Chapter 8
ActivityGoodbye Freshman Chemistry
Hello Real World
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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
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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
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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?
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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
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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.
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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
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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.
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)305/(1
51.0log
2
z
Where z is the ion charge, is the ionic strength and is the hydrated radius
Hydrated Radius ()
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You could also look up the value of the activity coefficient from Table 8-1 in the text.
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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?)
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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)
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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?
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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