Chapter 12
Aqueous Equilibrium
Common Ion Effect
Learning Outcomes and Prereqs
Pre-reqs:
Identify how changing concentrations of reactants or products shift
an equilibrium (Le Chatelier’s Principle)
Learning Outcomes
Identify the effect of adding a conjugate base of an acid on pH and
ionization.
Identify the effect of adding a conjugate acid of a base on pH and
ionization.
Reminder
Le Chatelier’s Principle
If you apply stress to a system it shifts to relieve the
stress.
Reaction shifts left or right to relieve the stress.
Analogy-
The common ion effect
If you have a solution of an acid and its conjugate
base the ionization is lessened by the presence of the ion.
Same is true for a base and its conjugate acid
Conjugate acid/base added through it’s salt.
So….what does that mean? Examples?
The common ion effect
Adding a conjugate lessons the effect of ionization.
Examples?
Formic acid and sodium formate
H
Ammonia and Ammonium Chloride
Why do we care?
If ionization is decreased how does this affect the OH- and
H+ concentrations?
H
Question: If HCOO- is added, what happens to the
concentration of H+? What does that do to pH?
[H+]:
It decreases
pH:
It increases
But does it really matter?
An example to illustrate:
Determine the pH of a 0.20M NH3 solution. Then determine the
pH of a solution that is 0.20M NH3 and also 0.30M NH4Cl
Review
If you add the conjugate base of an acid:
the ionization will be lessened
This means less [H+] so a higher pH
If you add the conjugate acid of a base:
The ionization will be lessened
This means less [OH-] so a lower pH.
Henderson Hasselbalch
Equation
Learning outcomes
Derive the Henderson Hasselbalch Equation
Use the Henderson Hasselbalch equation to solve a
problem.
Identify when we are allowed to use this, and when we have
to use the previous methods of solving acid/base problems.
Henderson-Hasslebalch:
Creating the generic equation
Henderson-Hasslebalch:
Problem solving tips
Use this for buffer problems only.
Be sure you are using the Ka NOT the Kb
As with all equations, get good at manipulating them.
Only use if weak acid approximation is valid
Example: Using the HendersonHasslebalch equation
A solution is made that is 0.20M CH3COOH and
0.30M CH3COONa are mixed. Find the pH.
0.30
𝑝𝐻 = 4.75 + 𝑙𝑜𝑔
0.20
𝑝𝐻 = 4.93
Review
The Henderson Hasselbalch equation comes from
solving a generic acid problem where both an acid and
its conjugate base are present.
You can use this equation to solve problems whenever
the 5% approximation is valid (note: you don’t have to
show me, but you do need to check in case it doesn’t
work)
Introduction to Buffers
Motivation and
Learning Outcomes
Identify why our body needs buffers, and what happens if
something causes that system to fail.
Define buffers.
Identify how buffers work to keep pH changes small
Identify if two components form a buffer
Real life importance check:
pH of blood=7.4
Acidosis <7.35
Some products of your metabolism:
Ammonia
Urea (a base)
Uric acid
CO2 (acidic when reacting with water)
Alkalosis> 7.50
Urea
Alkalosis and
Acidosis
Metabolic
Alkalosis and
Acidosis
Respiratory
Hypoventilation
increased CO2
acidosis
Hyperventilation
decreased CO2
alkalosis
Real life importance check:
pH of blood=7.4
Acidosis <7.35
Alkalosis> 7.50
Products of your metabolism:
Ammonia
Urea (a base)
Uric acid
CO2 (acidic when reacting with water)
So how can your body
control your pH so well?
Urea
Buffers
Acid and its conjugate base, or a base and its
conjugate acid
Essentially the systems we’ve talked about in the
common ion effect questions
It works by converting a strong acid into a weak acid,
or a strong base into a weak base.
A strong base can’t exist in solution with a weak acid
A strong acid can’t exist in solution with a weak base
Activity Part 1:
(I’ll need some volunteers)
Buffers
Acid and its conjugate base, or a base and its conjugate
acid
Basically the systems we’ve talked about in the
common ion effect questions
It works by converting a strong acid into a weak acid,
or a strong base into a weak base.
A strong base can’t exist in solution with a weak acid
A strong acid can’t exist in solution with a weak base
Activity Part 2:
(I’ll need some volunteers again)
Acids react with bases
Bases react with acids
Real life importance check:
pH of blood=7.4
Alkalosis> 4.5
Acidosis <7.35
Products of your metabolism:
Ammonia
Urea (a base)
Uric acid
CO2 (acidic when reacting with water)
Urea
So how can your body control your pH so well?
