2
Equilibrium
1. The Equilibrium State
In the previous (stoichiometry) section we assumed that
reactions proceeded until one or more of the reactants was
completely consumed; for many reactions this works well.
However, in some reactions there is a significant back
reaction. An example is the reaction of nitrogen and
hydrogen to form ammonia. Note the double arrow, which
indicates the occurence of both a forward and back reaction.
N2(g) + 3 H2(g)  2 NH3(g)
Here nitrogen and hydrogen are gradually consumed and
ammonia is formed until, as shown at the dashed line on
the right, the concentrations cease to change. At this point
the rates of the forward reaction and the reverse reaction
are equal. The system is then said to be at equilibrium.
Figure 1. A reaction reaching
equilibrium
2. The Equilibrium Constant
The extent to which a reaction proceeds before it reaches equilibrium is characterized by an expression
known as the mass action expression. This is set equal to a quantity known as K, the equilibrium
constant.
The mass action expression is a fraction with the products on the top and the reactants on the bottom,
each raised to an exponent equal to its coefficient in the balanced equation. Thus for the formation of
ammonia (above):
[NH3]2
K=
[N2][H2]3
In expressions like the one above, concentrations of gases are ususally given as pressures, with units of
atmospheres. However the brackets shown above indicate that the concentrations are in moles per liter.
Brackets should not be used for pressures, but sometimes they are.
It bears repeating that equilibrium is a dynamic and not a static process. In the rection A  B, A is
constantly being converted to B and B is constantly being converted to A. The equilibrium
concentrations are those at which the rates of the forward and back reactions are equal. What makes
equilibrium seem static is that since the rates of the forward and back reactions are equal, the reactant
(and product) concentrations do not change.
3. Gas Phase Equilibrium Calculations
Consider the ammonia equilibrium, shown above. Start with N2 = 0.35 atm; H2 = 0.25 atm; and NH3 =
0.40 atm. To help set up the calculation, we will use something called an ICE diagram.
-1-
An ICE diagram gives the pressure or concentrations of each material present in the reaction. It gives
the initial concentration, the change in concentration and the concentration at equilibrium.
In most cases the Initial concentrations
N2
H2
NH3
are given; the Changes are obtained
from the stoichiometry of the reaction;
Initial
0.35 atm.
0.25 atm.
0.40 atm.
the Equilibrium values are obtained by
adding the Change to the Initial.
Change
-x
-3x
+2x
Thus the mass action expression would
be written as:
𝐾=
Equilibrium
0.35 - x
0.25 - 3x
0.40 + 2x
[𝑁𝐻3 ]2
(.40 + 2𝑥)2
=
[𝑁2 ][𝐻2 ]3 (. 35 − 𝑥)(.25 − 3𝑥)3
Of course this is mathematically rather messy. So let's try another equilibrium calculation which works
out more easily.
Consider the reaction
2 A + B  2 C.
The system starts with 0.50 atm. of
A, 0.50 atm. of B and 0.00 atm of C.
If the system ends up with 0.20 atm
of C, what is the value of K? First we
set up an ICE table. Here the changes
are determined from the fact that 0.20
atm. of C are formed.
A
B
C
Initial
0.50
0.50
0
Change
-0.20
-0.10
+0.20
Equilibrium
0.30
0.40
0.20
Now we use the equilibrium pressures from the ICE chart to calculate the value of K.
[𝐶]2
. 22
𝐾=
=
= 1.1
[𝐴]2 [𝐵] . 32 × .4
Now consider the reaction
H2(g) + I2(g)  2 HI(g)
The system starts with [H2] = [I2] = 0.50 atm. and no HI. If K = 0.0100 we calculate:
[HI]2
K = [H ][I ]
2 2
(2x)2
0.010 = (0.50 - x)(0.50 - x)
Take the square root of both sides
Initial
Change
Equilibrium
2x
0.10 = 0.50 - x
2x = 0.050 - 0.10 x and x = 0.024
-2-
H2(g)
I2(g)
HI(g)
0.50
0.50
0
-x
-x
+2x
0.50 - x
0.50 - x
2x
4. The Mathematics of Equilibrium Constants
Although units are not usually given in equilibrim constants, most do have units associated with them.
