OBJECTIVES:
Learn about Acids and Bases
INTRODUCTION:
Dissociation of water
The dissociation of water to form H+ and HO- is shown below.
O
H
H
H
H2O
+
O
H
H
H2O
H2O
H
+
O
H3O
H
+
H
H
+
+
+
O-
HO-
HO-
The dissociation of water has equilibrium constant (Kw). It looks like this:
Kw = [Products]/[Reactants] = [H3O+ ][HO- ]/[H2O]2
…..…[H2O] is very very very very large in comparison to [H3O+ ] and [HO- ]. Any change in
its concentration is negligibly small. Its value is practically constant. So it can be ignored most
of the time. A more convenient expression for Kw ignores the value of [H2O] :
Kw = [H3O+ ][HO- ]/ 1
[H2O] is essentially constant so we give it a value of 1
Kw = [H3O+ ][HO- ]
see, there’s no [H2O]
or more commonly
Kw = [H+ ][HO- ]
Kw has an itty bitty teensy weensie value of:
Kw = 0.00000000000001 = 1 x 10-14
In other words, the concentration of reactants is much much much much greater than products.
There is way more H2O than H+ or HO-.
pH, Acidic, Basic and Neutral
Neutral
Pure water is neutral. Neutral means that the concentration of H+ and HO- ions are equal.
Neutral means this: [H3O+] = [ -OH]
Since Kw = [H+ ][HO- ] = 0.00000000000001 = 1 x 10-14
and [H+ ] = [HO-1] when neutral.
than [H+ ] = 1 x 10-7
[HO-1] = 1 x 10-7
Kw = [1 x 10-7 H+] x [1 x 10-7 –OH] = 1 x 10-14
try it on your calculator, it’s fun.
Acidic and Basic Solutions
Aqueous acidic and basic solutions (solutions that are not neutral) have different values for [H+]
and [HO-]. Acidic solutions have more [H+] than [HO-]. Basic solutions have more [OH] than
[H+]. BUT [H+ ] x [HO- ] must still equal Kw !!!
Example:
Acidic (pH 3):
Kw =
[1 x 10-3 H+]
Acidic (pH 4):
Kw =
[1 x 10-4 H+] [1 x 10
Acidic (pH 5):
Kw =
[1 x 10-5 H+] [1 x 10
Acidic (pH 6):
Kw = [1 x 10
Kw = [1 x 10-7
****Neutral (pH 7):



[1 x 10-11 –OH]
-10 –
OH]
-9 –OH]
-6
H+]
Kw = [1 x 10-8 H+]
Basic (pH 9):
-9
+
Kw = [1 x 10 H ]
Basic (pH 10):
-10
+
Kw = [1 x 10 H ]
Basic (pH 11):
Kw = [1 x 10
H+]
= 1 x 10-14
= 1 x 10-14
[1 x 10-8 –OH] = 1 x 10-14
[1 x 10-7 –OH] = 1 x 10-14
H+]
Basic (pH 8):
-11
= 1 x 10-14
-6 –
[1 x 10
OH]
[1 x 10-5 –OH]
= 1 x 10-14
= 1 x 10-14
[1 x 10-4 –OH] = 1 x 10
-14
[1 x 10-3 –OH] = 1 x 10
-14
Neutral means: [H+ ] = [HO-1]
Acidic means: [H+ ] > [HO-1]
Basic means: [H+ ] < [HO-1]
[H+ ] = [HO-1]
the [H+ ] = 0.0000001 or 1 x 10-7
the [H+ ] < 0.0000001 or 1 x 10-7
the [H+ ] > 0.0000001 or 1 x 10-7
pH = 7
pH < 7
pH >7
 An acid is a compound that increases the [H+] concentration when it is added
to water.
 A base is a compound that increases the [HO-] concentration when it is
added to water.
We can express how acidic(or basic) a solution is by reporting its H+ concentration. However,
numbers like 0.0000001 and 1 x 10-7 are not pHun to write. So instead we use the pH scale. It’s
just another way to report the concentration of H+. Believe it or not, it’s more convenient to
express very large or very small numbers such as [H+] this way. The little “p” in front of “H”
says to take the negative log of the H+ concentration.
