Chapter 8

ActivityGoodbye Freshman Chemistry

Hello Real World

Activity

Equilibrium up to this point has been based on a very simple model.

Basically that ions (a species with a high density of charge) do not interact other than the interaction of interest.

Activity Real Solutions are rather complex in

their nature. Ion interact with all ions in solution

and other polar species. Ions will interact with water since

water is a polar substance. Ions will have these water

molecules in their ‘waters of hydration’.

Just how much water?

Compound Tightly bound water

CH3CH2CH3 0

CH3OCH3 1

CH3CN 3

CH3COO- 5

NH3 9

NH4+ 12

Activity

In essence this ionic atmosphere shields the ions from each other which will diminish the interaction between the ions.

i.e. For a Ksp interaction the compound will be more soluble.

Solubity

KT(s) = K+ (aq) + T- (aq)

Potassium Tartarate

Glucose

Activity How do we quantify this ionic

atmosphere. We use a term called ionic strength.

It can be calculated

i

ii zczczc 2222

211 2

1.....)(

2

1

Activity

Let’s do an example. What is the ionic strength of a solution that is

0.010 M Na2SO4 and 0.050 M NaCl We will prepare a Table and ask our first

question. What ions are present?

Activity

Ion Conc Charge Charge^2

Product

Na+

SO42-

Na+

Cl-

Sum

0.5*Sum

0.010 M Na2SO4 and 0.050 M NaCl

What are the concentrations?

Activity

Ion Conc Charge Charge^2

Product

Na+ 0.02

SO42- 0.01

Na+ 0.05

Cl- 0.05

Sum

0.5*Sum

0.010 M Na2SO4 and 0.050 M NaCl

What are the charges and charges squared?

Activity

Ion Conc Charge Charge^2

Product

Na+ 0.02 +1 1

SO42- 0.01 -2 4

Na+ 0.05 +1 1

Cl- 0.05 -1 1

Sum

0.5*Sum

0.010 M Na2SO4 and 0.050 M NaCl

Let’s Finish the Math

Activity

Ion Conc Charge Charge^2

Product

Na+ 0.02 +1 1 0.02

SO42- 0.01 -2 4 0.04

Na+ 0.05 +1 1 0.05

Cl- 0.05 -1 1 0.05

Sum 0.16

0.5*Sum 0.08 M

Activity It gets even more complex. Not all

soluble salts will dissociate into the individual species. We get another form that will exist in solution. This is called and Ion Pair

MgSO4 (s) + H2O = Mg2+(aq) + SO42-(aq) + MgSO4(aq)

The MgSO4(aq) is the ion pair.

Activity You can see from Appendix J in the

book the equilibrium constants for this pair formation. Mg2+(aq) + SO4

2-(aq) = MgSO4(aq) Log K = 2.23

Which means that about ~93% will be in the ion pair form if the formal concentration of MgSO4 is 1M.

Activity

Ion Pairing has important implications in areas such as drug transport and delivery. Any drug species must be in the proper form to give its activity.

Activity So how do we find the “Activity of

an ion in solution”.

][CA cc Where A is the activity of species C is “activity coefficient”[C] is the concentration in moles per liter

Activity Let’s revisit our equilibrium expression. aA + bB = cC + dD

Which we write

bB

baA

a

dD

dcC

c

bB

aA

dD

cC

BA

DC

AA

AAK

][][

][][

Activity Lets revisit an equilibrium we have

seen several times. The solubility of silver sulfide.

SAgSAgsp SAgAAK ][][ 222

Activity

How do we find the value for the “activity coefficient” ?

For ionic strengths from 0.1M and down we can look up the value on a Table or we can calculate from the Extended Debye-Huckel equation.

Activity

)305/(1

51.0log

2

z

Where z is the ion charge, is the ionic strength and is the hydrated radius

Hydrated Radius ()

Activity

You could also look up the value of the activity coefficient from Table 8-1 in the text.

Activity

What if you want an intermediate value between two listed ionic strengths.

You will need to do a linear interpolation. (Do you remember this from using log tables in high school?)

Activity

Effect of Ionic Strength, Ion Charge and Ion Size on Activity Coefficient (0 to 0.1M) increases decreases z increases the faster the departure

from from unity smaller then the greater activity

effects

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Neutral Molecules Not charged so no ionic interaction so

we assume is 1 Gases

Not charged and again we can assume that is 1. A = P(bar)

Activity (In concentrated Solutions)

Activity

What is the pH of very, very pure water? Kw = AHAOH = 1.0 x 10-14

AH = 1.0 x 10-7 Thus pH = 7.00

What is the pH of 0.10 M NaCl? Would it still be 7.00?

Activity

Kw = H[H] OH [OH] Look up the for H+ and OH- on

Table 8-1. H = 0.83 and OH = 0.76 [H+] = [OH-] So Solving for [H+] = 1.26 x 10-7

pH = - log AH = -log (0.83)(1.26 x 10-

7) = 6.98