Turn Test Corrections in to the Box
You will need a calculator for these notes!
When salts, acids, or bases are added
to water they ionize. For example:
NaCl(s) → Na+(aq) + Cl-(aq)
Could an equilibrium be established for
this reaction? What happens if you add
a lot of salt to a small glass of water?
What would you be able to see if it was
at equilibrium? What would you not see
if it was at equilibrium?
How would this be written at
equilibrium? NaCl(s)  Na+(aq) + Cl-(aq)
Are there any salts that would never
come to equilibrium? What about acids
or bases? Why?
For acids and bases to set up an
equilibrium they must be weak by
definition. When they come to
equilibrium (in water) we can write
equilibrium expressions for them.
In general for a weak, monoprotic acid:
HA  H+ + A[H+][A-]
so Ka =
[HA]
Why “Ka”?
But acids must be in water, so it can
look like this:
HA + H2O  H3O+ + AKa =
[H3O+][A-]
[HA]
[H2O] usually considered = 1
A solution of nicotinic acid was found
at equilibrium to contain [HA] = 0.049M,
[H3O+] = [A-] = 8.4 x 10-4 M. What is the
Ka and pKa?
Starting with a 0.0100 M solution of
acetic acid, when added to water 4.2%
ionizes. What is the Ka and pKa?
The pH of a 0.15 M solution of
chloroscetic acid (ClCH2COOH) is 1.92.
What is the value of Ka?
What are the equilibrium
concentrations if the Ka of 0.10 M HOCl
is 3.5 x 10-8?
Calculate the percent ionization of a
0.10 M acetic acid solution.
If you haven’t caught it by now:
The higher the Ka value (less negative
exponent) the lower the pKa value
(closer to 0).
Also, the “rule of thumb” for the
simplification of the quadratic only
works if Ka is “x 10-4” or a larger
negative exponent. When compared to
the [HA]!
Calculate the [OH-] for a 0.20 M solution
of NH3. Kb is 1.8 x 10-5.
Calculate the pH for a 0.20 M solution
of NH3. Kb is 1.8 x 10-5.
Calculate the % ionization for a 0.20 M
solution of NH3. Kb is 1.8 x 10-5.
Polyprotic acids can give off more than
one H+, so it is considered to occur in
steps, each with their own Ka value.
For example:
H3PO4  H+ + H2PO4H2PO4-  H+ + HPO42HPO42-  H+ + PO43-
Ka1 = 7.5 x 10-3
Ka2 = 6.2 x 10-8
Ka3 = 3.6 x 10-13
So if the concentrations were wanted,
we would need to solve all three
reactions and then add up the results.
(Not hard, but definitely time consuming.)
Would anyone like a shortcut?
Ka3 = 3.6 x 10-13 and (remember) Kw = 1 x10-14,
so the contribution of the third equation to
the total of the acid is only a little larger than
water’s own contribution.
Specifically, if we started with 0.10 M H3PO4,
the third ionization would only add
0.00000000000000000093 M H+ to the total.
Rule of Thumb: “x 10-9” and larger negative
exponents indicate that reaction does not
contribute in any noticeable way to the total
(the additional amount will be rounded away).
Calculate the concentrations in 0.10 M
H2SO4. Ka2 = 1.2 x 10-2
When salts ionize it is possible one of
the ions formed would react with water
in a side reaction, but this reaction
could change the pH.
Bronstead Lowry definitions of acids
and bases give us conjugate acid-base
pairs. Strong acids/bases give weak
conjugate bases/acids as products, and
weak acids/bases give strong conjugate
bases/acids.
It is the conjugates of weak
acids/bases that are the ones that
typically react with water.
For nonexample:
NaCl → Na+ + Cl-
Na+ would be the cation of NaOH (a strong
base) so it would be a weak acid, and would
not react with water.
Cl- would be the anion of HCl (a strong acid)
so it would be a weak base, and would not
react with water.
Strong base - weak acid example:
NaC2H3O2 → Na+ + C2H3O2-
Na+ would be the cation of NaOH (a strong
base) so it would be a weak acid, and would
not react with water.
C2H3O2- would be the anion of HC2H3O2 (a
weak acid) so it would be a strong base, and
would react with water, like this:
C2H3O2- + H2O  HC2H3O2 + OHWhat type of solution is the result? What
would be the pH? Why would we use Kb?
Strong acid - weak base example:
NH4Cl → NH4+ + Cl-
NH4+ would be the cation of NH4OH (a weak
base) so it would be a strong acid, and would
react with water, like this:
NH4+ + H2O  NH3 + H3O+
Cl- would be the anion of HCl (a strong acid)
so it would be a weak base, and would not
react with water.
What type of solution is the result? What
would be the pH? Ka or Kb?
Once the side reaction is known, the
amount of acid or base formed can be
calculated just like we did earlier.