pH = -log[H+] or pH = log[H+]-1 or pH = log(1/[H+])
The “log” (pronounced: logarithm) is base 10. In case you didn’t know, here’s how logarithms
and inverse logs work:
Number
Regular Logs
5
100000 =10 ; log105 = 5
Number
100005 = 10 5;
Inverse Logs such as “pH”
log(1/105) = log10-5 = -5
10000 =104 ;
1000 =103 ;
100 =102 ;
10 =101 ;
1 =100 ;
0.1 =10-1 ;
0.01 =10-2 ;
0.001 =10-3 ;
0.001 =10-4 ;
0.0001 =10-5 ;
log104 = 4
log103 = 3
log102 = 2
log101 = 1
log100 = 0
log10-1 = -1
log10-2 = -2
log10-3 = -3
log10-2 = -4
log10-2 = -5
100004 = 10 4;
10003 = 10 3;
1002 = 10 2;
10 = 10 1;
1 = 10 0;
0.1 = 10-1 ;
0.01 = 10-2 ;
0.001 = 10-3 ;
0.0001 = 10-4 ;
0.00001 = 10-5 ;
log(1/104) = log10-4 = -4
log(1/103) = log10-3 = -3
log(1/102) = log10-2 = -2
log(1/101) = log10-1 = -1
log(1/100) = log100 = 0
log(1/10-1) = log101 = 1
log(1/10-2) = log102 = 2
log(1/10-3) = log103 = 3
log(1/10-4) = log104 = 4
log(1/10-5) = log105 = 5
These ones on the left are more important for our
discussion of acids and bases
One pH unit is an order of magnitude. pH is a logarithmic scale and each unit corresponds to a
factor of ten. Here’s how you should interpret pH:
 A solution that is pH 1 has 10 times more [H+] than a solution that has a pH = 2
 A solution that is pH 1 has 100 times more [H+] than a solution that has a pH = 3
 A solution that is pH 1 has 1000 times more [H+] than a solution that has a pH = 4
 A solution that is pH 1 has 10000 times more [H+] than a solution that has a pH = 5




A solution that is pH 8 has 1/10th the [H+] that a pH = 7 solution has.
A solution that is pH 8 has 1/100th the [H+] that a pH = 6 solution has.
A solution that is pH 8 has 1/1000th the [H+] that a pH = 5 solution has.
A solution that is pH 8 has 1/10000th the [H+] that a pH = 4 solution has.
You’d think there should be another scale called “pOH” for basic solutions and you’d be
right. But only losers use the pOH scale. All the cool people use the pH scale even when they
are talking about basic solutions. This is because Kw must equal [HO-] x [H+]. So, if you know
the [H+] concentration than it is easy to find the [HO-] concentration and vice versa.
If you know pH it is easy to find pOH and vice versa. This is because pKw = 14
and pKw = pH + pOH
example. What is the [OH] concentration for a solution that has a pH of 9.5?
Kw = [H+][HO-] = 1.0 x 10-14
pH = 9.5
[H+] = 10-9.5 = 3.16 x 10-10
Rearrange to solve for [OH]
[HO-] = Kw /[H+]
[HO-] = (1 x 10-14) /(3.16 x 10-10)
[OH]= 3.16 x 10-3 mol OH/L
Kw = 1.0 x 10-14
Kw = [H+][HO-] = 1.0 x 10-14
Solving for [HO-] we get:
[HO-] = 1.0 x 10-14/[H+]
Finding pOH from pH.
pKw = -log Kw = -log (1.0 x 10-14)
pKw = 14 the pKa of water is always 14.
What is the pOH of a solution if it has a pH of 4?
pKa = pH + pOH
pOH = pKa - pH
pOH = 14 - 4
pOH = 10
Basic solutions have low [H+] concentrations and high pH.
Titration
Titration is a method for experimentally determining the concentration of an acid(or base) in
solution. In a typical acid base titration, a solution containing a known concentration of base is
added slowly to a solution who’s acid concentration is unknown. To determine the amount of
acid in the solution, base is added slowly until the acid solution has been neutralized. At this
point, the moles of base added are equal to the moles of acid in the unknown solution. If more
base is added after neutralization the pH of the solution will rise quickly.
HCl
+
NaOH
H2O
+
NaCl
An indicator is typically added to the solution that is to be titrated. An indicator is some
molecule that changes color when a certain pH has been reached. In this experiment we will use
phenolphthalein as our indicator. Phenolphthalein turns pink at pH 8-9. Phenolphthalein will
turn pink and indicate when we have neutralized the acid.
In this experiment we will perform two titrations. The first involves titration of a strong
acid(HCl) with a strong base(NaOH). The second involves titration of a weak acid(acetic acid)
with a strong base(NaOH).
Neutralization Reactions & Titration
Neutralization reactions are the reaction between acids and bases that produce water and a salt.
