Acid/Base Strength and Ionization Constants
-
Do you recall using the Relative Strengths of Acids and Bases table to determine which of
two acid or base species is the stronger one? Did you wonder about the basis for the
rankings?
Strength versus Concentration
The strength of an acid or a base is a function of the extent to which it produces
hydronium or hydroxide ions in solution. Strong species produce stoichiometric
equivalents (100% ionization) of hydronium or hydroxide while weak species produce less
than a stoichiometric equivalent (< 100%) of hydronium or hydroxide.
-
The concentration of a species, or more precisely the molar concentration, is the moles of
solute per litre of solution. The higher the number of moles of solute dissolved, the higher
the molar concentration.
Are strength and concentration interchangeable terms? The answer is no.
Examples:
It is possible for a solution to be strong yet dilute. For example, 1.0 x 10-4 mol/L HCl is a
relatively dilute solution, but since 100% of the HCl molecules ionize to form hydronium
and chloride ions, it is a strong acid.
-
It is also possible for a solution to be weak yet concentrated. For example, a 17.4 mol/L
ethanoic acid solution is very concentrated, but since few of the ethanoic acid molecules
ionize to form hydronium and ethanoate ions, it is a weak acid.
Identifying Strong/Weak Species
-
There are number of tests that can be used to distinguish a stronger species from a weaker
one, including:
 rate of reaction with active metals (acids only)
 indicator test
 pH test
A)
Reaction Rate Test
In general, stronger acids react more vigorously with active metals (i.e. produce more
hydrogen gas per unit of time) than weaker ones.
It is not the most effective method because there are too many other variables like bond
type and bond strength that come into play. Measuring pH is a better test.
B)
Universal Indicator/pH paper
A universal indicator can be used to measure pH. When a drop of a universal indicator
solution or a strip of paper impregnated with indicator is added to a solution, the indicator
exhibits a certain colour depending on the pH of the solution. You can map the colour to a
specific pH value using a colour code/pH chart that comes with the indicator.
Example:
Two 0.100 mol/L solutions are tested using pH paper. Solution A is found to have a pH of
1.5 while Solution B has a pH of 3.0. (pH = - log [H3O+])
pH 1.5 converts to a [H3O+] of 0.03 mol/L while pH 3.0 converts to [H3O+] of 0.001
mol/L.
Conclusion: Solution A is a stronger acid than Solution B.
* The solution that ionizes to produce more hydronium is the stronger acid.
Note: it is important to compare solutions of equal molar concentration when comparing
their strengths using pH tests.
C)
pH Meter
The colors of universal indicator/pH paper correspond to pH values in 0.5 unit increments
at best, so use of these indicators can result in significant random error.
pH data collected using electronic pH meters are more accurate and more precise.
* Let's reconsider the above example:
Two 0.100 mol/L solutions are tested using a pH meter. Solution A has a pH of 1.394
while Solution B has a pH of 3.238.
pH 1.394 converts to a [H3O+] of 0.0404 mol/L while pH 3.238 converts to [H3O+] of
0.000578 mol/L.
Conclusion: Solution A is the stronger acid.
-
The conclusion is the same as before, but if the differences in pH were smaller, you would
not be able to distinguish between the two species using pH paper.
pH tests are effective in determining the relative strengths of bases too! What about cases
when the solutions being compared have different molar concentrations?
Example:
A 1.06 x 10-4 M NaOH solution is found to have a pH of 10.025 while a 2.00 M NH3
solution has a pH of 11.766.
Recalling that stronger bases have higher pH, does this mean that ammonia is a stronger
base than sodium hydroxide? The answer is no.
pH is a function of both the molar concentration and the strength of a solution. pH only
identifies the stronger species when solutions of similar molar concentration are tested.
-
The logical question becomes how do we compare strengths when molar concentrations
are not the same? Well, you could compare percent ionization values, but since percent
ionization values only apply to solutions of specific molar concentration (i.e. they are
concentration dependent), the answer is by comparing K values!
Ka , Kb and Acid/Base Strength
-
Most acids, like ethanoic acid, are weak. They do not completely ionize in water to produce
a stoichiometric equivalent of hydronium ions. This means aqueous ethanoic acid
molecules are in equilibrium with hydronium and ethanoate ions.
CH3COOH (aq) + H2O (l)  H3O+ (aq) + CH3COO - (aq)
For any system at equilibrium, the law of chemical equilibrium results in a K expression.
Recalling that you only include the equilibrium concentrations of gaseous and aqueous
species in equilibrium constant expressions, the K expression for ethanoic acid in water is:
Ka = [H3O+][CH3COO -]
[CH3COOH]
where the subscript "a" indicates that an acid ionizes in water to form hydronium. Ka is the
acid ionization constant or the equilibrium constant for the ionization of a weak acid in
water to produce hydronium and the conjugate base of the acid.
-
An equilibrium constant expression may also be written for a weak base like ammonia in
water.
NH3 (aq) + H2O (l)  NH4+(aq) + OH -(aq)
Kb = [NH4+][OH -]
[NH3]
-
The subscript "b" indicates that a base hydrolyses water to form hydroxide. Kb is the base
ionization constant or the equilibrium constant for the ionization of a weak base in water
to produce hydroxide and the conjugate acid of the base.
Ka and Kb values indicate acid strength and base strength respectively.
For example, a higher Ka value indicates higher hydronium ion concentrations or greater
ionization. The equilibrium position lies further to the right when Ka is higher.
A higher Kb value is interpreted in much the same way except that it means a higher
hydroxide ion concentration.
The first step in calculating acid or base ionization constants is writing the Ka or Kb
expression. You will rely on your Brønsted-Lowry acid-base equation skills to do this.
Example:
Write the ionization constant expression for an aqueous sodium cyanide solution.
Answer
Plan a strategy.
- Use the five step method to write the Brønsted-Lowry acid-base equation.
- Write the ionization constant expression.
Step 1: Write the Brønsted-Lowry acid-base equation.