Consider, for example:
[NH3]2
K=
[N2][H2]3
Here K will have the units of atm-2 or kPa-2.
However, suppose that the concentrations were given in moles/L instead of atm. Now the value of K
must be in units of L2/mol2. Thus a gas phase equilibrium can have two different equilibrium constants.
The term KP is used when the units are atmospheres (pressure) and the term KC is used when the units
are moles/L (concentration). If your concentrations are in one set of units and your equilibrium constant
uses another, K will have to be changed. This can be done with the equation:
KP = KC (RT)Δn
Here the gas constant, R, is 0.08206 L atm/mol  K (or 8.31 L kPa/mol), T is the Kelvin temperature,
and Δn is the change in the number of moles of gas as the reaction is written. For the formation of
ammonia, Δn is -2. This formula is given in the AP Chemistry formula sheet.
In addition to changing the units of equilibrium constants, you must also know when and how to
combine them. Suppose we want to combine the reactions:
A+BC
K = 50
A + C  2D
K = 300
If we combine (add) the two reactions we get a third reaction:
2A+B2D
What is K for the combined reaction?
The rule is that when you add reactions, you multiply the equilibrium constants. So K for the third
reaction would be 15,000. If you subtract the second reaction from the first, then you would divide the
first K by the second. If you double a reaction, you square its K. If you reverse a reaction, the new K is
the reciprocal of the K for the forward reaction. Thus:
H2 + X2  2 HX
K = 2500
½ H2 + ½ X2  HX
K = 50
Although we say that the products are on the top of the mass action expression, an equilibrium does not
have a reactant or product since it goes in both directions. What we actually mean is that the
compounds on the right side of the equation are on the top and the compounds on the left side of the
equation are on the bottom.
To summarize what is used in the mass action expression:
- the concentration of a solute is given in moles/L
- the concentration of a gas is either moles/L or atmospheres
- the concentration of an undissolved solid is exactly one.
-3-
5. Le Chatelier's Principle
Le Chatelier's Principle says that "if a stress is applied to a system at equilibrium, the system adjusts to
reduce the stress." After you are accustomed to using Le Chatelier's Principle, it seems so obvious that
one wonders why it had to be given a name. Yet it often shows up on the multiple choice portion of
every AP Chemistry test.
Consider the reaction:
N2 + 3 H2  2 NH3 + heat
Adding ammonia to a mixture of nitrogen, hydrogen and ammonia at equilibrium will place a stress on
the system. The system will relieve this stress by converting some of the ammonia to nitrogen and
hydrogen.
Adding nitrogen to the system will cause the equilibrium to shift to the right, consuming some of the
added nitrogen as well as some of the hydrogen you started with.
If the temperature is raised, the system will remove some of the excess heat by shifting to the left. If the
volume of the system is decreased, the equilibrium will shift to the right, toward fewer moles of gas, in
order to remove some of the excess pressure.
A common "trick" is to ask about raising the pressure by adding a nonreactive gas (e.g. argon).
Although the total pressure is increased, the partial pressures of the reactants are not affected and the
equilibrium is not shifted. Except for changing the temperature, none of the changes we have discussed
affect the value of K. Yet it would require a change of K in order to alter the concentrations of any of
the reactants.
Consider the equilibrium in which an excess of solid naphthalene (moth balls) is in equilibrium with its
vapor. How is the vapor concentration affected by adding more naphthalene? It isn't! Whether a system
contains more or less solid, the concentration of the solid is still considered to be one. It hasn't changed.
6. Solubility Equilibria
When a soluble or slightly soluble salt is added to water it dissolves initially. Eventually, however, the
solution reaches its capacity and further dissolution stops. What has happened is that the rate at which
the salt dissolves has become equal to the rate at which the dissolved salt precipitates. We say that the
solution is now "saturated." The system has reached equilibrium.