The neutralization reaction between strong acids such as HCl and strong bases such as NaOH
can be represented as:
HCl(aq)
+ NaOH(aq)
HOH + NaCl(aq)
Acid
base
water
salt
Strong acids and bases completely dissociate in water.
H+ + Cl-
HCl(aq)
-
OH + Na+
NaOH(aq)
~100%
~100%
So we can substitute H+ for our acid and HO- for our base and rewrite the neutralization reaction
for strong acids and bases as:
H+
+
HO-
HOH
When there are equal amounts of acid and base present the solution is neutral. We can calculate
this using the dissociation constant (Kw) for water.
Kw = [H+][HO-]
if [H+] = [HO-]
than
Kw = [H+]2
[H+] = √Kw = √1.0 x 10-14
[H+] = 1.0 x 10-7
pH = -log [H+]
pH = -log (1.0 x 10-7)
pH = 7 neutral
When there are unequal amounts of acid and base there will be an excess concentration of H+ or
HO-.
If there is excess acid the solution will be acidic. Calculating the pH is straight forward. We
assume that the base reacts completely with the acid and is consumed. The acid remaining after
the neutralization reaction is obtained by subtracting the moles of base consumed form the initial
number of moles of acid. The moles of acid remaining are equal to the number of moles of H+
in solution. The concentration of [H+] is obtained by dividing the moles of H+ by the total
volume of the solution.
Example: 1mol of HCl and 0.25mol of NaOH are dissolved in 1.5L of water. What is
the pH?
The NaOH will neutralize some of the HCl. There will be some HCl remaining since it is
in excess. The amount of HCl remaining after neutralization is:
1.0mol HCl – 0.25mol NaOH = 0.75mol HCl
The concentration of HCl is equal to the [H+] concentration. The concentration of HCl
is:
[HCl]= [H+] = (0.75 mol )/(1.5L)
=0.5 M
+
The pH = -log[H ]
pH = -log0.5
pH = 1.48
When there is excess base we first find the [OH] concentration. If we know the [HO-] we can
calculate the [H+] and the pH. We can find the moles of excess base by subtracting the number
of moles of acid. We than use the expression for Kw to solve for [H+]. Once we have [H+] we
can calculate the pH.
Example: 0.25mol of HCl and 1.0mol of NaOH are dissolved in 1.5L of water. What is
the pH?
The HCl will neutralize some of the NaOH. There will be some NaOH remaining since it
is in excess. The amount of NaOH remaining after neutralization is:
1.0mol NaOH – 0.25mol HCl = 0.75mol NaOH
The concentration of NaOH is equal to the [OH] concentration. The concentration of
NaOH is:
[NaOH]= [HO] = (0.75 mol )/(1.5L)
=0.5 M
+
Kw = [H ][HO-]
Rearranges to:
[H+] = Kw /[HO-]
[H+] = (1.0 x 10-14) /(0.5)
[H+] = 2.0 x 10-14
The pH = -log[H+]
pH = -log(2.0 x 10-14)
pH = 13.7
The neutralization of weak acids (or bases) is slightly more complicated. This is because they do
not completely dissociate in water. Since weak acids do not completely dissociate to form H+ we
cannot replace the acid concentration with H+ in our calculations. We’ll see some examples of
these in the experiment.
H3C
O
C
H
K = small
O
H
H
O
H
H3C
O
C
O-
H
O
+
H
Acetic acid is a weak acid. It’s dissociation to form H+ ions and acetate ions is shown above.
The equilibrium constant for the reactions is very small (Ka = 1.76 x 10-5). In other words, the
left side of the equation is favored and very little H+ forms. To calculate the pH we need to use
the equilibrium expression for the weak acid (we can’t use Kw= [H+][HO-]).
The equilibrium constant for the dissociation of acetic acid can be expressed as:
Ka = [CH3CO2-][H+]
[CH3CO2H][H2O]
We can ignore the [H2O] concentration since it is essentially constant.
Ka = [CH3CO2-][H+] = [CH3CO2-][H+]
[CH3CO2H] x 1
[CH3CO2H]
We can rearrange this to solve for [H+]
[H+] = Ka [CH3CO2H]
[CH3CO2-]
If we take the inverse log of this equation we will have an expression for the pH of the weak
acid.
-log[H+] = -log Ka
pH = pKa
+
-log [CH3CO2H]
[CH3CO2-]
-logKa = pKa
log [CH3CO2-]
[CH3CO2H]
This equation is called the “Henderson-Hasselbach” formula. It can be used to calculate the pH
od a buffer solution or the pH of a weakly acidic solution that is being titrated with a strong base.
