Transcript Acids bases and buffers - Grand Junction High School

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Arrehnius: acids produce H+ and bases produce OHions.
Bronsted-Lowry: acids donate H+ bases accept H+
Lewis acids accept electron pairs and bases donate
electron pairs.
Water can act as an acid or a base.
Ha(g)acid + H2O(l) base ↔ H3O+(aq) conj. acid + A- (aq) conj. base
A conjugate acid-base pair consists of two substances
related to each other by the donating and accepting
of a single proton.
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HF(aq) + H2O(l) ↔ F-(aq) + H3O+(aq)
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ACID
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HF(aq) + H2O(l) ↔ F-(aq) + H3O+(aq)
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WA
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HCl(aq) + H2O(l)  Cl-(aq) + H3O+(aq)
BASE
Conj. Base
BASE
SB
Base
weak base
Conj. Acid
ACID
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Strong acid
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CH3COOH(aq) + H2O(l) ↔ CH3COO-(aq) + H3O+(aq)
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Weaker acid
weaker base
acid
stronger base
stronger acid
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If the H2O is a much stronger base than A-, (has a
larger affinity) then the equilibrium constant will
be far to the right; most of the acid dissolved will
be in ionized form.
Conversely, if A- is a much stronger base than H2O,
the equilibrium position will lie far to the left.
Ka = [H3O+][A-] = [H+][A-]
[HA]
[HA]
Where Ka is called the acid dissociation constant.
A Ka value > 1 favors the products!
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Acid Dissociation (Ionization) Reactions
Write the simple dissociation (ionization)
reaction (omitting water) for each of the
following acids:
Hydrochloric acid
HCl (aq) ↔ H+(aq) + Cl-(aq)
Acetic Acid:
HC2H3O2(aq) ↔ H+(aq) + C2H3O2-(aq)
Ammonium ion:
NH4+(aq) ↔ H+(aq) + NH3(aq)
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C6H5NH3+
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C6H5NH3+ ↔ H+(aq) + C6H5NH2(aq)
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[Al(H2O)6]3+ (aq)Use the meaning of Ka for this.
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[Al(H2O)6]3+ (aq) ↔ H+(aq) + [Al(H2O)5]2+(aq)
Bronsted-Lowry can also be used for reactions in the
gas phase.
NH3(g) + HCl(g) ↔ NH4Cl(s) where the proton is
donated to the ammonia
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A strong acid is one for which the equilibrium lies far
to the right.
A strong acid will yield a weak conjugate base (weaker
than water) because it has a weak affinity for a proton.
When strong acids like HCl(aq) are in solution, their
position of the dissociation equilibrium lies so far to
the right that [HCl] cannot be measured accurately.
This prevents an accurate calculation of Ka .
Thus: Ka = [H+][Cl-]
[HCl]  very small and highly uncertain
Organic acids are usually weak acids and have a
carboxyl group – COOH.
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HCl
HBr
HI
HClO4
H2SO4
HNO3
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Group I and II metals
(except Mg and Be)
and their hydroxides
or oxides.
Example: NaOH or
Na2O
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Acids with oxygen present in the polyatomic ion part
will be stronger then those that do not have the
oxygen. Why?
Cl—O—H
O—Cl—O—H
O2—Cl—O—H
O3—Cl—O—H
All of these have an oxygen attached to a hydrogen
that will dissociate to form H+ ions leaving the rest of
the molecule in solution. The more hugely
electronegative oxygens there are, the more easily the
bond is broken for the H+ to dissociate.
Acidity and alkalinity are measured with a
logarithmic scale called pH. Here is why: a strongly
acidic solution can have one hundred million million
(100,000,000,000,000) times more hydrogen ions than
a strongly basic solution! The flip side, of course, is
that a strongly basic solution can have
100,000,000,000,000 times more hydroxide ions than a
strongly acidic solution. No one wants to deal with all
those zeros so:
pH = - log [H+] pOH = - log[OH-] pH + pOH = 14
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Arrange the following species
according to their strengths as
bases.
F-, Cl-, NO2-, CN-.