It uses buffers!!!!
Buffers in the Body:
We discussed how CO2 can become acid in the body
This equilibrium is as follows, if both are present it’s a buffer
Buffers in the Body:
How do we regulate CO2 levels?
Breathing levels
What happens to the CO2 if we breath
shallowly?
Respiratory acidosis
Why is this bad?
Many reasons: here is one:
Hemoglobin in red blood cells binds either
O2 or H+
𝐻𝑏𝐻 + + 𝑂2 ⇌ 𝐻𝑏𝑂2 + 𝐻 +
Hypoventilation
increased CO2
acidosis
Hyperventilation
decreased CO2
alkalosis
Which of the following are buffers:
KNO2/HNO2
yes
KCl/HCl
no
KHSO4/ H2SO4
no
Na2HPO4/NaH2PO4
Yes
KCN/HCN
Yes
Na2SO4/NaHSO4
Yes
NH3/NH4NO3
Yes
Review
Buffers prevent large changes in pH caused by the addition of
acids or bases.
This is needed in many systems. Our blood buffer system uses
buffers to keep the pH in a very narrow range.
Buffers work by having both an acid and its conjugate base, or a
base and its conjugate acid present.
Therefore if an acid is added, it reacts with the base
If a base is added, it reacts with the acid
This keeps the pH change small.
Buffer Calculations.
Learning outcomes
Use the Henderson-Hasslebalch equation to solve for
the pH of buffers.
Use the Henderson-Hasslebalch equation to solve for
the pH of buffers after adding an acid or base.
Test your knowledge of buffers by answering a couple
of conceptual questions.
Buffers
Calculate the pH of the solution that results from mixing 0.200L
of 0.050M (CH3)2NH with 0.320L of 0.040M (CH3)2NH2Cl.
(Kb=5.88x10-4)
You try:
Calculate the pH of the solution that results from mixing 72.0mL
of 0.015M (CH3)2NH with 20mL of 0.200M (CH3)2NH2Cl
The pH of blood plasma is 7.40, Assuming the principal buffer
system is HCO3- and H2CO3 find the ratio of HCO3- and
H2CO3. (Ka=4.2x10-7)
Adding acids and bases
Acids react with bases
Bases react with acids
Buffers when adding a base or acid
It works by converting a strong acid into a weak acid, or a strong
base into a weak base.
A strong base can’t exist in solution with a weak acid
A strong acid can’t exist in solution with a weak base
Notice NOT equilibrium arrow!!!
Use “new” concentration of acid or base to do equilibrium calculations in
the same manner as before.
Steps for buffer problem with addition of
acid or base.
Step 1: Decide if it is a buffer
Step 2: Write reaction of buffer with strong acid or base
Step 3: (if needed) do ICF chart (my take on ICE, its not an
equilibrium its “final”) to find amount of weak acid or base
Suggested, do this in mols
Step 4: fill into the HH equation if assumptions are valid (Ka/Kb
is low, concentrations high) or use ICE chart and Ka or Kb ect…as
we did with other equilibrium problems.
If using HH equation use moles
If using Ka or Kb equation use molarity
Buffers when adding a base or acid Examples:
Calculate the pH of 1.00L of the buffer 1.00M
CH3COONa and 1.00M CH3COOH. Ka=4.2x10-7
A) before
and after the addition of *
B) 0.080 mol NaOH
or
C) 0.120 mol HCl
*assume no change in volume
***Note: You’ll notice your book always uses an ICE chart, filling into K, solving for X, finding
pH protocol, I’m not sure why they use this after having taught you the easier way to do it.
Choose whichever you like most.
Example Questions:
We said the normal ratio of HCO3- :H2CO3 is 20:1, Is
your blood better at buffering added acid or added base?
Review
For most cases you can use the Henderson-Hasslebalch
equation for solving buffer problems.
(When the 5% rule is valid, you need to test this for
buffers, but you don’t need to show me, otherwise use
an ice chart).
If you add an acid or a base to a buffer, first react the
species away ( usually using moles) then solve a HH
equation.
Chemistry and life segment
The biology here will not be tested (however you better
know the difference between an alcohol and a carboxylic
acid since we talked about that earlier in the class).
Little side note while we are on
biological pH topics:
Hangovers: Dispelling
another myth
“Acetic acid causes acidosis
and this is why you get a hangover”
Hangovers:


Metabolic pathwayAcetaldehyde can cause migraine strength headaches and
vomiting.
Acetic acid is harmless
Alcohol decreases the Anti-diuretic hormone (adh)
Increases urine output, Increases dehydration
Depletes electrolytes leading to metabolic acidosis!
