Mn+ ] [ L ]m

[ M(L)xn+ ]

Page 10: 05 aqueous solutions

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

6Fe2+ + Cr2O72- + 14 H3O+ 6Fe3+ + 2Cr3+ + 21H2O

KREDOX =

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

[Fe3+]6 [Cr3+]2

[Fe2+]6 [Cr2O72-][H3O+]14

KA3 KA2 KA1

[H3O+] [PO43-]

[HPO42-]

KA3 =

[H3O+] [HPO42-]

[H2PO4-]

KA2 =

[H3O+] [H2PO4-]

[H3PO4] KA1 = 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]

KA1

[H3O+]

KA3

[H3O+]

[PO43-]

[HPO42-]

KA2

[H3O+]

[HPO42-]

[H2PO4-]

=

=

=

Page 11: 05 aqueous solutions

[H2PO4-]

[H3PO4]

7.5 x 10-2

10-7

4.8 x 10-13

10-7

[PO43-]

[HPO42-]

6.2 x 10-8

10-7

[HPO42-]

[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-] + [PO43-]

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-]

[HPO42-] = 0.62 [H2PO4

-]

1.0 = [H2PO4-] + 0.62 [H2PO4

-] = 1.62 [H2PO4

-]

[H2PO4-] = 1.0 / 1.62 = 0.617 M

[HPO42-]

[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