Biophysical Chemistry 2

7/29/2019 Biophysical Chemistry 2

1/18


2/18

Law of mass action

0 Consider the reaction

A + B C + D

0 A/B reactants C/D products0 Indicates that the reaction is reversible and

can reach equilibrium

0 At constant temperature the rate of a reaction is

proportional to the active mass of the reactants(mol/L)


3/18

A + B C + D

0 Therefore we can write an expression for theequilibrium constant

Keq = [products]

[reactants]

Keq = [C][D]

[A][B]

0 Activity of a substance refers to the effectiveconcentration of that substance

0 In dilute solutions activity is considered to be almostequal to concentration hence concentration will beused


4/18

0 According to the law of mass action:

At equilibrium the product of the concentrations of

the products divided by the product of the

concentrations of the reactants is a constant known

as the equilibrium constant

0 If the concentration of at least one other component is

changed then the concentration of another one also

changes

0 This is in order to meet the conditions of the

equilibrium as defined by Keq


5/18

0 Consider the reaction between ethanoic acid andethanol

0 Write the reaction

moles CH3COOH + C2H5OH CH3COOC2H5 + H20Initial a b - -

Change -x -x +x +x

Final a-x b-x x x

0 If the total volume of the mixture is V litres thenconcentration is:

CH3COOH = (a-x)/V mol/L

C2H5OH = (b-x)/V mol/L

CH3COOC2H5 = x/V mol/LH20 = x/V mol/L

0 Substitute into the equilibrium expression


6/18

Oswalds Dilution law

0 Provides a relationship between the dissociation

constant and the degree of dissociation for a weak

acid

0 Describes the equilibrium for a weak electrolyte using

the mass law

0 The degree of dissociation of a weak acid is the

fraction of the no. of moles of the acid that are

converted into ions


7/18

0 Consider the dissociation of 1 mole of the weak

electrolyte HAmoles HA H+ + A-

Initial 1 - -

Change - + +

Final 1-

0 Concentrations:

HA = (1-)/V mol/L

H

+

= /V mol/LA- = /V mol/L


8/18

Using the law of mass action to write an expression for

the dissociation constant:

K = [H+][A-]

[HA]

K = x

(1-)V

0 For weak electrolytes only little dissociation occurs

0 Therefore (1-) becomes approximately equal to 1

K = x = 2

(1-)V 1V= (K.V)


9/18

0 Therefore as long as the value of K remains constant

(at constant temperature) then the value of the

dissociation constant is proportional to V

0 For the same number of moles, an increase in V

implies that the solution is more dilute

0 This law can not be applied to strong electrolytes

0

Strong electrolytes completely ionise in aqueoussolutions


10/18

0 Example:

0 Calculate the degree of dissociation for a 0.1 M

solution of ethanoic acid given that Ka = 1.7x10-5

Mand the total volume is 10 L.

0 Solution:

CH3COOH H+ + CH3COOH

-

Mols at equi.1-x x x

Applying Ostwalds dilution law

x = (1.7x10-5 x 10) = 0.013 mols

0 Initial number of moles = MV = 0.1 mol/L x 10 L = 1

mol

0 Degree of dissociation = (0.013 / 1) x 100 = 1.3%


11/18

0 Example: The concentration of H+ ions in 0.10 M

solution of a weak acid is 1.0 10-5 mol/L. Calculate

the dissociation constant of the acid.0 Solution: HA H

+ + A-

Concentration (mol/L)

Initial 0.1 0 0

Equilibrium (0.1-1.010-5) 1.010-5 1.010-5

0 [HA] can be taken as 0.1 M as 1.0 10-5 is very small.

0 Applying law of mass action.

0 K = [H+][A-]/[HA] = (1.010-5) 1.0 10-5/0.10

= 1 10-9


12/18

Ion product of water

0 Water is a weak electrolyte and is capable of

dissociating slightly to form H+ and OH- ions

2H2O H3O+

+ OH-

H2O H

+ + OH-

0 Using the second reaction Keq is written as:

Keq = [H+][OH-]

[H2O]0 Experimental evidence has shown that Keq at 25C is

1.8x10-16 mol/L


13/18

0 Since only a few molecules of water dissociate, theconcentration of H2O is considered as constant

0 Pure water has a concentration of 1000 g/L

0 Therefore in pure H2O

[H2O] = (1000 g/L)/ 18 g/mol = 55.5 mol/L

Keq = [H+][OH-]

[H2O]

Keq[H2O] = [H+

][OH-

] *constant [H2O]Kw = [H

+][OH-]

Kw = 1.8x10-16 mol/L x 55.5 mol/L = [H+][OH-]

Kw = 1x10-14 = [H+][OH-]

Kw is the ion product of water


14/18

0 Therefore in pure water [H+] and [OH-] can be

calculated using:

[H+][OH-] = 1x10-14

x2

= 1x10-14

x = 1x10-7 M

0 Note [H+] and [OH-] are reciprocally related

0 A high concentration of one implies a low

concentration of the other


15/18

pH

0 Concentrations of [H+] are usually very low

0 Need a convenient way of expressing the [H+]

0 Introduce log10

to simplify the expression

pH = -log [H+]

0 pH is a logarithmic function

0 If pH decreases by 1 unit (i.e from 5 to 4) then the[H+] has increased tenfold (from 10-5 to 10-4)

0 An increase in pH would therefore mean a reductionin the [H+]

pH < 7 is acidic pH = 7 is neutral pH > 7 is basic


16/18

0 Example:

Calculate the pH of a solution having a [H+] of 6.28x10-6

mol/L.

0

Solution:pH = -log [H+]

pH = -log (6.28x10-6)

pH = 5.20


17/18

0 Recall

[H+][OH-] = 1x10-14

0 Taking log at both sides

log[H+] + log[OH-] = log (1x10-14) = -14

0 Then multiply through by -1

-log[H+] (log[OH-]) = 14

This simplifies to

pH + pOH = 14


18/18

0 Example:

Calculate the pH of a 0.1M solution of NaOH

0 Solution: Complete dissociation

NaOH Na+ + OH-

[OH-] = [NaOH] = 0.1M

pOH = -log [OH-]pOH = -log (0.1) = 1

Using H + pOH = 14

pH = 14 pOH

pH= 14 -1 = 13

Documents

Biophysical Chemistry 2