Water is a stronger base than the
conjugate base of a strong acid
but is a weaker base than the
conjugate base of a weak acid.
Cl- < H2O< conjugate bases of
weak acids.
Weakest bases  strongest bases
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You can order the remaining
conjugate bases by
recognizing that the strength
of an acid is inversely
related to the strength of its
conjugate base.
Ka for HF>Ka for HNO2>Ka
for HCN
Base strengths increase as
follows:
F- < NO2- < CNCl- <H2O <F- <NO2 - <CN-
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A substance is amphoteric if it can behave as either an
acid or a base.
Water is amphoteric: 2H2O(l) ↔ H3O+(aq) + OH-(aq)
Autoionization of water involves the transfer of a
proton from one water molecule to another to
produce hydroxide and hydronium ions.
Pure water at 250C has ion concentrations of:
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[H3O+] = [OH-] = 1.00 x 10-7 M
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Kw = [H3O+] [OH-] = 1.00 x 10-14 M
At 600C, the value of Kw is 1 x 10-13
a. Use Le Chatelier’s principle to predict whether the
reaction 2H2O(l) ↔ H3O+(aq) + OH-(aq) is exothermic or
endothermic.
b. Calculate [H+] and [OH-] in a neutral solution at
600C.
a. Kw increases from 1 x 10-14 at 250C to 1 x 10-13 at
600C. Le Chatelier’s principle states that if a system at
equilibrium is heated, it will adjust to consume
energy. Since the value of Kw increases with
temperature, we must think of energy as a reactant,
so the process must be endothermic.
b. At 600C, [H+] = [OH-] = 1.00 x 10-13 M
For a neutral solution, [H+] = [OH-] = √1.00 x 10-7 M
= 3.0 x 10-7 M
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Calculate [H+] or [OH-] as required for the
following solutions at 250C, and state whether
the solution is neutral, acidic or basis:
1.0 x 10-5 M OH1.0 x 10-7 M OH10.0 M H+
1.0 x 10-9 M solution is basic
1.0 x 10-7 M solution is neutral
1.0 x 10-15 M solution is acidic
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pH (percentage of H+) is a log scale based on 10.
pH = - log [H+]
[H+] = 1.0 x 10-7 M
pH = - (-7.00) = 7.00
Sig figs: the number of decimal places in the log is
equal to the number of sig figs in the original
number. So the equation above is correct (2 sig figs
to the right of the decimal pt.)
pOH = - log[OH-]
pK = - log K
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Calculate the pH and pOH for each of the
following solutions at 250C:
A. 1.0 x 10-3 M OH B. 1.0 M OHA. [H+] = Kw ÷ [OH-] = 1.0 x 10-11M
pH = - log [H+] = -log(1.0 x 10-11) = 11.00
pOH = - log [OH-] = - log (1.0 x 10-3) = 3.00
B. pH = 0.00
pOH = 14.00
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The pH of a sample of human blood was measured to be
7.41 at 250C. Calculate pOH, [H+] and [OH-] for the
sample.
pOH + pH = 14.00
pOH = 14.00 – 7.41 = 6.59
pH = -log[H+]  7.41 = -log[H+] or log[H+] = -7.41
Need to know the antilog of -7.41
Antilog (n) = 10n
- pH = log[H+]
[H+] = antilog (-pH)
[H+] = antilog(-7.41) = 10-7.41 = 3.9 x 10-8
[OH-] = antilog (pOH) = antilog -6.59 = 10-6.59= 2.6 x 10-7
a.
b.
Calculate the pH of 0.10 M HNO3
Calculate the pH of 1.0 x 10-10 M HCl
HNO3 is a strong acid and will have H+, NO3- and H2O in
solution.
b.
There are virtually no HNO3 molecules in solution and the
number of OH- will be very small since the H+ ions from the
acid will drive the equilibrium to the left:
c.
2H2O(l) ↔ H+(aq) + OH-(aq)
d.
[H+] > [OH-], so [OH-] < 10-7M
e.
The sources of H+ are from the HNO3 and water. The
number of H+ ions from the water are negligible so the only
important source of H+ comes from the acid:
f.
[H+] = 0.10 M and pH = -log(0.10) = 1.00
The HCl is in such as small concentration that it has no effect and
the pH will be the same as pure water = 7.00
a.
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Calculate the pH of a 1.00 M solution of HF (Ka = 7.2
x 10-4)
Step one: list major species in the solution
HF and H2O
Choose the species that can produce H+ and write
the balanced equations for the reactions producing
H+
HF(aq) ↔ H+(aq) + F-(aq) Ka = 7.2 x 10-4
H2O(l) ↔ H+(aq) + OH-(aq) Ka = 1.0 x 10-14
Both of these can produce H+ but typically one
source can be singled out, in this case the Ka of HF(aq)
is much bigger then water so we’ll ignore the water.