Since salts break into their component ions when they dissolve, this must be seen in the mass action
expression. Consider, for example, the dissolution of lead(II) chloride in water:
PbCl2(s)  Pb2+(aq) + 2 Cl–(aq)
(Note that you don't get Cl2 from PbCl2). The appropriate mass action expression would be:
[Pb+2][Cl-]2
K = [PbCl ]
2
However, since lead chloride is a solid, it will have a concentration of unity and thus can be removed as
a term in the equation. The equilibrium constant for a dissolution reaction is normally referred to as the
"solubility product" and is abbreviated KSP. Thus:
Ksp = [Pb+2][Cl-]2
-4-
In addition to the solubility product, a second quantity is often used in solubility calculations. This is
the molar solubility, the solubility of the substance in moles per liter. The simplest solubility
calculations are those which involve the conversion of the molar solubility to solubility product and the
solubility product to the molar solubility. Be certain you know the difference!
Given that the KSP of lead chloride is 1.6  10–5, what is its molar solubility? If the molar solubility of
lead chloride is x, then in a saturated solution the concentration of Pb2+ is x and the concentration of Cl–
must be 2x. Plugging these terms into the mass action expression gives:
Ksp = (x)(2x)2
1.6  10–5 = 4x3
solubility = x = 0.016 M
Be careful that when you calculate the value of (2x)2 you remember to square both the "2" and the "x."
An astonishing number of students make mistakes here.
Given that its molar solubility is 1.0  10–15, what is the KSP of Bi2S3?
[Bi+3] = 2  1.0  10–15 = 2.0  10–15 M
[S-2] = 3  1.0  10–15 = 3.0  10–15 M
Therefore:
Ksp = [Bi+3]2[S-2]3 = [2.0  10–15]2[3.0  10–15]3 = 108  10–75 = 1.1  10–73
To make it easy for the grader to give you partial credit, always write out the mass action
expression and then plug in the numbers.
7. The Common Ion Effect
If a 0.10 M sodium fluoride solution is used to dissolve calcium fluoride, substantially less calcium
fluoride will dissolve than would have dissolved in pure water. This is caused by fluoride, an ion which
is common to both compounds. This phenomenon is therefore known as the common ion effect. Let us
calculate how much calcium fluoride actually dissolves.
If the molar solubility of calcium fluoride is x, we have:
[Ca+2] = x and [F-] = 0.10 + 2x
This gives the following mass action expression:
Ksp = 4.0  10–11 = (x)(0.10 + 2x)2
We can solve the cubic equation or we can make the simplifying assumption that 0.10 >> 2x. If we
make this assumption, as we usually do, the 2x can be neglected and the equation becomes:
4.0  10–11 = (x)(0.10)2
x = [Ca2+] = 4.0  10–9 M
Was neglecting 2x a valid assumption? The criterion we will use is called "the 5% rule." If a variable is
less than 5% of the constant to which it is added, the variable can be neglected. Is
2x = 2  4.0  10–9 less than 5% of 0.10? It certainly is!
-5-
Most of the difficulties which you will have with solubility products can be summarized in the
following questions. Why did we double the x but not double 0.10? When do you double the fluoride
concentration? The answer is..............never. You are not doubling the fluoride concentration. You are
doubling x, the molar solubility, in order to get the fluoride concentration.
8. Selective Precipitation
In these problems two different compounds precipitate from the same solution. There are all sorts of
questions which can be asked. Consider, for example, a solution containing 0.010 M Ag+ solution and
0.010 M Ba2+. You are told that for silver chromate Ksp = 9.0  10–12 and for barium chromate Ksp =
8.5  10–11.
1) Na2CrO4 is added gradually and with stirring. Assuming that the chromate solution does not affect
the volume, which compound precipitates first? We will imagine the chromate starting at zero and
gradually rising until one compound — silver chromate or barium chromate — starts to precipitate. To
do this we calculate the chromate concentration at which the precipitation starts.