Sodium cyanide, NaCN, is an ionic compound. It dissociates in water to form Na+ and
CN- ions, so the species present in the mixture are Na+, CN-, and H2O.
Identify acids/bases and compare their strength:
CN- is a stronger base than water.
Water is the only species present that can act as an acid. (Strongest acid)
Neither H3O+ (strong acid) nor OH- (strong base) is involved so the reaction is
non-stoichiometric.
Therefore:
CN - (aq) + H2O (l)  HCN (aq) + OH - (aq)
Step 2: Write the ionization constant expression.
Kb = [HCN][OH -]
[CN -]
Closing Points
-
Ka and Kb are just special forms of the equilibrium constant expression.
-
Ka values for common acids are provided in the right-hand column of the Relative
Strengths of Acids and Bases table.
-
Kb values for conjugate bases in the strengths of acids and bases table can be calculated
using Kw and Ka of the acid:
Kb = Kw
Ka
Ka and Kb Calculations
-
You can compare the strengths of two acids by looking up their percent ionization values;
however there is a catch. The footnote in the table points out that the percent ionization data
refers only to 0.100 M solutions for the species in the table. That's because percent
ionization is concentration dependent.
-
If you are attempting to compare the relative strengths of two acids that have different
molar concentrations, you need to use different reference data than percent ionization.
That's where the equilibrium constant, K, becomes useful.
-
Recall that K values tell you something about the position of an equilibrium: the smaller the
value of K, the further to the left it lies. For acids, smaller K values indicate weaker acids.
-
You can rule out reactions between strong acids/bases and water because these are
quantitative reactions or those that go to completion. (Very large Ka or Kb)
-
Ka and Kb calculations follow the same general patterns established in Unit 1.
Calculating a K value Given Equilibrium Concentration Data
Example:
A cleaning solution is prepared by dissolving ammonia in water. The equilibrium concentrations
of ammonia and hydroxide found to be 0.35 mol/L and 0.0025 mol/L respectively. Calculate Kb.
Answer
Plan a strategy
- Write the Brønsted-Lowry equation and list the given data.
- Write the Kb expression.
- Since only two of the three equilibrium concentrations in the expression are given, you need
to find the third (ammonium) using stoichiometry.
- Substitute the equilibrium concentration data into the Kb expression and solve for Kb.
- Communicate the answer.
Step 1: Write the equilibrium equation and list the given data.
NH3 (aq) + H2O (aq)  NH4 + (aq) + OH- (aq)
[0.35]eq
[x]eq
[0.0025]eq
Step 2: Write the expression for Kb.
Kb = [NH4+][OH -]
[NH3]
Step 3: Find the missing equilibrium concentration (Use mole ratios to calculate concentrations.
Must compare a product concentration to another product because there is not 100%
ionization from reactant to product)
[NH4+] = [OH-] x 1 mol NH4+
1 mol OH= 0.0025 M OH- x 1 mol NH4+ = 0.0025 M NH4+
1 mol OHStep 4: Substitute the values and solve for Kb.
Kb = [NH4+][OH -] = (0.0025)(0.0025) = 1.78 x 10 -5 = 1.8 x 10 -5
[NH3]
(0.35)
The equilibrium constant for ammonia is 1.8 x 10-5.
Using Ka or Kb to Calculate Equilibrium Concentrations
Example:
A propanoic acid solution has Ka of 1.3 x 10-5 and an equilibrium concentration of 0.050 mol/L.
Calculate the concentration of hydronium ions at equilibrium.
Answer
Plan a strategy
Write the Brønsted-Lowry equation and identify the wanted and given information.
Write the Ka expression, rearrange it to solve for x, and substitute the given information.
Step 1: Write the equation and list the given and wanted information.
Let x = [H3O+] = [C2H5COO-] because the mole ratio for these species is 1:1. In other
words, for each mole of H3O+ formed there is one mole of C2H5COO- formed.
C2H5COOH (aq) + H2O (l)  H3O + (aq) + C2H5COO - (aq)
[0.050]eq
[x]eq
[x]eq
Step 2: Write and rearrange Ka expression and solve.
Ka = [H3O +][C2H5COO -]
[C2H5COOH]
Ka =
[x][x]
[C2H5COOH]
x2 = (Ka)([C2H5COOH])
=
x2
[C2H5COOH]
Ka = 1.3 x 10 -5
x = √(Ka)([C2H5COOH]) = √(1.3 x 10 -5)(0.050) = 8.0622 x 10 -4 = 8.1 x 10 -4
The equilibrium concentration of hydronium is 8.1 x 10-4 mol/L.
This problem could be extended to find the pH of the solution. This can be done by applying the
pH formula:
pH = - log [H3O +] = - log (8.1 x 10 -4) = 3.09
Using an ICE Table
In some cases you know the original concentration of a species, but not its equilibrium
concentration. In these cases you have to use an ICE table.
Example:
A nitrous acid solution has an initial concentration of 0.25 mol/L. At equilibrium the hydronium
concentration is 0.013 mol/L. Calculate Ka.
Answer
Plan a strategy.
- Write the Brønsted-Lowry equation and set up an ICE table. Use x to represent the change of
the species with the smallest mole coefficient in the ionization equation.
- Find change in [H3O+].
- Calculate the change in [NO2-] using the change in [H3O+].
- Calculate the change in [HNO2] using the change in [H3O+].
- Write the Ka expression and substitute the equilibrium values to solve for Ka.
Step 1: Set up ICE table
I
C
E
HNO2 (aq) +
0.25
-x
0.25 -x
H2O (l)
 H3O+ (aq)
~0
+x
0.013
+ NO2 - (aq)
0
+x
x
Step 2: Calculate change in [H3O+].
Since the initial concentration of hydronium is negligible (0.0000001) in relation to the
equilibrium concentration, the change in [H3O+] is 0.013 M.
Step 3: Calculate the change in [NO2-].