We will use it to calculate the pH of an acetic acid solution to which a NaOH solution has been
added.
pKa, pronounced “PeeKayAye” is the inverse log of an equilibrium constant. PeeKayAyes
follow the same rules as pH. pKa is often used as an indication of acid strength. Strong acids
have low pKa’s. Weak Acids have larger pKa’s. Here’s how you should interpret pKa’s:
 An acid with a pKa = 0 is 10 times more acidic than an acid with a pKa = 1
 An acid with a pKa = 1 is 10 times more acidic than an acid with a pKa = 2
 An acid with a pKa = 2 is 10 times more acidic than an acid with a pKa = 3
 An acid with a pKa = 3 is 10 times more acidic than an acid with a pKa = 4
 Etc.
The pKa of acetic acid is:
pKa = -log Ka
= -log (1.76 x 10-5)
= 4.754
We can use the Henderson-Hasselbach to calculate the pH of an acetic acid solution at any point
in a titration with a strong base. Here’s how you do it:
The dissociation of acetic acid in water to form [H+] is very small. So, when you titrate(add base)
there is very little H+ for the base to neutralize. Instead, the base(HO-) will react with the acetic
acid.
dissociation of acetic acid in water
H3C
O
C
H
K = small
O
H
H
O
H3C
H
O
C
H
O-
O
+
H
reaction of acetic acid with NaOH(a strong base)
H3C
O
C
K = BIG
O
+
Na -O H
H
Acetic acid
Sodium Hydroxide
H3C
O
C
O- Na
+
H
O
H
Acetate anion
For every molecule of NaOH added, 1 molecule of acetic acid will be neutralized. ALSO for
every molecule of acetic acid neutralized, one molecule of acetate anion will be produced.
This means that when one mole of Base is added to your solution essentially 1mole of acetic acid
is lost and add 1 mole of acetate ion is created.
Example. 100ml of 1.0M NaOH has been added to a 500ml 1.0M Acetic acid solution. What is
the pH?
We will use the Henderson Hasselbach:
pH = pKa
+
log [CH3CO2-]
[CH3CO2H]
The pKa = 4.754. To calculate the pH we need the [CH3CO2-] and [CH3CO2H] concentrations.
We can find these by calculating the moles of each after the NaOH has been added and than
divide by the total volume.
Concentration of [CH3CO2H]:
Moles CH3CO2Hinitial = M x V
= 1.0M x 0.5L
= 0.5mol
Moles NaOHadded = M xV
=1.0M x 0.1L
= 0.1 mol NaOHadded
notice that volume is in Liters
Moles CH3CO2Hafetrl = mole CH3CO2Hinitial - mol NaOHadded
= 0.5mol – 0.1mol
= 0.4 mol CH3CO2H
[CH3CO2H] = mol CH3CO2H/ Total Volume
= 0.4mol CH3CO2H/ 0.6L
= 0.667M CH3CO2H
total volume = 0.5L +0.1L
Concentration of [CH3CO2-]:
We will make the assumption that the initial [CH3CO2-] iniyial concentration is negligibly small
and that all of the CH3CO2- in solution comes from the reaction of NaOH with CH3CO2H. We
can do this because the Ka for acetic acid is very small(Ka = 1.76 x 10-5). This means that in a
1.0M solution of acetic acid there is 0.0000176mol of CH3CO2- in it. That’s a very tiny number
so assuming that [CH3CO2-] iniyial ~ zero is not too bad of an approximation. We could not
make this approximation if Ka was larger.
So assuming that
Moles of CH3CO2-initial = 0
and
Moles of CH3CO2-after = moles of NaOHadded
Moles of CH3CO2-after = MNaOH x VNaOH
= 1.0M x 0.1L
= 0.1 mol CH3CO2[CH3CO2 ] = mol CH3CO2-after/ Total Volume
total volume = 0.5L +0.1L
= 0.1 mol CH3CO2 /0.6L
= 0.167M CH3CO2FINALLY: Now that we know[CH3CO2H] and [CH3CO2-] we can calculate the pH:
pH = pKa
+
pH = 4.754
+
log [CH3CO2-]
[CH3CO2H]
log (0.167)
(0.667)
pH = 4.754
+ 0.602
pH = 5.357
TaDaaaah!