Titration Introduction
Learning outcomes
Describe the experimental set up of a titration.
Identify reasons for doing a titration.
Describe the basics of what is occurring during a
titration.
Which reactions are occurring in the beaker.
Using a titration curve, determine what is being
titrated (acid or base) and what the titrant is (acid or
base).
Experimental Set up
Burette= known: Titrant
Unknown: In Flask
Known: In “burette” (tall
thin volumetric glassware)
Typically “strong”
For acid/base titration:
unknown can be acid or
base, but burette must have
the opposite.
Flask/beaker=unknown: Analyte
Why?
We know what is in the
burette (titrant)
We know how much of the
known we add
We can find out the
concentration of the
unknown (analyte)!
Acidic Analyte
© Atkins and Jone
Basic Analyte
(c)Atkins and Jone
Titrations: Exercise
http://users.wfu.edu/ylwong/chem/titrationsimulator/
You should play with this a bit on your own as well.
Review
Titrations are used to determine the concentration of an
unknown acid or a base.
The unknown (analyte) goes in a flask beneath a burette
which contains a species of known concentration (titrant).
You can identify whether the analyte is an acid or base by
measuring the pH (on a titration curve, by seeing the initial
pH).
For acid base titrations there will be both an acid and a base
Titrant=acid, analyte= base OR Titrant=base, analyte=acid
Strong Acid/Strong Base
Titrations
Learning Outcomes
Calculate the pH of a strong acid or base titration at
various points:
Before the titration has started
Before the equivalence point
At the equivalence point (spoiler, its always 7 for
strong/strong)
After the equivalence point
Use the amount of titrant added to reach equivalence to
calculate the concentration, mol, or grams of analyte
General Problem Solving
Techniques
Neutralization Reactions:
React H+ and OH- to completion
𝐻 + (𝑎𝑞) + 𝑂𝐻 − (𝑎𝑞) → 𝐻2 𝑂(𝑙)
Calculate moles of H+ and OH-, and find excess
concentration.
Use pH= -log[H+] or pH=14+log[OH-]
Dance Chemistry Explanation
http://cast.es.uci.edu/tltc/Temp/Dance%20Chemistr
y/Acid%20Base%20Titration.mp4
Example: Strong/Strong
Work style #1
Example: If you titrate 100 mL of 1M HCl and 100 mL of 2M
NaOH, what is the pH?
0.100 mol
0.200 mol
-0.100 mol
-0.100 mol
0 mol
0.1 mol
Example: Strong/Strong
Example: If you titrate 100 mL of 1M HCl and 100 mL of 2M
NaOH, what is the pH?
Work style #2
OH-= 2mol/L*0.100L=0.2mol
H+= 1mol/L*0.100L=0.1mol
Which one goes to 0?
H+
How much of the excess is left over?
0.1 mol OHHow do we find pH?
14+log[OH-]=
14+log(0.100mol/0.200L)=13.7
Equivalence
True for all titrations:
At equivalence: mol of titrant added=mol of analyte initially
Example: Using volume to equivalence to
find information about analyte
A 0.120 g Sample of a base is added to 100. mL of water, it is titrated
with 3.00mL of 1.00M HCl, what is the molecular mass of the base?
(𝑚𝑜𝑙𝑠𝑏𝑎𝑠𝑒)𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = (𝑚𝑜𝑙𝑠𝑎𝑐𝑖𝑑)𝑎𝑑𝑑𝑒𝑑
𝑚𝑜𝑙
𝑚𝑜𝑙𝑠𝑎𝑐𝑖𝑑 = 1.00
∗ 0.003𝐿 = 0.00300 𝑚𝑜𝑙
𝑙
𝑚𝑜𝑙𝑠𝑎𝑐𝑖𝑑 = 𝑚𝑜𝑙𝑠𝑏𝑎𝑠𝑒 = 0.00300 𝑚𝑜𝑙
𝑔𝑟𝑎𝑚𝑠
0.120𝑔
𝑔
𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 =
=
= 40.0 𝑚𝑜𝑙
𝑚𝑜𝑙
0.00300𝑚𝑜𝑙
Review
Calculating the pH of a strong acid or base titration simply
requires finding how much H+ or OH- is in excess and solve
as we did our “mixing” problems.
Because initial mols of analyte=added mols of titrant you
can determine the mols of starting analyte using what you
know about how much titrant was added
Use this to solve for concentration, molecular mass
ect….using all previous stoichiometry methods.
Weak/Weak Titrations
Learning outcomes
Describe the basics of what is occurring during a
titration.