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Calculating pH of weak acids involves setting up
an equilbrium. Use a RICE table and start with a
balanced equation, setting up Ka, find initial
concentrations, changes and final concentrations.
You will find that for acid dissociations:
[H+][A-]
x2
Ka = [HA] = [HA] – x
This will either be a quadratic problem OR
sometimes you can “neglect” x value in the
denominator if the Ka is really small. How can you
tell? Compare the original Ka to 100Ka . If the
initial concentration is large by comparison, you
can neglect subtracting the x term.
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[H+][A-]
Ka = [HA]
=
x2
1.8 x 10-5 = [1 x 10-4] – x
Neglect x? 100 (1.8 x 10-5) =
.0018 which is too close to
.0001 so you must use the
quadratic for this one.
x2 + 1.8 x 10-5x – 1.8 x 10-9 =
3.44 x 10 -5 M
pH = -log 3.44 x 10-5 = 4.46
R
HC2H3O2
H+
C2H3O2-
.0001
0
0
-x
+x
+x
.0001 – x
+x
+x
I
C
E
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Write the equilibrium expression for the
dominant equilibrium:
Ka = 7.2 x 10-4 = [H+][F-] ÷ [HF]
To solve, list initial concentrations before any
HF dissociates:
[HF]0 = 1.00M [H+] = 10-7 ≈ 0 and [F-] = 0
HF
H+
F-
1.00
0
0
-x
+x
+x
1.00 – x
+x
+x
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Ka = 7.2 x 10-4 = (x)(x) ÷ 1.00 – x
This can be solved with the quadratic formula
but since the Ka for HF is so small, HF will only
dissociate slightly and x is expected to be really
small. (Ka = .00072 x 100K = .072 while the
initial concentration is 1.00—get rid of the x in
the denominator!)
So: Ka = 7.2 x 10-4 = (x)(x) ÷ 1.00
x = √7.2 x 10-4 = 2.7 x 10-2
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The hypochlorite ion (OCl-) is a strong oxidizing
agent often found in household bleaches and
disinfectants. It is also the active ingredient that
forms when swimming pool water is treated with
chlorine. In addition to its oxidizing abilities, the
hypochlorite ion has a relatively high affinity for
protons (it is a much stronger base than Cl-, for
example) and forms the weakly acidic
hypochlorous acid (HOCl, Ka = 3.5 x 10-8).
Calculate the pH of a 0.100 M aqueous solution of
hypochlorous acid.
They gave you Ka, search for x (RICE table)
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Ka = [H+][A-]
[HA]
3.5 x 10-8 = x2 ÷ .100 –x
You can neglect the x in the
.100 – x because the Ka is so
super small.
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R
HA ↔
H+
A-
I
0.100
0
0
C
-x
+x
+x
E
.100 - x
+x
+x
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x = √ 3.5 x 10-9
x = 5.92 x 10-5
- log 5.92 x 10-5 = 4.23
pH = 4.23
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Calculate the pH of a solution that contains
1.00 M HCN (Ka = 6.2 x 10-10) and 5.00 M HNO2
(Ka = 4.0 x 10-4) Also calculate the
concentration of cyanide ion (CN-) in this
solution at equilibrium.
They gave you K, which acid do you choose?
Now what do you need to do????
HA ↔ H+ + A-
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R
HA
H+
A-
I
5.00 M
0
0
C
-x
+x
+x
E
5-x
+x
+x
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4.0 x 10-4 = x2 ÷ 5.0 – x
Is K small enough to
neglect x? Why yes,
yes it is…
x = √20. x 10-4
x = .0447 = [H+]
What about CN-?
Solve for [A-]
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[H+] = 0.0447
Ka = [H+][A-]
[HA]
[A] = (6.2 x 10-10)(1.00)
0.0447
= 1.4 x 10-8
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Calculate the pH of a solution that contains
1.00 M HCN (Ka = 6.2 x 10-10) and 5.00 M HNO2
(Ka = 4.0 x 10-4). Also calculate the
concentration of CN- in this solution at
equilibrium.
They give you Ka , you find x!
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4.0 x 10-4 = x2 / 5.00 –x
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Can you neglect x? Yep.
x = √.002 = .0447 M H+
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pH = - log .0447 = 1.35
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CN- = solve for [A-]
[A-] = (6.2 x 10-10(1.00M)
.0447
1.4 x 10-8 M CN-
R
HA
H+
A-
I
5.00 M
0
0
C
-x
+x
+x
E
5.00 - x
+x
+x
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Lactic acid (HC3H5O3) is a waste product that accumulates in
muscle tissue during exertion, leading to pain and a feeling
of fatigue. (Cramps) In a 0.100 M aqueous solution, lactic
acid is 3.7% dissociated. Calculate the value of Ka for this