ONE LAST THING ON pKa
As was stated earlier, the pKa is an indication of an acid’s strength. The pKa of an acid can be
determined experimentally by titration. Take a look at the Henderson Hasselbach when
[CH3CO2H] and [CH3CO2-] are equal:
pH = pKa
+
log [CH3CO2-]
[CH3CO2H]
If [CH3CO2H] = [CH3CO2-]
than
[CH3CO2-] = 1.0
[CH3CO2H]
pH = pKa
pH = pKa
+
+
log (1.0)
0
The pH is equal to the pKa when [CH3CO2H] = [CH3CO2-]. This condition is satisfied when
the acid is 50% titrated. We can experimentally measure the pKa by measuring the pH at this
point. It is easy to find this point on a titration curve. At the equivalence point (where the big
jump in pH occurs) the acid is 100% titrated. At half this volume, the acid is 50% titrated and
[CH3CO2H] = [CH3CO2-]. Below is the titration curve for acetic acid. Its equivalence point
is at 20 ml. At half titration (10ml) the pH is 4.75. Pretty cool.
Titration curve for Acetic Acid
14
13
12
11
10
pH
9
8
7
6
5
4
3
0
5
10
15
20
25
30
35
ml 1.0 NaOH
= 20 ml; the Equivalence Point; a.k.a. end point; 100% titrated
= ½ titration. 50% titrated. pH = pKa = 4.75
40
45
Procedure:
PART 1: Playing with Acids and Bases: Learning about pH and Acid Base Strength
PART 1 A: HCl a STRONG ACID (Dissociates completely to form H+ and Cl-)
1) Add 100 ml of .1M HCl to a 100ml graduated cylinder.
Calculate the pH.
Mol HCl = mol H+ produced
[HCl]initial = [H+]
pH = -log [H+]
Measure the pH with the pH meter.
Record your results in the “Observation” section. Sweet.
Decant 10ml of this solution into a 10ml graduated cylinder.
2) 10 fold dilution:
Decant the contents of the 10ml graduated cylinder into a 100ml graduated cylinder.
Fill the graduated cylinder to 100ml with water. This is a 10 fold increase in volume (from 10ml
to 100 ml). Now its 1/10th as concentrated.
Calculate the Molarity of HCl using: Mnew = (MinitVinit )/Vnew
Calculate its pH.
Measure the pH with the pH meter.
Record your results in the “Observation” section. Sweet.
Decant 10ml of this solution into a 10ml graduated cylinder.
3) 10 fold dilution:
Decant the contents of the 10ml graduated cylinder into a 100ml graduated cylinder.
Fill the graduated cylinder to 100ml with water. Now its 1/100th as concentrated as the 0.1M
solution.
Calculate the Molarity of HCl using: Mnew = (MinitVinit )/Vnew
Calculate its pH.
Decant the solution into a 100 ml beaker.
Measure the pH with the pH meter.
Record your results in the “Observation” section. Sweet.
Decant 10ml of this solution into a 10ml graduated cylinder.
4) 10 fold dilution:
Decant the contents of the 10ml graduated cylinder into a 100ml graduated cylinder.
Fill the graduated cylinder to 100ml with water. Now its 1/1000th as concentrated as the 0.1M
solution.
Calculate the Molarity of HCl using: Mnew = (MinitVinit )/Vnew
Calculate its pH.
Decant the solution into a 100 ml beaker.
Measure the pH with the pH meter.
Record your results in the “Observation” section. Sweet.
PART 1 B: NaOH: STRONG BASE (Dissociates completely to form HO- and Na+)
DO IT AGAIN WITH: 0.1 M NaOH
1) Add 10 ml of 0.1M NaOH to a 100ml graduated cylinder.
Calculate the pH. To get pH we need [H+] BUT NaOH is forming HO-. We can use Kw to find
the [H+].
Kw = [H+ ][HO- ] = 1 x 10-14
Solving for [H+] we get
[H+] = Kw/ [OH]
[H+] = (1.0 x 10-14)/ [OH]
pH = -log [H+]
Decant the solution into a 100 ml beaker.
Measure the pH with the pH meter.
Record your results in the “Observation” section. Sweet!
Pippette 10ml (using a 10 ml pipette) of this solution back into the graduated cylinder.
2) 10 fold dilution:
Fill the graduated cylinder to 100ml with water. This is a 10 fold increase in volume (form 10ml
to 100 ml). Now its 1/10th as concentrated.
Calculate the Molarity of NaOH using: Mnew = (MinitVinit )/Vnew
Calculate its pH.
Decant the solution into a 100 ml beaker.
Measure the pH with the pH meter.
Record your results in the “Observation” section. Sweet!