For BaCrO4
Ksp = [Ba2+][CrO42–]
8.5  10–11 = (0.010)(x)
x = [CrO42–] = 8.5  10–9 M
For Ag2CrO4
Ksp = [Ag+]2[CrO42–]
9.0  10–12 = (0.010)2(x)
x = [CrO42–] = 9.0  10–8 M
The answer is barium chromate because it starts to precipitate at a lower chromate concentration.
2) What is the concentration of the first cation to precipitate when the second cation has just started to
come out of solution?
Since the first cation to precipitate is barium,the second cation must be silver. We previously calculated
that when silver chromate starts to precipitate the chromate concentration will be 9.0  10–8 M. What
barium concentration is consistent with that? We can calculate that as follows:
Ksp = [Ba2+][CrO42–]
8.5  10–11 = [Ba2+](9.0  10–8)
[Ba2+] = 9.4  10–4 M
9. Mixtures of Two Solutions and Bounce Back
A common type of solubility problem involves the mixing of two solutions which might or might not
form a precipitate. The first question you will have to answer is whether a precipitate forms.
Suppose, for example, we mix 40 mL of 6.5  10–3 M Pb(NO3)2 with 60 mL of 0.30 M KI solution.
Knowing that for PbI2 the solubility product, KSP = 1.4  10–8, does a precipitate form? To answer that
question we must first calculate the concentrations of Pb2+ and I– after dilution.
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If we assume that the volumes of the two solutions are additive, which we always do, then the final
volume of the solution will be 100 mL. Thus, after dilution:
40 𝑚𝐿
[𝑃𝑏 2+ ] = 6.5 × 10−3 ×
= 2.6 × 10−3 𝑀
100 𝑚𝐿
60 𝑚𝐿
= 0.18 𝑀
100 𝑚𝐿
Then take the concentrations of the diluted ions and use them to calculate Q, the equilibrium quotient.
You may recall the Q has the same form as K but consists of concentrations as they really are and not
the equilibrium values. So at equilibrium Q = K.
[𝐼 − ] = 0.30 𝑀 ×
Q = [Pb2+][I–]2 = [2.6  10–3][0.18]2 = 8.42  10–7
Since Q > K, that is 8.42  10–7 > 1.4  10–8, a precipitate forms.
Next calculate the concentrations of ions after the precipitation has occured and the precipitate has been
removed. It would not work to set up an equation like:
1.4  10–8 = (2.6  10–3 - x)(0.18 - 2x)2
This gives x = 2.6  10–3 and, therefore, [Pb2+]= 0. Certainly the lead concentration is close to zero, but
it is not exactly zero and the exact value is what we need to calculate. The calculation is done by a
method called "bounce back."
In bounce back we assume that the concentration of limiting reactant, in this case Pb2+, really does go
all the way to zero. If necessary we correct the concentration of the other ion, iodide in this case. Then
we calculate how much of the slightly soluble salt redissolves. The concentration of iodide is
0.18 - (2.6  10–3)/2 = 0.17 M. So we calculate the concentration of Pb2+ using the mass action
expression.
1.4  10–8 = [Pb2+][I–]2 = [Pb2+](0.17)2
[Pb2+] = 4.8  10–5 M
You may also be asked the concentrations of the other ions, K+
and NO–3 in this case. To calculate them, go back to the initial
concentrations after dilution. If after dilution but before reaction
[Pb2+] = 2.6  10–3, then [NO–3] = 2  2.6  10–3 = 5.2  10–3.
Similarly the K+ concentration in KI solution will be equal to the
I– concentration before precipitation occurs. Thus [K+] = 0.18 M.
“Bounce back,” which you
use when two solutions are
mixed to form a precipitate,
has three steps.
1. Dilution
2. Reaction
3. Equilibration
10. Strong Acids and Bases
Examples of strong acids are H2SO4, HNO3, HCl, HBr, and HI. (There are a few others, e.g. HClO4,
which we will not discuss.) Since strong acids dissociate completely in water, the H+ concentration (or
the H3O+ concentration) is equal to the concentration of acid. The exception is sulfuric acid which,
since it is diprotic, gives an H+ concentration which is as twice the acid concentration (or would be
twice the concentration if the second proton came off completely).