acid.
[H+][A-]
[H+]
Ka = [HA]
∴
[HA] x 100 = 3.7%
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Solve
[H+] / .100 = .037 M
[H+] = .0037
[H+] = [A-] so both are .0037
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[.0037]2/[.100] = 1.4 x 10-4
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Calculate the pH for a 15.0 M solution of NH3
(Kb = 1.8 x 10-5)
Balanced equation:
NH3 + HOH  NH4+ + OHKa expression:
[NH4+][OH-]
Ka =
[NH3]
Make a RICE table.
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R
NH3
NH4+
OH-
I
15.0 M
0
0
C
-x
+x
+x
E
15.0 – x
+x
+x
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Can you neglect x?
K b = 1.8 x 10-5 x 100 = .0018
Initial concentration = 15.0 M
BIG difference-no x
x = √2.7 x 10-4 = 0.0164M OH- log .0164 = 1.8 pH 1.80
∴ 14.00 – 1.80 = pH of 12.20
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Calculate the pH of a 1.0 M solution
of methylamine (Kb = 4.38 x 10-4)
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4.38 x 10-4 = x2/1.0 M= .0209 M OH-
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pOH = - log .0209
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pOH = 1.68
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pH = 14.00 –1.68= 12.32
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Diprotic acids have more than one H+ that can
dissociate. This is done in stages with the first
ionization Ka1 the most significant. The
subsequent Ka2, Ka3 etc. dissociations yield
negligible concentrations of H+.
Because a diprotic like H2SO4 is a strong acid in
its first dissociation and a weak acid in its
second, you need to consider both if the
concentration is less than 1.0 M.
The quadratic equation is needed for this type
of problem.
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Calculate the pH of a 5.0 M H3PO4 solution and
the equilibrium concentrations of the species
H3PO4, H2PO4-, HPO42-, PO43Ka = 7.5 x 10-3 = x2 / 5.00 – x
0 = x2 + 7.5 x 10-3x - .0375
x = .19
∴ pH = -log .19 = 0.72
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[H+][HPO42-]
2. Ka2 = [H2PO4-]
= 6.2 x 10-8
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∴ [HPO42-] = (6.2 x 10-8)[0.19] / [0.19] = 6.2 x 10-8M
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3. Ka3 = [H+][PO43-]
[HPO42-]
4.8 x 10-13 = [PO43-][0.19] / 6.2 x 10-8
= 1.6 x 10 -19M PO43-
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Calculate the pH of a 1.0 M H2SO4 solution.
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- log 1.0 = 0.00 (2 sig fig)
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Calculate the pH of a 1.0 x 10-2 M H2SO4
solution.
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Salts produced from an acid-base reaction are not
always neutral. Some produce solutions with a pH
other than 7.
If both the acid and the base are strong, the salt is
neutral. Weak acid and weak bases make neutral
salts but ONLY if Ka = Kb
Basic salts: salts formed from the cation of a strong
base reacting with the anion of a weak acid are
basic.
Acidic salts: salts formed from the cation of a
weak base and a strong acid are acidic.
If both cation and anion contribute to the pH,
compare Ka to Kb. If Kb is larger, basic and if Ka is
larger, acidic.
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Look at the salt and decide which acid and base
parts it came from.
Strong wins!
If you predict basic, write ↔ OHIf you predict acidic, write ↔ H+
The remaining ion of the salt is the reactant
along with water.
Write water as HOH to make it easier to see
how the hydroxide or hydrogen ion was
formed.
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Predict whether an aqueous solution of each of
the following salts will be acidic, basic, or
neutral:
NaC2H3O2 + HOH  NaOH + HC2H3O2 basic
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NH4NO3 + HOH  NH4OH + HNO3 acidic
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Al2(SO4)3 + 3HOH  2Al(OH)3 + 3H2SO4 acidic
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Calculate the pH of a 0.30 M NaF solution. The Ka
value for HF is 7.2 x 10-4
Where do the species originate?
NaOH and HF
Strong or weak base/acid?
Write the equation.
Strong base/weak acid: F- + HOH ↔ OH- + HF
Make a RICE table:
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R
F-
HOH
OH-
HF
I
.30
------
0
0