Nitrous acid ionizes to form hydronium and nitrite in a 1:1 ratio, therefore the change in
[NO2-] = the change in [H3O+] = 0.013 M.
Step 4: Calculate the change in [HNO2]
Likewise since the mole ratio of nitrous acid to hydronium is 1:1, the change in [HNO2] =
the change in [H3O+] = 0.013 M. So, [HNO2]eq = 0.25 - 0.013 = 0.237
Step 5:
Ka = [H3O+][NO2 -] = (0.013)(0.013) = 7.1308 x 10 -4 = 7.1 x 10 -4
[HNO2]
(0.237)
The equilibrium constant for nitrous acid is 7.1 x 10-4.
Using pH measurements to find Ka or Kb
pH measurements can be used to provide the data needed to calculate Ka or Kb. Then the
compounds can be ranked for strength using the K value.
Example:
An aqueous methanoic acid solution has an initial molar concentration of 0.75 mol/L. The pH of
the solution at equilibrium is 1.95. Calculate Ka of methanoic acid.
Answer:
Step1: Write the ionization equation and list the given information.
I
C
E
HCOOH(aq) + H2O(l)  H3O+(aq) + HCOO -(aq)
0.75 M
~0
0
-x
+x
+x
0.75 - x
x
x
pH=1.95
Step 2: Calculate [H3O+]
[H3O+] = 10 -pH
= 10 -1.95 = 0.0112201 M
Step 3: Update the ICE table.
Since the initial concentration of hydronium was 0 and we now know that the equilibrium
concentration is 0.0112201 M, we know the value of x = 0.0112201 M.
I
C
E
HCOOH(aq) + H2O(l)  H3O+(aq) + HCOO -(aq)
0.75 M
~0
0
-0.0112202
+0.0112202 +0.0112202
0.7387798
0.0112201
0.0112202
Step 4: Write the Ka expression, substitute in concentration values and solve.
Ka = [H3O+][HCOO -] = (0.0112202)(0.0112202) = 1.704 x 10 -4 = 1.7 x 10 -4
[HCOOH]
0.7387798
So, the acid ionization constant for methanoic acid is 1.7 x 10 -4.
Example:
An aqueous 0.50 mol/L sodium sulfite solution is prepared. The pH at equilibrium is 10.45.
Calculate the Kb for the sulfite ion.
Answer:
Step 1: Write the ionization equation and list the given information.
SO32-(aq) + H2O(l)  HSO3-(aq) + OH-(aq)
I
0.50
0
~0
C
-x
+x
+x
E
0.50-x
x
x
pH=10.45
Step 2: Calculate [OH-].
pOH = 14.00 - pH
= 14.00 - 10.45 = 3.55
[OH-] = 10 -pOH = 10 -3.55 = 0.0002818 M
Step 3: Update the ICE table.
Since the initial concentration of hydroxide was 0 and we now know that the equilibrium
concentration is 0.0002818 M, we know the value of x = 0.0002818 M.
I
C
E
SO32-(aq) + H2O(l) HSO3-(aq) + OH-(aq)
0.50
0
~0
-0.0002818
+0.0002818 +0.0002818
0.4997181
0.0002818 0.0002818
Step 4: Write the Kb expression, substitute in the equilibrium concentration values, and solve.
Kb = [HSO3-][OH-] = (0.0002818)(0.0002818) = 1.5895 x 10 -7 = 1.6 x 10 -7
[SO32-]
0.4997181
So, the base ionization constant for the sulfite ion is 1.6 x 10 -7.
The Quadratic Equation
In the problems so far, the rules for significant digits with addition and subtraction have not been
followed. If they had been, the initial and equilibrium concentrations for various reactants would
have been the same.
This occurs when the initial concentration is at least 500 times larger than the K value for the acid
or base. This is irrelevant when the equilibrium concentrations are given for the weak acid or
base, but what about when this equilibrium concentration is not given?
In cases where [initial] < 500, then the quadratic equation must be used to solve for unknowns
K
such as [H3O+], [OH -], pH, or pOH.
In cases where [initial] > 500, then the change in the concentration of the reactant is ignored and
K
the following assumption is made: [initial] = [equilibrium] to solve for the unknown variable.
Example:
A 0.022 mol/L aspirin (acetylsalicylic acid) solution is made by dissolving a 500.0 mg
tablet in about 125 mL of water. Aspirin formula: C8H7O2COOH. Given Ka for aspirin is
3.3 x 10 -r, calculate the pH of the aspirin solution.
Answer:
Step 1: Write the acid ionization equation and set up the ICE table.
I
C
E
C8H7O2COOH(aq) + H2O(l) H3O+(aq) + C8H7O2COO -(aq)
0.022
0
0
-x
+x
+x
0.022 - x
x
x
Step 2: Write the Ka expression and substitute ICE table values into the expression.
Ka = [H3O+][C8H7O2COO -]
[C8H7O2COOH]
3.3 x 10 -4 =
(x)(x)
(0.022-x)
Step 3: Check the [initial]/K ratio.
0.022 = 66.7
3.3 x 10 -4
Since the initial concentration of the aspirin solution is less than 500 times greater than the
Ka value, the assumption that [aspirin]I = [aspirin]EQ cannot be used. The problem has to
solved by rearranging the Ka expression to the quadratic equation form and then solving the
equation.
Rearrange the Ka expression:
3.3 x 10 -4 =
(x)(x)
(0.022-x)
(3.3 x 10-4)(0.022-x) = x2
7.26 x 10-6 - 3.3 x 10-4x = x2
x2 + 3.3 x 10-4x - 7.26 x 10-6 = 0
The terms of the quadratic equation are then substituted into the general form of the
solution for a quadratic equation.