Which reactions are occurring in the beaker
Which equilibria are present as the titration is progressing.
Using a titration curve, determine what is being titrated
(weak/strong and acid/base) and what the titrant is (weak/strong and
acid/base)
Determine the pH at any point during a titration.
Determine information about the analyte using what
you are given about the titrant.
Activity Part 3: Titrations
Types of Titrations
Buret
Holds
“Titrant”
Known
concentration
Strong Acid/Strong Base
Strong Base/Strong Acid
Weak Acid/Strong Base
Weak Base/Strong Acid
Weak Acid/Weak Base
Erlenmyer
Holds
“analyte”
Unknown
concentration
Weak Base/Weak Acid
Example Problem:
We’ll do this as we discuss how to do the problem so you can
see it done:
A 10.0 mL solution of 0.300M NH3 is titrated with a
0.200M HCl solution. Calculate the pH after the following
additions of the HCl solution
0.0mL, 10.0 mL, 15.0mL, 30.0mL
First establish what type of
titration you are dealing with.
Step One: Establish what titration you are dealing with
Lets walk through them all together:
figures © Atkins and Jones
Titrations
Very complicated problems, be systematic!
Two similar approaches that work well to these
problems are detailed in the following slides. Pick
you’re a favorite way, you really don’t need to know
both.
In all cases you must know what type of solution you
have and how much of each species. We will do one
problem each way.
“weak” Titrations: Approach #1
(Typical approach taught, but honestly students tend to not like this way)
Step One: Establish what titration you are dealing with,
1B) if it helps draw a sketch of the graph
Step Two: Find the Equivalence Point
2B) if it helps label this on the graph axis
Step Three: Based on the question decide what area of the
titration you are in and how the problem is treated
i.e. weak acid only, weak base only, buffer, strong acid only,
strong base only.
Step 4: Treat problem just as we did in simpler previous
problems, just because it’s a titration doesn’t make it different.
Step Two: Find the Equivalence Point
(c) Atkins and Jones
A 10.0 mL solution of 0.300M NH3 is titrated
with a 0.200M HCl solution. Calculate the pH
after the following additions of the HCl solution
0.0mL, 10.0 mL, 15.0mL, 30.0mL
We do this because it tells is which portion of the
graph we are in at any given point. Often you will
be asked for it. It is an important point.
Finding the Equivalence Point
Unnecessary if using approach #1 but you’ll often be
specifically asked for the equivalence point
What is the equivalence point?
If you start with acid- its when the moles of base
to the moles of initial
acid
If you start with base- its when the moles of acid
to the moles of initial
added is equal
added is equal
base
Initial Molesacid=Final molesbase OR Initial Molesbase=Final molesacid
Find using MaVa=MbVb (concentration must be in mol/volume)
Step Two: Find the Equivalence Point
(c) Atkins and Jones
A 10.0 mL solution of 0.300M NH3 is titrated
with a 0.200M HCl solution. Calculate the pH
after the following additions of the HCl solution
0.0mL, 10.0 mL, 15.0mL, 30.0mL
We do this because it tells is which portion of the
graph we are in at any given point. Often you will
be asked for it. It is an important point.
𝑚𝑎 𝑉𝑎 = 𝑚𝑏 𝑉𝑏
(0.200𝑀)𝑉𝑎 = 0.300𝑀 (10.0𝑚𝐿)
𝑉𝑎 = 15.0𝑚𝐿
Based on the question decide what area of the titration
you are in and how the problem is treated
Not required if using approach 1: however, its extremely
useful to be able to be able to recognize the regions and you
will be asked about the regions and the regions based on given
equivalence points.
Beginning: weak base
Buffering region: buffer
Equivalence point: weak acid
15mL
All base is used up and doesn’t
play a role, now it is strong acid
problem
Titrations
Step 4 (both approaches): Treat problem just
as we did in simpler problems
We will do this on the document camera!
Step 4: Treat problem just as we did in simpler problems,
0.0 mL HCl added
Only weak base is present, treat as weak base problem
Since kb will be used, you’ll need to use concentration
15mL
Step 4: Treat problem just as we did in simpler problems,
10.0 mL HCl added
In buffering region:
Use HH equation: so use mols
(you can use concentration but it is more complicated
due to changing volumes)
15mL
Step 4: Treat problem just as we did in simpler problems,
15.0 mL HCl added
At equivalence point,
Same amount of strong acid added as weak base to start
This means all the weak base has been converted to a
weak acid
Treat as weak acid, use ka be sure to adjust concentration
(using m1v1=m2v2) use concentration since you are using
the k
15mL
Step 4: Treat problem just as we did in simpler problems,
30.0 mL HCl added
All of the weak base was changed into weak acid, and
then more strong acid was added.