Decant 10ml of this solution into a 10ml graduated cylinder.
3) 10 fold dilution:
Fill the graduated cylinder to 100ml with water. This is a 10 fold increase in volume (form 10ml
to 100 ml). Now its 1/100th as concentrated as the initial solution.
Calculate the Molarity of NaOH using: Mnew = (MinitVinit )/Vnew
Calculate its pH.
Decant the solution into a 100 ml beaker.
Measure the pH with the pH meter.
Record your results in the “Observation” section. Sweet!
Decant 10ml of this solution into a 10ml graduated cylinder.
4) 10 fold dilution:
Fill the graduated cylinder to 100ml with water. This is a 10 fold increase in volume (form 10ml
to 100 ml). Now its 1/1000th as concentrated as the initial solution.
Calculate the Molarity of NaOH using: Mnew = (MinitVinit )/Vnew
Calculate its pH.
Decant the solution into a 100 ml beaker.
Measure the pH with the pH meter.
Record your results in the “Observation” section. Sweet!
PART 1 C: WEAK ACID
DO IT AGAIN WITH: 10 M Acetic acid.
1) Add 10 ml of 10M Acetic acid to a 100ml graduated cylinder.
Calculate the pH just as you did with HCl.
Mol Acetic Acid = mol H+ produced
[CH3CO2H ]initial = [H+]
pH = -log [H+]
Decant the solution into a 100 ml beaker.
Measure the pH with the pH meter.
Record your results in the “Observation” section. CRAAAAP!
Pippette 10ml (using a 10 ml pipette) of this solution back into the graduated cylinder.
2) 10 Fold Dilution.
Fill the graduated cylinder to 100ml with water. This is a 10 fold increase in volume (form 10ml
to 100 ml). Now its 1/10th as concentrated.
Calculate the Molarity of NaOH using: Mnew = (MinitVinit )/Vnew
Calculate its pH.
Decant the solution into a 100 ml beaker.
Measure the pH with the pH meter.
Record your results in the “Observation” section. CRAAAAP!
Pippette 10ml (using a 10 ml pipette) of this solution back into the graduated cylinder.
3) 10 Fold Dilution.
Fill the graduated cylinder to 100ml with water. This is a 10 fold increase in volume (form 10ml
to 100 ml). Now its 1/100th as concentrated.
Calculate the Molarity of NaOH using: Mnew = (MinitVinit )/Vnew
Calculate its pH.
Decant the solution into a 100 ml beaker.
Measure the pH with the pH meter.
Record your results in the “Observation” section. CRAAAAP!
Pippette 10ml (using a 10 ml pipette) of this solution back into the graduated cylinder.
4) Flip out because it’s not working.
Acetic acid is a weak acid. It has a Ka = 1.76 x 10-5 . The strong acid(HCl) and strong
base(NaOH) we did previously had HUGE equilibrium constants. They essentially dissociated
completely in water so that for every mole of HCl (or NaOH) you add you get a mole of H+ (or
HO-).
[HCl] = [ H+] for a strong acid
[NaOHl] = [ OH] for a strong base and [H] = Kw / [OH]
This isn’t true for acetic acid. It only partially dissociates so the moles of Acetic acid in
solution is not equal to moles of H+.
[CH3CO2H] ≠ [ H+]
So what do we do now? USE THE “ICE” Method. ( InitialChangeEquilibrium). We need to use the
equation for the equilibrium constant and our initial acid concentration to calculate the [H+] and
pH.
Ka = 1.76 x 10-5
CH3CO2H
Initial
Change
Equilibrium
[CH3CO2H]
10
-x
10 - x
H
+
[ H+]
0
+x
+x
+
CH3CO2[CH3CO2-]
0
+x
+x
-x is the amount of CH3CO2H that dissociates to form H+ and CH3CO2- at equilibrium.
+x is the amount of H+ that forms from CH3CO2H. [ H+] = +x at equilibrium.
+x is the amount of CH3CO2- that forms from CH3CO2H. [ H+] = +x at equilibrium.
Notice that [CH3CO2-] = [ H+]
+
Ka = [CH3CO2 ][ H ] = 1.76 x 10-5
[CH3CO2H]
Since [CH3CO2-] = [ H+]
Ka = x ∙ x
= 1.76 x 10-5
(10 - x)
Ka =
x2
= 1.76 x 10-5
(10 - x)
If x is much much smaller than the initial [CH3CO2H] we can ignore it and get a close
approximation to the actual answer. Ignoring “x” in the bottom, we get:
Ka =
x2
10
= 1.76 x 10-5
x = (1.76 x 10-5)∙(10)
x = 1.327 x 10-2 and x = [ H+]
Now you can calculate the pH
[ H+] = 1.327 x 10-2
pH = -log [ H+]
pH = -log (1.327 x 10-2)
pH = 1.877 TADAAAAH!