General Form:
x = -b ± b2 - 4ac
2a
Substitution:
x = -3.3 x 10-4 ± (3.3 x 10-4)2 - 4(1)(7.26 x 10-6)
2(1)
= -3.3 x 10-4 ± 2.8931 x 10-5
2
= -3.3 x 10-4 ± 5.3787 x 10-3
2
= 2.5243 x 10-3 or -2.8543 x 10-3
Since a negative value for concentration is not possible, only one solution is possible, and
the concentration of hydronium at equilibrium (x) is 2.5 x 10-3 mol/L.
Step 4: Calculate the pH.
pH = - log [H3O+] = - log (2.5 x 10-3) = - (-2.597859) = 2.60
So, the pH of the aspirin solution is 2.60.
Step 5: Check your answer.
Estimate the value for [H3O+] by assuming [aspirin]I = [aspirin]EQ.
x2 = 3.3 x 10-4
0.022
x2 = (3.3 x 10-4)(0.022)
x = √7.26 x 10-6
x = 2.6944 x 10-3 = 2.7 x 10-3
Since 2.7 x 10-3 is pretty close to the value produced using the quadratic equation, you can
be sure that you correctly performed the operations associated with the quadratic equation.
Example:
Calculate the pH of a 0.50 mol/L ethanoic acid solution. Ka is 1.8 x 10-5.
Answer:
Step 1: Write the ionization equation and set up the ICE table.
CH3COOH (aq) + H2O (l)  H3O+(aq) + CH3COO -(aq)
I
0.50
0
0
C
-x
+x
+x
E
0.50 - x
x
x
Step 2: Write the Ka expression and substitute ICE table values into the expression.
Ka = [H3O+][CH3COO -]
[CH3COOH]
1.8 x 10-5 =
x2
0.50-x
Step 3: Check the [initial]/K ratio.
0.50
= 28000
1.8 x 10-5
Since the initial concentration of ethanoic acid is 28000 times greater than the K value, you
can use the assumption that [CH3COOH]I = [CH3COOH]EQ.
So, the Ka expression becomes: 1.8 x 10-5 =
x2
0.50
Rearrange the expression to solve for x:
(1.8 x 10-5)(0.50) = x2
x = √(1.8 x 10-5)(0.50)
x = 3.0 x 10-3
Step 4: Calculate the pH.
pH = - log [H3O+] = - log (3.0 x 10-3) = -(-2.5228788) = 2.52
So, the pH of the ethanoic acid solution is about 2.52.
Percent Ionization
A K value indicates the extent to which an acid or a base ionizes and this in turn gives you
information about the strength of an acid or a base. However, percent ionization (or percent
reaction with water) also indicates relative strength of an acid or a base.
You can use information about hydronium or hydroxide concentration and the initial concentration
of an acid or a base, respectively, to calculate the percent ionization.
For example, using the data from the previous example, the percent ionization of a 0.50 mol/L
ethanoic acid solution is:
% ionization = [H3O+] x 100% = 3.0 x 10-3 x 100% = 0.60 %
[acid]I
0.50
Compare that value to the Relative Strengths of Acids and Bases table. What is the percent
ionization for a 0.10 mol/L ethanoic acid solution?
Unlike Ka , percent ionization values are a function of the initial concentration of a weak acid.
Percent ionization actually decreases with increasing weak acid concentration! That’s why it is
important to compare acid strength using Ka values.
Indicators
Litmus paper is something you are undoubtedly familiar with at this point in your schooling. The
phrase "acids turn blue litmus red and bases turn red litmus blue" has probably been repeated in
science courses since your days in elementary school.
Litmus is an indicator. It has two forms: a weak acid form (HLt) which reflects pink light in low pH
conditions and a conjugate base form (Lt-) which reflects purplish-blue light in higher pH
conditions. The colour of litmus that you see is a function of the form of the indicator present in
solution.
Each weak acid form of an indicator has a unique affinity for its proton (H+). The pH range at
which an indicator molecule changes from the weak acid form to the conjugate base form is a
function of the ease with which the proton can be removed from the weak acid form. This is the
property that makes indicators useful in acid-base chemistry because indicators can be used to
reveal information about the pH of a solution.
For example, phenolphthalein is colourless below pH 8.2 and deep pink above pH 10.0.
If you add a drop of phenolphthalein solution to a solution and you see a deep red colour, then you
know that the pH of the solution is something above pH 10.0. However, if you see no colour, then
the pH of the solution is below pH 8.2.
A slight pink colour tells you that the pH is somewhere between pH 8.2 and 10.0 because similar
amounts of both forms of the indicator will be present in this pH range.
Symbol Convention
The conjugate base form of an indicator is a polyatomic anion. We use a two letter abbreviation to
represent the conjugate base form. For example, Ph- and Lt- are the symbols used for the conjugate
base forms of phenolphthalein and litmus respectively. The generic symbol is In-.
The symbol for the weak acid form of an indicator contains one or more hydrogen atoms. HPh and
HLt are the symbols for the weak acid forms of phenolphthalein and litmus respectively. The
generic symbol for the weak acid form of any indicator is Hin.
See the Acid Base Indicators Chart.
Indicators and Acid-Base Equilibria
You can write a net ionic equation for the reaction between a weak acid form of an indicator and
water:
Hin (aq) + H2O (l)  H3O+(aq) + In-(aq)
where In- represents the conjugate base form of an indicator. What should the equation for litmus
look like?
Since the two forms of an indicator have different colours, a colour change represents a shift in the
position of an indicator equilibrium. For example, imagine that you have a beaker of ammonia to
which you have added one drop of phenolphthalein indicator. Since the pH of the ammonia is
fairly high, say 10.5, the colour you should see is deep pink - the colour of the conjugate base form,
Ph-.
The pink colour indicates that the position of the indicator equilibrium lies to the right favouring
the formation of the conjugate base form, Ph-.
HPh (aq) + H2O (l)  H3O+(aq) + Ph-(aq)
colourless form
deep pink form
According to Le Châtelier’s principle, the position of this equilibrium can be shifted to the left by
adding an acid. The sequence of events unfolds this way:
1. the addition of an acid increases the hydronium ion concentration ([H3O+]),
2. the hydroxide ions present in the solution neutralize the hydronium ions, and as a
result, the pH of the solution decreases,
3. the few Ph- ions in the solution also react with the added hydronium ions favouring the
reaction that produces the weak acid form of the indicator (HPh) which is colourless.