The weak acid doesn’t have much of an effect as the
strong acid
Treat as strong acid problem
Need to figure out concentration though, some of it
was used to convert the weak base to weak acid, subtract
that
Find volume of HCl past equivalence, use m1v1=m2v2 to
find new concentration and fill in to pH=-log(H+)
15mL
“weak” Titrations: Approach #2:
(This is not the traditional method taught, however most students appear to prefer it)
Step One: Establish what titration you are dealing with,
Step Two: Write and equation that describes how the titrant (in
buret) and analyte (in beaker) react.
Step Three: Make ICF chart and react system to completion
(till one of reactants equals zero).
Step Four: Look at what is left over. What type of solution do
you have?
weak acid only, weak base only, buffer, strong acid only,
strong base only.
Step 4: Treat problem just as we did in simpler problems, just
because it’s a titration doesn’t make it different.
Example Problem:
Repeating the problem as you would with approach number
2:
A 10.0 mL solution of 0.300M NH3 is titrated with a
0.200M HCl solution. Calculate the pH after the following
additions of the HCl solution
0.0mL, 10.0 mL, 15.0mL, 30.0mL
Note:
A weak acid titrated with a strong
base is very similar to before: You’ll
do one of these in discussion.
Review
Titrations of a weak analyte require an understanding of what
part of the titration curve you are in before solving for a pH.
In both methods for solving you are required to find out what
type of solution you are dealing with after equilibrium has been
reached.
In method 1 this is done by comparing the volume to equivalence
In method 2 this is done by reacting the species to completion
using stoichiometry.
The equivalence point can be used to solve for initial moles of
titrant exactly like we did in strong acid and base calculations
(we did not do an example for weak acids/bases, but it’s the
same process)
Titration Example
Questions
Learning outcomes
Describe the basics of what is occurring during a titration
and answer questions based on this knowledge: For example:
Which reactions are occurring in the beaker
Which equilibria are present as the titration is progressing.
Answer questions about how initial conditions change the
equivalence point, and how percent purity of the analyte change
the equivalence point
Identify different types of titrations and describe why they are
different.
Estimate (< or > 7) the pH of the equivalence point of a titration
Identify the pKa by looking at a titration curve.
Question 1:
If you are doing a strong acid/strong
base or strong base/strong acid
titration is there a buffer area? Why?
No!!!
If a strong acid and base are combined
they neutralize each other (turning
into water) rather than create a weak
acid or base which can then dissociate.
Question 2
If a generic weak acid “HA” is titrated
with a strong base NaOH, what is the
major species at each labeled point?
A) HA
B) HA and AC) AD) A- +OH-
Question 3
What is the Ka of the acid? Estimate based
on the graph.
pH=pKa at half equivalence
About 3.5 ish.
Candy demo
Question 4
The analyte has 2 grams of base in 100mL of water.
How does the volume needed to reach equivalence change if the 2
grams are instead dissolved in 50mL of water?
No change
What if it is dissolved in 100mL of water?
No change
What if 1 gram is used instead?
It halves
What if 4 grams are used instead?
It doubles
Will the pH at equivalence be higher or lower than 7?
more acidic since all base is converted
to its conjugate acid. pH<7
Polyprotic acid titrations
Polyprotic acids:
Mostly conceptual since the
math is tedious
Learning Outcomes
Identify the major components present at each point in a poly
protic acid titration.
Identify the pKas of the acid AND be able to associate each Ka
with its appropriate equation.
Use volume to an equivalence point to determine the volume to
other equivalence points, and initial information about the
titration.
Note: We will not be doing any of the calculating pH for this, while technically it’s the
exact same as normal weak acid titrations it gets very tedious. If you are going into
analytical chemistry it may be worth spending some time learning how to do this, see me
in office hours for explanations.
Poly Protic Acids
A2-
HA-
HA-/A2-
H2A
H2A/HA-
Questions
A2-
HA-
What is the pKa associated with the removal of the
first H? (pKa1)
1.23
What is the Ka associated with the removal of the
second H? (pKa2)
HA-/A2H2A/HA-
H2A
4.19
Questions
A2-
HA-
HA-/A2H2A/HA-
H2A
If the volume of titrant to reach point D is 50.0
mL, what is the volume required to reach point B?
Point A?
B: 25.0 mL
A: 12.5 mL
If it takes 0.1 moles to reach the first equivalence
point, how many moles of analyte were there?