It might be helpful to know:
Kw = 1 x10-14 = KaKb
so if you know Ka and need Kb or viseversa you can calculate what you need
Weak acid - weak base:
If the salt of a strong base with weak
acid gave a basic solution, and
the salt of a strong acid with a weak
base gave an acidic solution, then
What happens if they are both weak?
if Ka = Kb, then neutral
if Ka > Kb, then acidic
(acid was stronger than base, but still weak)
if Kb > Ka, then basic
(base was stronger than acid, but still weak)
Let’s recap:
NaCl
Na+ + H2O ⇌ NaOH + H+
Cl- + H2O ⇌ HCl + OH-
Let’s recap:
Strong Base
NaCl
Na+ + H2O ⇌ NaOH + H+
Cl- + H2O ⇌ HCl + OH-
Let’s recap:
Strong Base = No Reaction
NaCl
Na+ + H2O ⇌ NaOH + H+
Cl- + H2O ⇌ HCl + OH-
Let’s recap:
Strong Base = No Reaction
NaCl
Na+ + H2O ⇌ NaOH + H+
Cl- + H2O ⇌ HCl + OHStrong Acid
Let’s recap:
Strong Base = No Reaction
NaCl
Na+ + H2O ⇌ NaOH + H+
Cl- + H2O ⇌ HCl + OH-
Strong Acid = No Reaction
Let’s recap:
Strong Base = No Reaction
NaCl
Na+ + H2O ⇌ NaOH + H+
Cl- + H2O ⇌ HCl + OH-
Strong Acid = No Reaction
Conclusion: Neither Acidic or Basic = Neutral
Let’s recap:
NH4Cl
NH4+ + H2O ⇌ NH4OH + H+
Cl- + H2O ⇌ HCl + OH-
Let’s recap:
Weak Base
NH4Cl
NH4+ + H2O ⇌ NH4OH + H+
Cl- + H2O ⇌ HCl + OH-
Let’s recap:
Weak Base = Yes Reaction
NH4Cl
NH4+ + H2O ⇌ NH4OH + H+
Cl- + H2O ⇌ HCl + OH-
Let’s recap:
Weak Base = Yes Reaction
NH4Cl
NH4+ + H2O ⇌ NH4OH + H+
Cl- + H2O ⇌ HCl + OHStrong Acid
Let’s recap:
Weak Base = Yes Reaction
NH4Cl
NH4+ + H2O ⇌ NH4OH + H+
Cl- + H2O ⇌ HCl + OH-
Strong Acid = No Reaction
Let’s recap:
Weak Base = Yes Reaction
NH4Cl
NH4+ + H2O ⇌ NH4OH + H+
Cl- + H2O ⇌ HCl + OH-
Strong Acid = No Reaction
Conclusion: as H+ is produced = acidic
Let’s recap:
NaC2H3O2
Na+ + H2O ⇌ NaOH + H+
C2H3O2- + H2O ⇌ HC2H3O2 + OH-
Let’s recap:
Strong Base
NaC2H3O2
Na+ + H2O ⇌ NaOH + H+
C2H3O2- + H2O ⇌ HC2H3O2 + OH-
Let’s recap:
Strong Base = No Reaction
NaC2H3O2
Na+ + H2O ⇌ NaOH + H+
C2H3O2- + H2O ⇌ HC2H3O2 + OH-
Let’s recap:
Strong Base = No Reaction
NaC2H3O2
Na+ + H2O ⇌ NaOH + H+
C2H3O2- + H2O ⇌ HC2H3O2 + OHWeak Acid
Let’s recap:
Strong Base = No Reaction
NaC2H3O2
Na+ + H2O ⇌ NaOH + H+
C2H3O2- + H2O ⇌ HC2H3O2 + OHWeak Acid = Yes Reaction
Let’s recap:
Strong Base = No Reaction
NaC2H3O2
Na+ + H2O ⇌ NaOH + H+
C2H3O2- + H2O ⇌ HC2H3O2 + OHWeak Acid = Yes Reaction
Conclusion: as OH- is produced = basic
Let’s recap:
NH4C2H3O2 – Your Turn…
Let’s recap:
NH4C2H3O2
NH4+ + H2O ⇌ NH4OH + H+
C2H3O2- + H2O ⇌ HC2H3O2 + OH-
Let’s recap:
Weak Base = Yes Reaction
NH4C2H3O2
NH4+ + H2O ⇌ NH4OH + H+
C2H3O2- + H2O ⇌ HC2H3O2 + OHWeak Acid = Yes Reaction
Let’s recap:
Weak Base = Yes Reaction
NH4C2H3O2
NH4+ + H2O ⇌ NH4OH + H+
C2H3O2- + H2O ⇌ HC2H3O2 + OHWeak Acid = Yes Reaction
Conclusion: we cannot yet say if it is acidic or basic!