5) Now do this for all your other Acetic Acid samples
PART 2: Titration and Neutralization Reactions.
PART 2a: Titration of a STRONG acid with a STRONG base.
1) Select a 100ml beaker. Right click on the beaker and select chemicals. Left click on 0.1M
HCl and add 20.0 ml to the beaker. Right click on the beaker and select “indicators…”. Add
3 drops of phenolphthalein.
2) Obtain a 50ml burette from the equipment menu. Right click on the burette and select
chemicals. From the chemical menu, select “0.1M NaOH”. Add 50ml of this solution to the
burette.
3) Right click on the beaker and select “pH meter”, “collect titration data” and “view titration
data”. This will produce a graph that plots pH vs. ml NaOH added.
4) Perform the titration. Click on “Start volume” in the titration dialog box. We are going to
add the 0.1M NaOH solution very slowly to the beaker containing 0.1M HCl. Addition of
NaOH will neutralize the HCl. The pH will rise sharply when all of the HCl is consumed
and than begin to plateau after further addition of NaOH. The inflexion point on the curve
marks the equivalence point. The equivalence point is where the [HCl] = [NaOH].
[HCl] < [NaOH]
example: In the titration curve illustrated,
20ml of 0.1M NaOH were required to reach
the endpoint. 20.0ml 0f 0.1M NaOH contain
0.002mol NaOH. Since it requires 1mol of
NaOH to neutralize 1mol of HCl than there
must have been 0.002mol of HCl in the
beaker.
End Point
a.k.a. Equivalence
point.
[HCl] > [NaOH]
[Acid] = [Base]
Neutralization of acid is
complete
5) Adjust the rate of addition of 0.1M NaOH to ~0.1ml/sec. Look at it go.
6) Copy and paste the graph into your observation section.
7) Perform long tedious boring calculations. We are going to calculate the pH at various points
during the titration and compare them to your experimental data. Fill in the chart in
observations.
a. Before the equivalence point the pH. Perform these calculations for the following
volumes of titrant: 5ml, 10ml 15ml 17.5ml 19.5 and record your results in your
observations. The pH Calculaton is exactly like that in Part 1a. BUT we first have
to calculate the concentration of HCl. During the titration, NaOH neutralizes the
HCl. So we have to subtract the moles of NaOH added from the moles of HCl present
in the beaker. The volume also changes during the titration so we have to account for
the dilution as well. Here’s an example:
After 5 ml of 0.1M NaOH has been added.
Moles HCl initially in beaker = M x V
= 0.1 x 0.02L
= 0.002 moles HCl before titration
Moles NaOH in 5ml of titrant = M x V
= 0.1 x 0.005
= 0.0005 mol NaOH
Moles HCl after 5ml NaOH added
= moles HClinitial – moles NaOHin 5ml
= 0.002 – 0.0005
= 0.0015 mol HCl
Concentration of HCl =(mol HCl)/Total Volume
Total Volume = 20ml +5ml = 25ml = 0.025L
= 0.0015molHCl/ 0.025L
= 0.06M HCl
pH after 5ml of NaOH
[HCl] =[H+] since HCl is a strong acid
pH = -log(0.06)
pH = 1.22
At the equivalence point. There is no excess HCl or NaOH. The products formed
during the neutralization are Na+ and Cl-. Neither of these are particularly acidic or basic
so the pH should be ~7.
b. After the equivalence point all. Perform these calculations for the following
volumes of titrant: 20.5ml, 25ml 30ml 35ml and record your results in your
observations. After the equivalence point all of the HCl has been neutralized and the
pH shot way up. Now there is an excess of NaOH. This does not mean that there is
no H+ in solution. It just means that the [H+] will be very small and the pH will be
high. Calculating pH after the equivalence point is exactly the same as that done in
PART 1b. BUT we first have to calculate the moles of NaOH present and than take
dilution into account. Here’s an example:
After 20.5 ml of 0.1M NaOH has been added.