How many moles (total added) will it take to reach
the second equivalence point? **
Analyte: 0.1 mol
Second: 0.2 mol
**Note: I’m doing this in mols to speed up the discussion, of course I could also give you this in
volume and concentration.
Review
Polyprotic acids titrations have two equivalence points,
and two half equivalence points.
This is because they have different Ka values for each
hydrogen.
The moles required to reach the first equivalence point
is equal to the moles of initial analyte.
The moles required to reach the second equivalence
point is equal to 2x the moles of initial analyte.
Indicators
Learning Outcomes
Define an indicator.
Choose an indicator for a titration when given a chart
of possible indicators, or the pKa of possible
indicators.
Choose it based either on the equivalence point of the
titration or based on the complete titration curve.
Define an end point, and explain the difference
between an equivalence point and an end point.
Indicators
Red Cabbage Indicator
Red Cabbage Indicator
Purple Cabbage Indicator
https://www.youtube.com/watch?v=K8opH7qprX8
Good: Universal indicator (like pH paper)
Bad: No sharp contrast like in other indicators such as
phenolthalein
End point vs equivalence point
Equivalence point.
When moles of titrant added = moles of analyte
initially
End point
When the color of the indicator changes color
We want these to be in the same place!!!!
A “good” indicator
Changes color at the equivalence point.
Aka- end point and equivalence point is the same place.
Which of these is a good
indicator?
Methyl Red
Phenolphthalein
Could you use Thymol blue?
Not perfectly but it could
work….
Using pKa.
Indicators change color when pH=pKa.
We want indicators to change color at the equivalence
point.
Therefore pKa of indicator≅ equivalence point of analyte.
Example: Picking Indicator Based on pKa.
pKa of indicator≅ equivalence point of analyte.
Your stock room has available Methyl
Orange (pka=3.47) , and phenolphthalein
(pka=9.3).
For the titration shown, which is best?
phenolphthalein
Review
Your indicator should change color at the equivalence
point.
Aka: the end point and the equivalence point should be
similar.
To make this happen, the pKa of your indicator should be
approximately equal to the pH at the endpoint of your
titration.
Due to the big jump in pH once the buffering region is past,
this normally gives a wide range of possibilities.
Introduction to Solubility
Equilibrium
Just another K
Learning outcomes
Define solubility equilibrium.
Relate solubility equilibrium to other forms of
equilibrium we have discussed.
Define Ksp and write equations for Ks given a slightly
soluble compound.
Define Molar Solubility
Solubility Equilibrium
Not all substances are soluble
Many substances are only slightly
soluble
Ksp the new version of K, same as all the
other
Products over reactants
Now reactants are solids not aqueous
though, are they included?
As with other it is a dynamic
equilibrium
Reminder on Dynamic Equilibrium
Constantly changing
Constant precipitation, and
dissolving
Total amount dissolved stays
constant. (after eq is reached)
Why doesn’t the solid matter in the
K equation here?
Solubility Product Examples
Molar SolubilityThe amount of a compound that is
soluble.
Mols/L
Calculated from Ksp
Usually designated “s”, I use x as we’ve
been doing with equilibrium constants.
Why? A handwritten s looks very much
like a handwritten 5.
Problem solving with
solubility equilibrium.
Learning outcomes
Luckily you have done most of this already with other equilibria
Given the Ksp we can find the concentrations of ions
Same as Ka, Kb, Keq ect…, only easier since there is no
denominator.
Given the concentration of an ion we can find the ksp
Same as Ka, Kb, Keq ect…, only easier since there is no
denominator
Use Molar solubility to solve for Ksp
Use Molar solubility to solve for concentrations of ions.
On your own
You have done most of this already with other equilibria
Given the Ksp we can find
the concentrations of ions
Same as Ka, Kb, Keq ect…,
only easier since there is
no denominator.
Given the concentration of
an ion we can find the ksp
Same as Ka, Kb, Keq ect…,
only easier since there is
no denominator
This is absolutely identical to every other
equilibrium we have done, only easier. I will let
you do these problems on your own and focus
on molar solubility. One note: most people
choose not to make an ice chart for these, that is
totally acceptable.
Molar Solubility: Examples
Basically the same as previous equilibrium problems but now using the definition of
molar solubility (The amount of a compound that is soluble in mols/L).
The molar solubility of MnCO3 is 4.2x10-6M What is the Ksp?
If Aluminum Sulfate Ksp is 9.83x10-10 what is the molar solubility?
Hint: Most wrong answers on Ksp questions are because its easy to forget to include the coefficients inside the
brackets, and easy to forget to exponentiate them given our time spent with acids and bases
Molar Solubility and Common
Ion Effect Examples
How many grams of CaCO3 will dissolve in 3.0x10-3 mL of 0.050 M
Ca(NO3)2 . The Ksp is 8.7x10-9.