C
-x
------
+x
+x

E
.30 - x ------
+x
+x
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Solve for Kb
1 x 10-14
7.2 x 10-4 = 1.39 x 10-11
Kb = [OH-][HF]
[F-]
1.39 x 10-11 = x2/.30 – x
√4.17 x 10-12 = 2.04 x 10-6
pOH = - log 2.04 x 10-6
pH = 14.00 – 5.69 = 8.31
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Calculate the pH of a 0.10 M
NH4Cl solution. Kb for NH3 =
1.8 x 10-5
R
I
Acidic or basic?
C
Calculate Ka
E
-5
5.56 x 10
Solve for H+ and then –log [H+]
for pH.
Ka = x2 / .10 – x (neglect x)
x = √5.56 x 10-11 = 7.46 x 10-6
[H+]
-log 7.46 x 10-6 = 5.13
NH4+
HOH↔ H+
NH3
0.10
----------
0
0
- x
----------
+x
+x
0.10 –x
----------
+x
+x
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Calculate the pH of a 0.010
M AlCl3 solution. The Ka
value for Al(H2O)62+ is 1.4 x
10-5
R
I
C
E
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Calculate the pH of a 0.010
M AlCl3 solution. The Ka
value for Al(H2O)62+ is 1.4 x
10-5
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Kb = 7.14 x 10 -10
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1.4 x 10-5 = x2 / 0.10 – x (neglect x)
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x = √1.4 x 10-2 = 3.74 x 10-4
- log 3.74 x 10-4 = 3.43
R Al
HOH H+
Al(H2O)2+
I 0.01
-------
0
0
C -x
-------
+x
+x
E 0.01 - x -------
+x
+x
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1. Calculate the mass of sodium hydroxide
needed to prepare 100 mL of a 0.100 M
solution.
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2. Calculate the mass of KHP (204.23 g/mol)
needed to react completely with 25 mL of a
0.100 M NaOH solution.
HP-(aq) + OH- (aq)  H2O(l) + P2-(aq)
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3. Calculate the molarity of a solution of
sodium hydroxide if 23.64 mL of this solution
is needed to neutralize 0.5632 g of KHP.
4. It is found that 24.68 mL of 0.1165 M NaOH
is needed to titrate 0.2931 g of an unknown
monoprotic acid to the equivalence point.
Calculate the molar mass of the acid.
Titration
14
12
10
8
6
4
2
0
0
5
10
12.5
15
pH
20
Column2
24
24.9
Column1
25
26
30
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A buffer is a solution of a
weak acid or base and its
salt (which is its
conjugate).
Buffers resist changes in
pH. Hugely important in
biological functions.
A buffer is a common ion
effect. Large buffer
capacity for the
mole:mole ratio of a
conjugate acid and base if
1:1 because it neutralizes
either an acid or a base.
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A buffer consists of both an
acid, a base and its conjugate.
If a small amount of strong
acid is added to the buffer,
there is a base component
ready and waiting to
neutralize the “invader”.
Any titration involving a
weak acid/base is a buffer
problem.
ONLY strong base/strong
acid titrations are not buffer
problems.
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Identify major species:
HC2H3O2 (weak acid)
Na+ (neither acid or base)
C2H3O2- (conjugate base of HC2H3O2)
H2O (very weak acid or base)
The acetic acid will control the pH:
HC2H3O2(aq) ↔ H+(aq) + C2H3O2- (aq)
[Acid]
[H+] = Ka [Base]
Go for it…
R
HC2H3O2
H+
C2H3O2-
I
0.50
0
0.50
C
-x
+x
+x
E
0.50 - x
+x
0.50+x