The position of the new equilibrium lies to the left.
How can you shift the equilibrium back to the right?
When dealing with acid base indicators, the best way to shift the position of the equilibrium
is to either add an acid (shifts left) or a base (shifts right).
The addition of a strong base like sodium hydroxide will cause a decrease in hydronium ion
concentration and the system will respond by decomposing the weak acid form of the
indicator (HPh) to replace the hydronium ions.
The result is an increase in [Ph-] and a return of the deep pink colour.
Does the addition of an indicator to an acid or base solution affect the pH of the solution? The
answer is no.
When an indicator is added to a solution, it does not change the hydronium or hydroxide
ion concentrations because the amount of indicator present is very small in relation to the
amount of substance to which it is added.
For example, one or two drops of phenolphthalein is enough to colourize a 50 mL sample
of base, but the volume the indicator is tiny relative to the total volume of base, so it has no
measurable effect on hydronium or hydroxide ion concentrations.
Sample Exercise 1
Phenol red indicator (HPr) has a yellow colour below pH 6.6 and a red colour above pH 8.0.
A. Write the net ionic equation for the reaction between phenol red (HPr) indicator and water.
B. Predict the position of the indicator equilibrium when a drop of the indicator is added to a
flask containing 50 mL of 0.100 M nitric acid.
C. Suggest a way to shift the position of the indicator equilibrium in the opposite direction.
Answer
Step A: Write the ionization equation for HPr.
HPr (aq) + H2O (l)  H3O+(aq) + Pr -(aq)
Step B: Apply LeChâtelier’s Principle.
The high concentration of hydronium ions in nitric acid will disturb the indicator
equilibrium.
The conjugate base form of the indicator, Pr -, will react with H3O+ in the nitric acid
solution to form the weak acid HPr. The position of the indicator equilibrium shifts to the
left.
The colour that should be observed in the flask of nitric acid is yellow - the colour of the
weak acid for of the indicatror.
Step C: Reverse the shift by adding hydroxide ions.
Adding a base like sodium hydroxide will neutralize the hydronium ions and reduce their
concentration. The indicator system will respond to the decrease in hydronium by
favouring the forward reaction. The [HPr] will decrease and the [Pr-] will increase The
solution will become red.
Using Indicators to Estimate pH
Before the invention of pH meters, scientists used combinations of indicators to estimate the pH of
a solution. They would place samples of a solution in to separate beakers and add an indicator to
each one. Using the known pH range for each indicator and colour charts, they could come up with
fairly precise estimates of pH.
Sample Exercise 2
What is the approximate pH range of a solution if it caused methyl red indicator to turn
yellow and phenol red indicator to turn yellow?
Answer
Methyl red is a yellow colour above pH 6.0, so the pH of the solution is equal to or greater
than pH 6.0.
Phenol red is a yellow colour below pH 6.6, so the pH of the solution is equal to or less than
pH 6.6.
Therefore, the approximate pH range for the solution is between 6.0 - 6.6.
Titration
What is Titration?
Titration is the progressive addition of one reagent to another. In acid-base chemistry, the reagents
are aqueous solutions, one of which is either a strong acid or a strong base.
A typical titration involves progressively adding a solution of known concentration (the titrant)
from a buret to a solution of unknown concentration (the sample) situated in a container below the
buret. The amount of titrant added is just enough to completely neutralize the sample.
An indicator is usually added to the sample to detect the equivalence point - or the point at which
chemically equivalent amounts of titrant and sample have reacted. The chosen indicator should
change colour at the pH which corresponds to the equivalence point, usually somewhere between
pH 5 and 9. During a titration, the point at which the indicator changes colour is called the
indicator endpoint.Typically, the goal of a titration procedure is to acquire information that
allows you to calculate the molar concentration of a solution of unknown concentration. However,
there are cases in which data collected in a titration can be used to determine the mass of an acid or
a base in a sample.
A buret is a piece of glassware that delivers a precise liquid volume. Upon detection of an
equivalence point in a titration, you record the volume of titrant delivered from a buret. This
information is then used in stoichiometric calculations to determine things like an unknown molar
concentration.
Primary Standards
HCl and NaOH are the most commonly used titrants in acid-base titrations. However, neither HCl
nor NaOH is a primary standard or a solution that can be prepared from a pure form of the solute to
a precise molar concentration..
Solutions of HCl and NaOH have to be standardized. A standardized solution is one whose
concentration is determined from data collected by titration.
HCl has to be standardized because it is a gas at room temperature. It is difficult to dissolve precise
amounts of any gas in water. Moreover, once dissolved, tiny amounts of HCl have the tendency to
escape from solution thereby reducing the concentration of dissolved HCl.
NaOH is a deliquescent or hygroscopic solid - it absorbs moisture from the air. That's why NaOH
pellets have a shiny or glassy appearance when you take them out of a container. When you weigh
out a sample of NaOH, you are obtaining the mass of NaOH plus the mass of the water it has
absorbed from the air. If you prepare a solution using the measured mass of NaOH, its molar
concentration will be slightly lower than expected because of the absorbed moisture.
Acids like HCl are standardized using a primary standard such as sodium carbonate (Na2CO3),
while bases like NaOH and KOH are standardized using potassium hydrogen sulfate (KHSO4) or
potassium hydrogen phthalate (KHC8H4O4).
Standardizing a Solution Using a Primary Standard
The standardization of solutions like HCl or NaOH is carried out using a titration apparatus. It is
important that you know how to properly use a buret before attempting a standardization
procedure.
In this example, the unknown will be the titrant, and the primary standard will be the sample.