Calculate the molar solubility of AgCl in a 1.00L solution containing 10.0g
of dissolved CaCl2 Ksp=1.6x10-10
Review
Ksp, like other equilibrium constants are products over
reactants (with exponents equal to the coefficients).
Because the reactants are solid, and are not included this makes
calculations much easier (seldom are there quadratics).
Molar solubility, being the amount that a solid can dissolve will
allow you to calculate the amount of ions dissolved (using
stoichiometric coefficients), and therefore the Ksp.
Of course this means that using Ksp, you can solve for the ions
dissolved (as done in previous sections) and use that to solve for
molar solubility.
Common Ion Effect
Learning Outcomes
Determine if mixing two solutions will cause a
precipitate to form.
Identify precipitate and spectator ions
Using previous problem solving techniques determine
the concentration of all ions in solutions
Common Ion Effect and Solubility
If two solutions which are both
soluble are mixed, but any
combination of those ions form an
insoluble compound a precipitate
may form.
Lets look at this happening!
Demo: CuCl2 and NaOH
soluble
soluble
soluble
insoluble
Mixing Solutions!
We can see if you mix two salt solutions if a precipitate will form.
This is like using Q to determine which way the reaction will go.
You can have two different salt
solutions that are perfectly
soluble, but when mixed form a
precipitate
See the picture for how this
happens
You can calculate if this will
happen
Q<K no precipitate
Q=K saturated solution
Q>K precipitate
Demo: CuSO4 and NaOH
Lets look at that last one a bit closer….
precipitate
You can calculate if this will happen
Q<K no precipitate
Q=K saturated solution
Q>K precipitate
Spectator
ions
Basically same concept as
Le Chatlier’s Priciple and
shifting right or left!
From earlier chapters
Precipitation Examples
If 20.0 mL of 0.10 M Ba(NO3)2 are added to 50.0 mL of 0.10M
Na2CO3 will BaCO3 precipitate? Ksp=8.1x10-9
Toughy:
A volume of 75 mL of 0.060 M NaF is mixed with 25 mL of
0.15M Sr(NO3)2. Calculate the concentrations in the final
solution of NO3- , Na+, and Sr2+, F-. (Ksp for SrF2=2.0x10-10)
Review
Two solutions with soluble ions can form a precipitate if
mixed.
A precipitate is formed if any combination of the ions are a
slightly soluble compound AND if the Q>K
If a precipitate is formed you can calculate how much is
formed using stoichiometry. (limiting reagent problem)
You can figure out how much of each ion is present.
Depending on the ion:
Treat as spectator ion
Excess reagent
Or find using Ksp
Selective Precipitation and
precipitating agents
Learning Outcomes
Use a selective precipitation scheme, and results from
an experiment based on it to determine which ions are
present in a sample. (note: do not memorize these
schemes, they will be given to you).
Calculate the amount of a precipitating agent needed
to precipitate ions at a given concentration.
Use the above knowledge to tell how pure compounds
precipitated by selective precipitation will be.
Selective Precipitation
Unknown 1:
Step 1: Add peroxide and
NaOH. No precipitate
present.
Step 2: Add HNO3 and
NH3: precipitate present.
Al(OH)3
Selective Precipitation
Unknown 2:
Step 1: Add peroxide and NaOH.
Precipitate present.
Step 2: Add H2SO4 precipitate
present.
Step 3: HNO3 and NaBiO3
Step 4: Add KNO3, precipitate
K3(Co(NO2)6)
Co2+
Selective Precipitation:
Difference in solubility
Two ions with different Ksp can be separated by slowly adding a
precipitating agent:
Example:
BaSO4 has a Ksp of 1.1x10-10
PbSO4 has a Ksp of 1.6x10-6
If a solution has dissolved barium and lead ions, which will
precipitate first if a small amount of sulfate is added?
Lower Ksp= lower solubility =BaSO4
Example:
Selective Precipitation:
Difference in solubility
BaSO4 has a Ksp of 1.1x10-10
PbSO4 has a Ksp of 1.6x10-6
At what concentration of sulfate would barium start to precipitate if the
solution contained 0.010M barium ions?
At what concentration of sulfate would lead start to precipitate if the
solution contained 0.010 M lead ions?
