Applications of Aqueous
Equilibria
Or
ICE Problems Part III
Common Ion
• Produced by both ___________________- while in
solution
• Effect of common ion can be predicted by
____________
• Will have an impact of pH
• Polyprotic acids can also experience the common ion
effect. What is the common ion?
• Problems will solve with ICE, but the initial
concentration of A- will not be zero b/c of the
common ion from the salt
Example:
A solution of __________. What is the common
ion and how does it impact equilibrium? What
happens to the pH of this solution in
comparison to a solution of HF only?
NaF  Na+ + FHF  H+ + FF- is the common ion. The equil. shifts left and
the pH would be less acidic.
Buffered Solutions
• Resists a change in ___ when either
__________ is added
• Common ion
• ______ is a buffered solution that maintains
the necessary pH for cells for survive
• Can be a weak ____ and salt (HF and NaF)
OR
a weak ____and a salt (NH3 and NH4Cl)
A buffered solution contains 0.50 M acetic acid and 0.50 M sodium acetate. Calculate the pH of this solution.
How Does a Buffer Work?
Buffered soln of HA and A- to which OH- is added --OH- + HA  A- + H2O
***OH- do not accumulate, they are replaced by A• pH is governed by [HA]/[A-] ratio and b/c such large
quantities have been used in comparison to the [OH-] –
the ratio changes very little
• [HA] and [A-] >> [OH-] added, thus they change
very little
Calculate the change in pH when 0.010 mol of solid NaOH is added to 1.0 L of the buffered solution from the
last example. Compare this to the pH change when 0.010 mol of NaOH is added to 1.0 L of water.
Buffered Soln of HA and A- is which H+ is added ________  HA***
*** H+ does not accumulate
• Same as w/OH-  [HA] & [A-] >> [H+]
so the ratio changes very little
Henderson-Hasselbalch Equation
• Useful when [A-]/[HA] ratio is known, but not
necessary to solve the problem.
pH = pKa + log
[A-]
[HA]
= pKa + log
[base]
[acid]
**Assume that _______ and _______ are the initial
concentrations.
Calculate the pH of a solution containing 0.75 M lactic acid (Ka – 1.4 x 10-4) and 0.25 M sodium lactate.
A buffered solution contains 0.25 M NH3 (Kb = 1.8 x 10-5) and 0.40 M NH4Cl. Calculate the pH of this solution.
Calculate the pH of the solution that results when 0.10 mol of gaseous HCl is added to 1.0 L of the buffered
solution from the previous example.
Few more buffer items
• For a particular buffering system, all solutions
that have the same ratio [A-]/[HA] will have the
same pH.
• All buffered solution have a ________________
present.
Buffer Capacity
• Represents the amount of H+ or OH- the buffer
can absorb w/o a significant change in pH
• To have a large capacity, the buffer must contain
a large concentration of buffering components
• Determined by the magnitudes of [HA] and [A-]
Last little buffer stuff
• Large changes in the ratio [A-]/[HA] will
produce large changes in pH.
• Optimal buffering occurs when [HA] = [A-],
most resistant to addition of H+ or OH• Best buffering when
[A-]/[HA] = 1 or pka = desired pH
Calculate the change in pH that occurs when 0.010 mol of gaseous HCl is added to 1.0 L of each of the following
solutions. Solution A – 5.00 M HC2H3O2 and 5.00 M NaC2H3O2
Solution B – 0.050 M HC2H3O2 and 0.050 M NaC2H3O2
Is is a buffer???
A. 0.1 M KOH and 0.1 M CH3NH3Cl
B. 0.1 M KOH and 0.2 M CH3NH2
C. 0.2 M KOH and 0.1 M CH3NH3Cl
D. 0.1 M KOH and 0.2 M CH3NH3Cl
Online Lecture #2 -Titrations
Titration and pH Curves
_________ – method used to determine the
amount of acid or base in solution
– Progress of the titration can be monitored by plotting
the pH of the solution as a function of the titrant
added
_________________ Titration –
H+ + OH-  H2O
– To determine the pH or [H+] at any point, the amount
of [H+] that remains must be divided by the total
volume of solution
Strong Acid-Strong Base Titration
Millimole – mmol, a thousandth of a mole
– 1 mmol = 1mol/1000 = 10-3 mol
– Molarity = mol = mmol solute
L mL solution
– # of mmol = volume(mL) X molarity
• Before Equivalence Point - [H+] = mmol H+
mL of soln
• Equivalence Point – [H+] = [OH-]
pH = 7.00
• After Equivalence Point – [OH-] = mmol OHmL of soln
Then [H+] from Kw
Example – Get some paper
Titration of 50.0 mL of .200 M HNO3 with .100 M
NaOH. Calculate the pH @ the following
intervals:
0 mL NaOH
100 mL NaOH
20 mL NaOH
200 mL NaOH
Make sure I show you the titration curve.