The steps in the standardization procedure are:
1. Rinse and fill a buret with the unknown.
2. Add the primary standard to an Erlenmeyer flask.

Option 1: Rinse a pipette with the primary standard solution and then use it to
transfer an exact volume like 20.00 mL into a clean Erlenmeyer flask.

Option 2: Add a precise mass of pure, solid primary standard to a clean Erlenmeyer
flask and add 15-20 mL of distilled water to dissolve it.
3. Add one or two drops of phenolphthalein indicator to the sample.
4. Add titrant to the sample by opening the stopcock of the buret while swirling the flask. At
the first hint of a colour change in the indicator, close the stopcock. Swirl the flask to return
the indicator to its original colour.
5. Add titrant one drop at a time until a slight colour change becomes permanent. Do not
overshoot the endpoint of the indicator.
6. Calculate and record the volume of titrant added.
7. Pour the sample into the sink and flush with plenty of water.
8. Clean and rinse the flask several times with distilled water to make sure it is as clean as
possible.
9. Repeat steps 2-8 three more times. Your goal is to get accurate results; in other words, your
goal is to observe the indicator endpoint using the same volume of titrant three times in a
row.
10. Clean all glassware and return materials to their designated locations. Make sure your
workspace is clean before you leave the lab.
Once you get three accurate results, average the volume of titrant added. Averaging reduces the
effects of random errors like misreading of the meniscus, overshooting the endpoint, or incorrectly
measuring the mass or volume of the sample. Then:
 write the balanced chemical equation for the acid-base reaction.
 and use the volume of titrant, the amount of the sample and the mole ratio to calculate the
molar concentration of the titrant.
A standardized solution can be used as the species of known concentration in other titration
procedures. Most titrations you will do in this course follow the procedure described above - it is a
good idea to learn those steps now so that you can perform them in the upcoming labs.
Closing Point
The quality of the data collected by carrying out titrations is dependent on a number of factors
including:
 the accuracy of the equipment used to prepare a primary standard
 the precision of the molar concentration of the primary standard
 the accuracy of the buret and other measuring devices used in a titration
 the accuracy of the indicator endpoint.
Acid - Base Stoichiometry
In Science 1206 and Chemistry 2202, you wrote full balanced equations for double replacement
reactions. In this lesson, you will use those skills along with your stoichiometry skills to make
predictions about the relative amounts of acid and base involved in neutralization reactions.
When you carry out a titration, the titrant and sample react quantitatively; that is, the reaction goes
to completion - there is negligible reverse reaction. This means that you can use the mole ratio to
determine the relative amounts of acid and base that react.
Strong Acid-Strong Base Stoichiometry
Stoichiometric calculations for quantitative reactions involve conversion factors and the mole
ratio. The conversion factors and their variants are the same ones you used in Chemistry 2202:
n= m
and
C= n
M
V
These are the steps for stoichiometric calculations related to titration.
1.
Write the balanced chemical equation for the reaction.
2.
Identify the "given" and "wanted" species based on the information provided in the
item.
- Generally, you are given enough information to calculate the number of
moles of the given species. (The given species is often, but not always, the
titrant.)
- The wanted species is usually identified as the species for which a mass or
molar concentration has to be calculated. It is also the species which tends
to have just one associated value, usually a volume. (The wanted species is
often, but not always, the sample.)
3.
Set up the mathematical solution to the item. This requires use of the mole
coefficients from the balanced chemical equation.
Let's begin your exploration of acid-base stoichiometry by doing a typical calculation associated
with standardization of a species using a primary standard.
Sample Exercise 1
A series of 20.00 mL samples of sodium hydroxide are titrated with an average of 5.47 mL of
0.100 mol/L sodium hydrogen sulfate. Calculate the molar concentration of the sodium hydroxide
solution.
Answer
Step 1: Write the balanced chemical equation.
Sodium hydroxide will undergo a double replacement reaction with sodium hydrogen sulfate.
NaOH (aq) + NaHSO4 (aq) –> H2O (l) + Na2SO4 (aq)
Step 2: Identify the given and wanted species.
The item provides a volume and molar concentration for sodium hydrogen sulfate. It is the given
species. It is also the titrant.
The item requires you to calculate the molar concentration of sodium hydroxide, so it is the wanted
species.
NaOH (aq) + NaHSO4 (aq) –> H2O (l) + Na2SO4 (aq)
20.00 mL
0.100 mol/L
5.47 mL
Step 3: Solve for the wanted value.
1.
Convert volume and molar concentration of NaHSO4 to moles using the formula:
VxC=n
0.00547 L x 0.100 mol/L = 0.000547 mol
2.
Apply the mole ratio:
n given x mole ratio = n wanted
0.000547 mol NaHSO4 x 1 mol NaOH = 0.000547 mol NaOH
1 mol NaHSO4
3.
Convert moles and volume of wanted to molar concentration: C = n/V
C = 0.000547 mol NaOH
0.02000 L
= 0.02735 mol/L = 0.0274 M NaOH
The standardized sodium hydroxide solution has a molar concentration of 0.0274 M.
The option you choose for calculating the wanted information is entirely up to you. Each option
has its own advantages. The formula method depends on the application of memorized formulas.
The dimensional analysis method involves fewer conversions and calculator inputs.
Sometimes standardization procedures involve the use of a solid mass of primary standard. For
example, hydrochloric acid is standardized by titrating it against solid anhydrous sodium
carbonate.
Sample Exercise 2
Alan and Allison needed standardized hydrochloric acid to carry out a titration. Their teacher
provided then with a stock solution labeled 1.0 M HCl. Knowing that stock solutions of HCl can
have variable molar concentrations, Alan and Allison decide to standardize the stock solution.
They titrated three 1.00 g samples of sodium carbonate with an average of 18.97 mL of stock
hydrochloric acid.
Calculate the molar concentration of the hydrochloric acid stock solution.
Answer
Step 1: Write the balanced chemical equation.
Sodium carbonate will undergo a double replacement reaction with hydrochloric acid to
produce carbonic acid and sodium chloride. The carbonic acid quickly decomposes to
water and carbon dioxide.