As Kb = 1.8 x 10-5 and Ka = 1.8 x 10-5 it is neutral,
otherwise the larger K wins! (larger = small -exponent
More Ionic Equilibria
(Acid-Base Interactions)
Much of life involves acids and bases,
including your body.
•What is in your stomach?
•What must your pancreas produce to avoid
dissolving your intestines?
•When carbon dioxide is dissolved in water
(blood) it forms carbonic acid. What must
also be in our blood to avoid dissolving our
vessels?
•There are people who get sick after drinking
a coke (coke contain acids). What must they
be lacking?
Much of life, then, involves buffers.
A buffer contains a conjugate acid-base pair
with both the acid and base in reasonable
concentrations. (By Bronstead-Lowry
definition.)
The acidic part would react with added base.
The basic part would react with added acid.
For example: acetic acid and sodium acetate
NaC2H3O2 → Na+ + C2H3O2HC2H3O2  H+ + C2H3O2If OH- were added, what would react?
OH- + H+  H2O How would pH be affected?
If H+ were added, what would react?
H+ + C2H3O2-  HC2H3O2 How would pH be
affected?
Notice that the addition of acetic acid and
sodium acetate both give off acetate ion.
NaC2H3O2 → Na+ + C2H3O2HC2H3O2  H+ + C2H3O2What would having this much acetate ion do
to the equilibrium of acetic acid?
What would this mean to the amount of acetic
acid that would ionize?
When a solution of a weak electrolyte (weak
acid or base or not very soluble salt) is altered
by adding one of the ions from another
source, the ionization of that weak electrolyte
is decreased. This is called the common-ion
effect.
Why is the common-ion effect just an
application of LeChatelier’s Principle?
NaC2H3O2 → Na+ + C2H3O2HC2H3O2  H+ + C2H3O2Why is it just the acetic acid that is affected?
What happens to the pH of the mixture
compared to the pH of acetic acid?
Let’s prove it: what is the pH of a 0.10 M
acetic acid solution if Ka = 1.8 x 10-5?
Let’s prove it: what is the pH of a 0.10 M
acetic acid and 0.20 M sodium acetate
solution if Ka = 1.8 x 10-5?
Shortcut anyone?
Thanks to the Henderson-Hasselbalch
equation…
[salt]
pH = pKa + log
[acid]
and by logic…
[salt]
pOH = pKb + log
[base]
What is the pH of a 0.10 M acetic acid and 0.20
M sodium acetate solution if Ka = 1.8 x 10-5?
Using the H-H equation…
The purpose of a buffer system in our bodies
or anywhere else is to minimize the change
that adding an acid or base causes to the pH.
For example: adding 0.010 mols of NaOH to…
1L water makes pH 7→ 12
0.1 M acetic acid makes pH 2.89 → 3.78
What about 1L of acetic acid/acetate buffer?
Start by finding the initial pH, and then the
change…
What is the original pH of a 1.00 L solution
that is 0.100 M acetic acid and 0.100 M sodium
acetate.
What is the new pH of a 1.00 L solution that is
0.100 M acetic acid and 0.100 M sodium
acetate if 0.010 mol solid NaOH is added. (No
volume change.)
Buffers are prepared by mixing solutions, but
there are several methods to do this.
Regardless of the method remember:
When mixing solutions the final volume is the
sum of the two solutions’ volume, so the
concentrations will need to be recalculated!
Calculate the pH of a buffer solution prepared
by mixing 200. mL of 0.10 M NaF and 100. mL
of 0.050 M HF. Ka = 7.2 x 10-4
Calculate the grams of NH4Cl that must be
used to prepare 500. mL of a buffer solution
that is 0.10 M in NH3 and has a pH of 9.15.
Calculate the pH of a solution obtained by
mixing 400. mL of a 0.200 M acetic acid
solution and 100. mL of a 0.300 M NaOH soln.
What is the concentration of a 25.0 mL of
ammonia solution if it took 30.0 mL of 1.0 M
HCl to titrate it to the equivalence point?
What will be the pH of the titrated ammoniaHCl mixture at the equivalence point?
What will be the pH of the titrated ammoniaHCl mixture at the half equivalence point?
What was the original pH of the ammonia?
Draw a titration curve for all of this.
Titration Curves and Buffers
Titration Curves graphically represent the
results on pH of adding acid to base or base
to acid. Points to note:
Equivalence Point - the point where [H+]=[OH-]
Indicators change at the end point, ideally
identical to the equivalence point.
Half Equivalence Point - the point at where the
pH = pKa, so Ka could easily be calculated
Strong Acid-Strong Base Curves
Where should the endpoint occur?
Where is the half equivalent point?
Strong Acid-Strong Base Curve
Weak Acid-Weak Base Curve
Triprotic Acid-Strong Base Curve