[H+] = 1.8 x 10-5 ( 0.50/0.50) = 1.8 x 10-5

pH = - log 1.8 x 10-5

pH = 4.74
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Calculate the pH of 0.500 L of a buffer solution
composed of 0.50 M formic acid and 0.70 M sodium
formate before and after adding 10.0 mL of 1.00 M
HCl. Ka of formic acid is 1.8 x 10-4
Before: [H+] = 1.8 x 10-4 [0.50] / [0.70] = 1.29 x 10-4
- log 1.29 x 10-4 = 3.89 pH
After: Go directly to moles then add .10 L (10
mL) to the acid and subtract .10 L from the base.
.50 mol/L x .500L = 0.25 moles
.70mol/L x .500 L = .35 moles
1.8 x 10-4 = (.25 mol + .01 = .26)
(.35 mol - .01 = .34)
[H+] = 1.38 x 10-4
- Log 1.38 x 10-4 = 3.86 pH
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Ka acetic acid = 1.8 x 10-5
[H+] = 1.0 x 10-5 (pH of 5.00)
[Acid] 1 x 10-5
[H+] = Ka [Base] = 1.8 x 10-5 = 1: 1.8 ratio
Mix 1 mole of acetic acid with 1.8 mol sodium acetate
in enough water to dissolve the salt.
** The amount of water in this case is not critical, only
the relative number of moles of A:B since both share
the same container and the total volume for both is
the same.

Calculate the pH of this solution:
1.0 x 10-4 Kw
1.8 x 10-5 Kb
= 5.56 x 10-10 Ka

[H+] = 5.56 x 10-10 ( [.40] /[.25]) = 8.89 x 10-10

- log 8.89 x 10-10 = 9.05 pH



Go to moles!!

WB: 0.25 M NH3 x 1.0 L = .25 mol - .10 mol = .15 mol

WA = .40 M NH4Cl x 1.0 L = .40 mol + .10 mol = .50 mol

∴ [H+] = 5.56 x 10-10 (.50 / .15) = 1.8 x 10-9

-log 1.8 x 10-9 = 8.73 pH

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a.
b.
c.
d.
chloroacetic acid:
propanooic acid:
benzoic acid:
hypochlorus acid:
Ka = 1.35 x 10-3
Ka = 1.3 x 10-5
Ka = 6.4 x 10-5
Ka = 3.5 x 10-8
First, predict which ones it could be.
If you said b or c you are correct.
Calculate the ratio of A/B required for each system to
yield a pH of 4.30. Which system works best? Hint:
the one closest to 1 has the best buffering capacity. It
is the best to neutralize one another with a 1:1 ratio.






pH 4.30 = 5.01 x 10-5 [H+] (inv log - pH)
A. 5.01 x 10-5
1.35 x 10-3 = .037
B. 3.9
C. .78
D. 1431



If you need to calculate moles from molarity it can
be done by M x V as long as your volume is in
Liters, otherwise you have to convert mL to L.
A simpler way will be to just multiply your
molarity by mL and rather than call it moles, it can
be called millimoles or mmol. It’s a lot faster and
you won’t forget to convert as long as you keep
track of your units.
For example, you have a base molarity of .100 and
a volume of 50. mL. Multiply .100 x 50. = and you
get 5 mmol.