2 HCl (aq) + Na2CO3 (aq) –> H2O (l) + CO2 (g) + 2 NaCl (aq)
Step 2: Identify the given and wanted species.
The item requires calculation of the molar concentration of hydrochloric acid, so it is the
wanted species.
The only information given for sodium carbonate is its mass. Convert mass to moles
using the molar mass. The molar mass can be determined from the chemical formula:
MNa2CO3 = 2(22.99) + 1(12.01) + 3(16.00) = 105.99 g/mol
2 HCl (aq) + Na2CO3 (aq) –> H2O (l) + CO2 (g) + 2 NaCl (aq)
18.97 mL
1.00 g
Step 3: Solve for the wanted value
Arrange the terms in a logical sequence so that the units cancel out leaving the unit of
the wanted term.
In this case, convert mass to moles, apply the mole ratio, and then convert moles and
volume to molar concentration. Note that 18.97 mL is converted to litres so that the
molar concentration value has the unit mol/L or M.
Formula Method:
Convert mass to moles: n Na2CO3 =
m
M
=
1.00 g
105.99 g/mol
=
0.0094348 mol
Apply the mole ratio to find the moles of wanted:
n HCl
=
n Na2CO3 x
2 mol HCl
1 mol Na2CO3
= 0.0094348 mol x
2 mol HCl
1mol Na2CO3
= 0.0188697 mol HCl
Calculate the molar concentration: C = n = 0.0188697 mol HCl = 0.995 mol/L HCl
V
0.01897 L
The standardized hydrochloric acid solution has a molar concentration of 0.995 mol/L.
Notice that the amount of water used to dissolve the calcium carbonate is not a factor is the
calculations.
Typical titration problems involve a strong acid and a strong base. The next sample exercise
illustrates how a standardized hydrochloric acid solution is used to collect data to determine
the molar concentration of a sodium hydroxide solution.
Sample Exercise 3
In a titration, a 20.00 mL sample of sodium hydroxide is titrated with an average of 14.35 mL
of standardized 1.06 M hydrochloric acid. Calculate the molar concentration of the sodium
hydroxide solution.
Answer
Step 1: Write the balanced chemical equation.
HCl (aq) + NaOH (aq) –> H2O (l) + NaCl (aq)
Step 2: Identify the given and wanted species.
The item identifies HCl as a standardized titrant. The molar concentration and the
volume of this species is provided. HCl is the given species.
The item asks you to calculate the molar concentration of sodium hydroxide. There is
just one piece of information provided for NaOH. Therefore, NaOH is the wanted
species.
HCl (aq) +
14.35 mL
1.06 M
NaOH (aq)
20.00 mL
–>
H2O (l) +
NaCl (aq)
Step 3: Set up the mathematical solution.
CW = 14.35 mL x 1.06 M x 1 mol NaOH
1 mol HCl
x
1
= 0.76055 M = 0.761 M NaOH
20.00 mL
The molar concentration of the sodium hydroxide solution is 0.761 mol/L.
Polyprotic acids (more on this later) are those that have the ability to donate more than one
proton in acid base reactions. The most common polyprotic acid is sulfuric acid (H2SO4). It is
sometimes called a diprotic acid because it can donate both of its hydrogen ions (protons).
Sample Exercise 4
Three 20.00 mL samples of sulfuric acid are titrated with an average volume of 8.74 mL of
standardized 1.16 M sodium hydroxide. Calculate the molar concentration of the sulfuric acid
solution.
Answer
Step 1: Write the balanced chemical equation.
Sulfuric acid will undergo a double replacement reaction with sodium hydroxide.
H2SO4 (aq) + 2 NaOH (aq)
–>
2 H2O (l)
+
Na2SO4 (aq)
Step 2: Identify the given and wanted species.
The item requires calculation of the molar concentration of sulfuric acid, so it is the
wanted species.
The molar concentration and the volume of sodium hydroxide are given, so NaOH is
the given species.
H2SO4 (aq) + 2 NaOH (aq)
20.00 mL
8.74 mL
–>
2 H2O (l)
+
Na2SO4 (aq)
1.16 M
Step 3: Solve for the wanted value.
Arrange the terms in a logical sequence so that the units cancel out leaving the unit of
the wanted term.
In this case, convert volume and concentration of given to moles, apply the mole ratio,
and then convert moles and volume of wanted to molar concentration.
C H2SO4
=
=
8.74 mL NaOH x 1.16 M x
0.25346 M H2SO4 =
1 mol H2SO4
2 mol NaOH
x
1
20.00 mL
0.253 M H2SO4
The sulfuric acid solution has a molar concentration of 0.253 mol/L.
There are stoichiometry items where you may be required to predict the volume of a reagent
involved in a quantitative acid-base reaction.
Sample Exercise 5
Calculate the volume of standardized 0.992 mol/L hydrochloric acid required to completely
neutralize 25.0 mL of 1.06 mol/L potassium hydroxide.
Answer
Step 1: Write the balanced chemical equation.
Hydrochloric acid will undergo a double replacement reaction with potassium
hydroxide.
HCl (aq)
+
KOH (aq)
–>
H2O (l)
Step 2: Identify the given and wanted species.
+
KCl (aq)
The item requires calculation of the volume of hydrochloric acid, so it is the wanted
species.
The molar concentration and the volume of potassium hydroxide are given, so KOH is
the given species.
HCl (aq) +
0.922 M
KOH (aq) –>
1.06 M
25.00 mL
H2O (l)
+
KCl (aq)
Step 3: Solve for the wanted value.
Arrange the terms in a logical sequence so that the units cancel out leaving the unit of
the wanted term.
In this case, convert volume and concentration of given to moles, apply the mole ratio,
and then convert moles and volume of wanted to molar concentration.