______ Acid-_____ Base Titrations
• Important – Even weak acids react to completion
w/ OH• When using a weak acid, the calculation of [H+]
after a strong base has been added requires an
ICE problem.
• The pH of the equivalence point of the weak
acid – strong base titration will always be greater
than 7.
• Equivalence point is defined by stoichiometry
not by the pH.
____ Acid-______ Base Titrations
• Halfway to the equivalence point the
[HA] = [A-]
• It is the amount of acid, not its strength, that
determines the equivalence point.
(mol OH- = mol H+)
• pH value at the equivalence point is affected
by acid strength, the weaker the acid the
greater the pH
Weak Acid-Strong Base Titrations
________ Problem Solving
1) ____________ Problem
• Determine the acid remaining and the conjugate base
• Before, Change, After (BCA problem)
2) __________ Problem
• Weak acid equilibrium to determine pH
• ICE
Weak Acid-Strong Base Titrations Example
Problem – Get some paper
Titration of 50.0 mL of .10 M HC2H3O2 with .10
M NaOH. Calculate the pH @ the following
intervals:
0.0 mL NaOH
50.0 mL NaOH
25.0 mL NaOH
75.0 mL NaOH
Ka of HC2H3O2 = 1.8 x 10-5
Make sure I show you the titration curve!
Weak Acid-Strong Base Titrations Example
Problem – Get some paper
Titration of 50.0 mL of .100 M HCN with .100 M
NaOH. Calculate the pH @
a. 8.00 mL of NaOH added
b. Halfway point of titration
c. Equivalence point of titration
Ka of HCN = 6.2 x 10-10
Weak Base w/ Strong Acid Titration Example
Problem – More paper
Titration of 100.0 mL of .050 M NH3 with .10 M
HCl. Calculate the pH at the following
intervals:
0.0 mL HCl 50.0 mL HCl
25.0 mL HCl 60.0 mL HCl
Make sure I show you the titration curve. Ask
me about a polyprotic curve as well!!
Acid – Base _________
• Marks the endpoint by changing color
• Endpoint and equivalence point are not
necessarily the same (but will be close)
• Complex molecules that are weak acids
– One color w/ H+ attached and another color when
H+ is removed
– Phenolpthalein – HIn – clear (acid)
In- - pink (base)
Online Lecture #3 – Solubility
Equilibria
Solubility Equilibria
Ionic solids when put into water dissociate into cations and
anions –
BaSO4(s)  Ba+2 + SO4-2
As the ion concentrations increase, they collide and reform
the solid –
SO4-2 + Ba+2  BaSO4(s)
An ________ is eventually reached –
BaSO4(s)  Ba+2 + SO4-2
For which an equilibrium expression can be written –
Ksp = [Ba+2][SO4-2] (leave out solid)
Ksp - solubility product constant
Calculate the Ksp of 1.0 x 10-15 M Bi2S3.
Calculate the solubility of Cu(IO3)2 that has a Ksp of 1.4 x 10-7.
Common Ion Effect and Solubility
•
The solubility of a solid is _______ if the
solution already contains ions common to the
solid
Relative Solubilities
1. If the salts being compared have the ____
number of ions, use the ksp to compare the
solubilities.
2. If the salts being compared have
_______numbers of ions, calculate the
solubilities from ksp and then compare.
Calculate the solubility of solid CaF2 (Ksp = 4.0 x 10-11) in 0.25 M NaF.
For each of the following pairs of solids, determine which solid has the smaller molar solubility.
A. CaF2(s), Ksp = 4.0 x 10-11 or BaF2(s), Ksp = 2.4 x 10-5
B. Ca3(PO4)2(s), Ksp = 1.3 x 10-32 or FePO4(s), Ksp = 1.0x 10-22