𝐾𝑠𝑝 = 𝐵𝑎2+ [𝑆𝑂42− ]
[𝑆𝑂42− ]
𝐾𝑠𝑝
=
𝐵𝑎2+
[𝑆𝑂42− ] =
[𝑆𝑂42− ]
=
1.1𝑥10−10
0.010𝑀
1.1𝑥10−8 M
𝐾𝑠𝑝 = 𝑃𝑏 2+ [𝑆𝑂42− ]
[𝑆𝑂42− ]
𝐾𝑠𝑝
=
𝑃𝑏 2+
−6
1.6𝑥10
[𝑆𝑂42− ] =
0.010𝑀
[𝑆𝑂42− ] = 1.6𝑥10−6 M
Example:
A chemist attempts to separate barium ions from lead
ions by using the sulfate ion as a precipitating agent.
What is the concentration of barium ions when the lead
sulfate begins to precipitate?
1) You need to find what sulfate concentration will precipitate each ion. We did this on
the last slide !
Lead
Barium
[𝑆𝑂42− ] = 1.1𝑥10−8 M
[𝑆𝑂42− ] = 1.6𝑥10−6 M
2) What do we know about barium when lead starts to precipitate? We don’t! BUT we
know the concentration of sulfate at that moment in time. SO…..Use the sulfate when
LEAD starts to precipitate to calculate the [Ba2+] concentration from the BaSO4 Ksp.
𝐾𝑠𝑝 = 𝐵𝑎2+ [𝑆𝑂42− ]
𝐵𝑎2+
𝐾𝑠𝑝
=
[𝑆𝑂42− ]
𝐵𝑎2+
1.1𝑥10−10
−6 𝑀
=
=
6.9𝑥10
1.6𝑥10−6 M
Review
Selective precipitation can be used when two ions have
different solubility rules where one precipitates and one
doesn’t.
There are schemes worked out to separate out lots of these
ions based on their solubility rules.
You can also separate out ions that both precipitate with
the same counter ion, BUT have different Ksp.
Remember though there will be some contamination.
The further away the Ksp are from each other, the less cross
contamination.
Ksp and tooth decay.
Learning outcomes
Combine materials from all chapters to:
Describe the reasons behind common advice given by
dentists about diet for preventing tooth decay.
Define how common tooth decay preventions work.
And yes, these are all based on solubility and acid/base
chemistry!!!!
Tooth Decay
How does teeth decay occur?
How does your body naturally prevent tooth decay?
Why does sugar increase tooth decay?
How do we artificially help prevent tooth decay and how
do these processes work?
How does teeth decay occur?
Tooth Decay
How does teeth decay occur?
How does your body naturally prevent tooth decay?
Why does sugar increase tooth decay?
How do we artificially help prevent tooth decay and how
do these processes work?
How does your body naturally prevent
tooth decay?
Saliva!
PO43- and HPO42-
Tooth Decay
Why does sugar increase tooth decay?
Bacteria create acidic metabolites!
Tooth Decay
How do we artificially help prevent tooth decay and how do these
processes work?
Tooth Brushing:
Fluoride Treatments
Sugar Free Gum:
Tooth Decay
How do we artificially help prevent tooth decay and how do these processes work?
Tooth Brushing:
removes food
Fluoride Treatments
Sugar Free Gum:
Increases Saliva
Review
Tooth decay is generally caused by acidic metabolites from bacteria
eroding enamel.
Important general concept: For things that react with acidic
compounds, acid increases solubility.
We combat not only by physically removing the food and bacteria
but altering the chemistry
Gum (sugar free) causes increased saliva which has buffering/basic
compounds.
Important general concept: buffers decrease the effect of acids
Fluoride treatments change the enamel of our teeth to a compound
with a lower Ksp.
Important general concept: lower Ksp means compounds are less soluble.
Complex Ion formation.
Note:
We are running behind so I’m not going to cover this
in class, you will not have homework on it and you will
not be tested on it. However I will include my notes on
it in case you are curious about what I have to say on it.
This is solely FYI. I am happy to answer questions on it
in office hours/facebook/through the survey if you
have any.
Complex Ion Formation.
Used to increase the solubility of a salt. Using a second
reaction. E.G.
How does this work?
“hides” ions away, allowing for more to be dissolved.
What is the new solubility of AgCl?
Lets add reactions and find the new K
Steps for solving complex ion
formation problems:
Step 1: Find K
Use Ksp and Kf
Step 2: Use K to solve solubility problem as previous
Ksp problems
ICE chart (if you want)
Fill into K equation.
Complex Ion Formation: Example
Calculate the molar solubility of AgCl in pure
water and 3.0 M NH3
First Pure water:
Complex Ion Formation: Example
Calculate the molar solubility of AgCl in pure
water and 3.0 M NH3
3.0 M NH3