VHCl = 25.00 mL KOH x 1.06 M x 1 mol HCl x
1 mol KOH
= 26.7 mL HCl
1
0.992 M
=
26.71371 mL HCl
So, 26.7 mL of standardized hydrochloric acid solution is required to completely
neutralize the potassium hydroxide.
Sometimes titrations are performed to determine the purity of an acid or a base.
Sample Exercise 6
Extra strength ASA tablets come in a 500 milligram size. Three acetylsalicylic acid tablets
were weighed and found to have an average mass of 499 mg. Each tablet was dissolved in
approximately 25.00 mL of distilled water.
Three titrations were performed using standardized 0.101 mol/L sodium hydroxide. An
average volume of 24.04 mL titrant was used in three trials.
Phenolphthalein indicator was used to detect the equivalence point in each trial. A slight pink
colour was detected after addition of about 24 mL of titrant in each trial.
Calculate the purity of the acetylsalicylic acid (C8H7O2COOH) tablets.
Answer
Step 1: Write the balanced chemical equation.
ASA will undergo a double displacement reaction with sodium hydroxide.
C8H7O2COOH (aq)
+
NaOH (aq)
–>
H2O (l)
+
NaC8H7O2COO (aq)
Step 2: Identify the given and wanted species.
The item requires calculation of the purity of ASA, so the amount of ASA must be
unknown. ASA is the wanted species.
The molar concentration and the volume of sodium hydroxide are given, so NaOH is
the given species.
C8H7O2COOH (aq)
+
NaOH (aq)
24.04 mL
–>
H2O (l)
+
NaC8H7O2COO (aq)
0.101 M
Step 3: Calculate the molar mass of the sample species (ASA).
Since the unknown in this item is the mass of acetylsalicylic acid, a molar mass value
will be needed in the mathematical solution.
M C8H7O2COOH
=
9(12.01) + 8(1.01) + 4(16.00) = 180.17 g/mol
Step 4: Solve for the wanted value.
Arrange the terms in a logical sequence so that the units cancel out leaving the unit of
the wanted term.
In this case, convert volume and concentration of given to moles, apply the mole ratio,
and then convert moles and volume of wanted to molar concentration.
m C8H7O2COOH = 0.02404 L
=
=
x
0.101 M NaOH
x
1 mol C8H7O2COOH
1 mol NaOH
0.4374599 g C8H7O2COOH
0.437 g C8H7O2COOH x 1000 mg/1 g
=
x
180.17 g
1 mol
437 mg
Step 5: Calculate the purity of the tablets.
The purity of ASA tablets is the mass of the ASA divided by the mass of the tablet. It
can be expressed as a percentage:
% purity
=
mass of ASA x 100%
mass of tablet
=
437 mg
499 mg
x
100 %
=
87.6%
The purity of the Aspirin tablet is 87.6 %.
Excess and Limiting Species
Throughout this lesson and the previous one, titration has been described as a procedure
carried out to find the volume of one solution needed to neutralize a measured amount of
another substance. The key part of any titration is detecting the equivalence point. This is
typically achieved using an indicator.
Equivalence points can also be detected using a pH meter. The technique involved in these
cases is a little different than what you have already seen. You still have to approach the
equivalence point carefully by adding a drop of titrant at a time, but you keep adding titrant
after the equivalence point is achieved!
Before the equivalence point is reached, the sample species is in excess. That is, all the titrant
being added is consumed (i.e. is limiting) and the sample still contains unreacted acid or base.
Beyond the equivalence point, the titrant is in excess and the sample is limiting.
Given information about the concentrations and volumes of the titrant and the sample, you can
calculate pH, pOH, [H3O+] or [OH-].
Sample Exercise 7
A 20.00 mL sample of 0.105 M hydrochloric acid is titrated with 28.55 mL of 0.102 M sodium
hydroxide. Calculate the pH of the mixture in the Erlenmeyer flask.
Answer
Step 1: Write the balanced chemical equation.
HCl (aq) +
20.00 mL
0.105 M
NaOH (aq)
28.55 mL
0.102 M
–> H2O (l)
+
NaCl (aq)
Step 2: Calculate the mole amount of each species.
n NaOH
n HCl
=
=
0.02855 L x 0.102 M = 0.0029121 mol
0.02000 L x 0.105 M = 0.00210 mol
Step 3: Identify the limiting and excess species.
Since the mole ratio is 1 mol HCl to 1 mol NaOH, and since the mole amount of NaOH
is greater than the mole amount of HCl, it seems reasonable to conclude that HCl is
limiting and NaOH is in excess. This can be confirmed by calculation:
n NaOH
=
0.00210 mol HCl x 1 mol NaOH = 0.00210 mol NaOH
1 mol HCl
Since only 0.00210 mol of NaOH is needed to neutralize the HCl sample, but the
available amount of NaOH is 0.0029121 mol, NaOH is in excess.
Step 4: Calculate the unreacted amount of the excess species.
By subtraction:
n in excess
= amount available - amount required
= 0.0029121 mol - 0.00210 mol = 0.0008121 mol
Step 5: Calculate the molar concentration of the excess species.
The molar concentration is calcualted using the moles of the excess species and the
total volume of the solution. in this case the total volume is equal to the volume of
titrant plus the volume of the sample or 49.55 mL.
C NaOH = n
V
=
0.0008121 mol
0.04955 L
=
0.0163895 mol/L
Step 6: Write the dissociation/ionization equation and calculate [ion].
NaOH (aq) –>
0.0163895 M
[OH -]
=
Na+ (aq)
+
OH - (aq)
?M
0.0163895 M NaOH
x
1 mol OH 1 mol NaOH
=
Step 7: Calculate the pH of the solution.
pOH = - log [OH -] = - log (0.0163895 M) = 1.785
pH = 14 - pOH = 14 - 1.785 = 12.215
The pH of the mixture in the Erlenmeyer flask is 12.215.
Titration Curves
See CDLI notes - attached
Polyprotic and Polybasic Species
See CDLI notes - attached
0.0163895 M OH -