Strong acid + strong base = stoichiometry problem
with a limiting reactant and excess reactant. The
excess reactant will be responsible for the pH.
No NaOH is added: It’s all about the acid.
- log .20 = pH 0.699
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The balanced equation for a strong acid/base is:
H+
+ OH-  H2 O
You have 50. mL of .200 M HNO3 so 50. x .2 = 10 mmol of acid
in the solution.
You are adding 10. mL of a .100 M NaOH so 10. x .10 = 1 mmol.
H+
+ OH-  H2 O
1
:
1 ratio reaction:
10 mmol
-1 mmol
9 mmol (excess)
1 mmol
(limiting)
Total volume = 60 mL
- log [H+] = pH
∴ - log [ 9 mmol/60 mL] = .824 pH






H+
+
OH-  H2 O
10 mmol
2 mmol
-2 mmol
8 mmol (excess)
Total volume = 70 mL
∴ - log [ 8 mmol/70 mL] = .942 pH

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
50.0 mL of NaOH has been added:
H+
+ OH-  H2 O
10 mmol
-5 mmol
5 mmol (excess)
5 mmol
Total volume = 100 mL
∴ - log [ 5 mmol/100 mL] = 1.301 pH


100.0 mL of NaOH has been added:
H+
+ OH-  H2 O

10 mmol
10 mmol

Acid and base have been equalized: pH = 7.000



200.0 mL NaOH has been added: H+
OH-  H2 O
10 mmol
20 mmol
10 mmol in excess

Total volume = 250 mL
∴ the pOH = - log [ 10 mmol/250 mL] = 1.398

pH = 14.000 – 1.398 = 12.602

+
CN- + HOH ↔ OH- + HCN


After 8.00 mL of 0.100 M NaOH is added.
[Acid]
[H+] = Ka [Base]



6.2 x 10-10
[5 mmol - .8 mmol]
[ 0 + .8 mmol]
= 3.255 x 10-9
- log 3.255 x 10-9 = 8.49
pH = pKa
∴ pH = -log 6.2 x 10-10 = 9.21
Mol acid = mol base = mol salt formed
5 mmol
5 mmol
5 mmol
Total volume = 100 mL
Ka = 6.2 x 10-10 so Kb = ?
R
CN-
I
5 mmol ------100 mL
0
0
Kb = 1.61 x 10-5
C
-x
-------
+x
+x
E
.05 – x
-------
+x
+x
x2
∴ 1.61 x 10-5 = .05 – x
(neglect x?)
x = √8 x 10-7 = 8.94 x 10-4 [OH-]
-log 8.94 x 10-4 = 3.05 pOH
14.00 – 3.05 = 10.95 pH
HOH OH-
HCN

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
You can use a titration curve
to determine the Ka.
Go to the midpoint of the
titration, called the half
equivalence point. (12.5 mL)
then to the corresponding
pH value (about 5)
Once the midpoint is
reached, [H+] = Ka since ½ of
the acid or base has been
neutralized and the
resulting solution in the
beaker is composed of the
half that remains as well as
the salt. Thus, the A/B term
is equal to one and pH = pKa
Ka = 1 x 10-5

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NH3 is a weak base
HCl is a strong acid
Expect acidic solution.
NH3 Kb = 1.8 x 10-5
So the pH will be
somewhere around 57 ish…
Methyl red.

Bromthymol blue has
a Ka = 1.0 x 10-7 is
yellow in its HIn form
and blue in its Inform. A few drops
are put in a strong
acid then titrated with
NaOH. At what pH
will the indicator
color change first be
visible?

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An indicator’s useful
range is equal to its
pKa +/- 1 to 7 +/- 1.
Ka = [H+][In-] / [Hin]
∴ [H+] = 1 x 10-6
∴ pH = 6.00
Check out the chart…
Bromthymol blue
starts at 6 and ends at
7.