Moles NaOH added = Vtitrated x M
= 0.0205L x0.1M
= 0.00205mol NaOH
Moles NaOH excess = mol NaOH - mol HClinit
= 0.00205mol NaOH - 0.002mol HCl
= 0.00005mol NaOH
Concentration NaOH = (mol NaOHexcess)/ Total Volume
Total Volume = 20ml +20.5ml = 40.5ml = 0.0405L
= 0.00005mol/0.0405L
= 0.01235 M NaOH
pH after 20.5 ml NaOH
Kw = [H+ ][HO- ] = 1 x 10-14
Solving for [H+] we get
[H+] = Kw/ [OH]
[H+] = (1.0 x 10-14)/ (0.01235)
[H+] = 8.1 x 10-12 M
pH = -log [H+]
pH = -log (8.1 x 10-12)
pH = 11.09
8) Compare your calculations with the experimentally determined pH values. You can find
these values by selecting “data” under the view menu on your titration curve. Enter these
values in your chart as well.
PART 2b: Titration of a WEAK acid with a STRONG base.
1) Select a 100ml beaker. Add 20.0 ml of 0.1M Acetic Acid and add 3 drops of
phenolphthalein.
2) Obtain a 50ml burette from the equipment menu and fill it with 50 ml of 0.1M NaOH.
3) Right click on the beaker and select “pH meter”, “collect titration data” and “view titration
data”.
4) Perform the titration. Adjust the rate of addition of 0.1M NaOH to ~0.1ml/sec. Look at it
go.
5) Copy and paste the graph into your observation section.
6) Perform long tedious boring calculations. We are going to calculate the pH at various points
during the titration and compare them to your experimental data.
7)
We are going to calculate the pH at various points during the titration and compare them to
your experimental data. Perform these calculations for the following volumes of titrant: 5ml,
10ml 15ml 17.5ml 19.5 and record your results in your Fill in the chart in observations.
a) Before the equivalence point: the pH will be weakly acidic. Acetic acid is a weak
acid and dissociates only slightly (K= 1.76 x 10-5) This means there is very little H+
to react with –OH during the neutralization. In fact, NaOH will react predominately
with the acetic acid and form acetate (CH3CO2- ). NaOH is a strong base and its
reaction with acetic acid will essentially go to 100% completion. For every molecule
of CH3CO2H neutralized by NaOH one molecule of CH3CO2- will form.
CH3CO2H + -OH
CH3CO2- + HOH
The dissociation of acetic acid is shown below. The addition of NaOH will consume
CH3CO2H and produce CH3CO2-. This will change the concentration of CH3CO2H and
CH3CO2 but will not change the equilibrium constant.
CH3CO2H
CH3CO2- + H+
The expression for the equilibrium constant for the dissociation of acetic acid is shown
below. According to LeChartier’s principle, the values of [CH3CO2H] and [CH3CO2] may vary
but the equilibrium constant will remain unchanged.
+
Ka = [CH3CO2 ][ H ] = 1.76 x 10-5
[CH3CO2H]
We can rearrange this equation and solve for [H+] concentration to get:
[H+] = Ka [CH3CO2H]
[CH3CO2-]
To get the pH we need to take the negative log.
-log [H+] = -logKa - log [CH3CO2H]
[CH3CO2-]
or
p[H+] = pKa + log [CH3CO2]
[CH3CO2H]
The pKa of an acid = -logKa
It is often used as a measure of
acid strength. The pKa of
acetic acid is 4.74
This is the famous “Henderson Hasselbach” equation. We can use it to calculate the pH
if we know [CH3CO2H] and [CH3CO2-] .
Here’s an example after 5.0 ml of 0.1M NaOH has been added.
The moles of CH3CO2H during the titration will be
Mol CH3CO2H = Mol CH3CO2Hinitial – mol NaOHadded
= 0.002molCH3CO2H - 0.0005 mol NaOHadded
= 0.0015 molCH3CO2H
And the moles of CH3CO2- formed will be
Mol CH3CO2- = Mol CH3CO2-initial + mol NaOHadded
Since acetic acid is a weak acid the moles CH3CO2-initial is very small and can be ignored.
Mol CH3CO2- = mol NaOHadded
= 0.0005
Now we can us the Henderson Hasselbach.
pH = 4.74 + log(0.0005/0.0015)
pH = 4.26
At the equivalence point: all of the CH3CO2H has been neutralized to CH3CO2-.
CH3CO2- is the conjugate base of acetic acid. The pH at the equivalence point will not be
7 because of the presence of CH3CO2-.
8)
b) After the equivalence point: all of the Acetic Acid will be consumed. The pH
calculation is identical to that performed in 7b above. Calculate the pH when 20.5,
25, 30 35ml of 0.1M NaOH have been added.
Compare your calculated pH values to your experimental data.