Similarly strong bases, the hydroxides of Group IA and Group IIA metals, dissociate completely. So in
a solution of potassium hydroxide, the hydroxide concentration is equal to the concentration of
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potassium hydroxide. In a solution of calcium hydroxide, Ca(OH)2, the hydroxide concentration is
twice the concentration of calcium hydroxide.
Because of their complete dissociation, calculating the pH of a strong acid or a strong base solution is
simple. Let us calculate, for example, the pH of a 2.0  10–4 M solution of HBr. The H+ concentration is
2.0  10–4 and the log of 2.0  10–4 is -3.70. (Recall that in a logarithm only the digits after the decimal
are significant.) Since pH is -log[H+], the pH of the solution is 3.70.
As an example of a strong base, suppose we have a solution which is 2.0  10–4 M in Ca(OH)2. This
leads to a hydroxide concentration of 4.0  10–4. The log of 4.0  10–4 is -3.40. However since this is
based on hydroxide ion and not hydrogen ion, the log corresponds to something we call the pOH and
not the pH. Here the pOH will be 3.40. Because [H+][OH–] = 1.00  10–14 we know that the log of H+
plus the log of OH– add up to 14 (i.e. pH + pOH = 14.0). Thus, pH = 14.0 - pOH = 14.00 - 3.40 = 10.60
11. Weak Acids and Weak Bases
Acids which are not listed among the strong acids (i.e. acids which are not H2SO4, HNO3, HCl, HBr, or
HI) are classed as weak acids. Bases which are not strong (i.e. bases other than the hydroxides of Group
IA and Group IIA metals) are weak. In contrast to a strong acid, which is completely dissociated in
aqueous solution, a weak acid dissociates only partly. That is, the dissociation of a weak acid or base is
an equilibrium. In order to calculate the results of that equilibrium, we need to use the equilibrium
constant.
Let us consider the dissociation of a 0.50 M HF, a weak acid which reacts as below.
HF  H+ + F–
The dissociation has an equilibrium constant, referred to as "Ka", equal to 7.2  10–4.
[𝐻 + ][𝐹 − ]
−4
𝐾𝑎 = 7.2 × 10 =
[𝐻𝐹]
The reaction starts with [HF] = 0.50 M and [H+] = [F–] = 0. If the amount of HF which dissociates is x,
then [HF] = (0.50 - x) and [H+] = [F–] = x. Therefore:
(𝑥)(𝑥)
7.2 × 10−4 =
(0.50 − 𝑥)
Clearing the fraction and combining like terms leads to a quadratic equation. Generally, however, it is
possible to make an approximation which allows us to avoid using the quadratic formula. If K is small,
x should be small. In fact it will probably be so small that (0.50 - x) will be equal to 0.50, and the x on
the bottom can be ignored. Neglecting x gives the following equation.
𝑥2
7.2 × 10−4 =
0.50
This is easily solved to give x = 0.019, which means that:
[H+] = [F–] = 0.019 M and
[HF] = 0.50 - 0.019  0.50 M
As before we use the "5% rule" to determine whether neglecting x was valid. Since 0.019 is less than
5% of 0.50, the assumption is valid. Of course in some problems you know the value of x and are using
it to calculate K. In such a case you don't neglect x — even if it is less than 5%.
-8-
Weak base calculations typically involve ammonia or a similar compound (e.g. trimethylamine,
N(CH3)3). Let us calculate the pH of a 0.10 M ammonia solution, which reacts with water:
NH3 + H2O  NH4+ + OH–
The equilibrium constant, Kb, is equal to 1.8  10–5. The equilibrium expression is, therefore:
[𝑁𝐻4+ ][𝑂𝐻 − ]
1.8 × 10−5 =
[𝑁𝐻3 ]
Notice that water is in the reaction but not in the mass action expression. Substituting the
concentrations gives:
𝑥𝑥
1.8 × 10−5 =
0.10 − 𝑥
Neglecting x and multiplying gives:
x2 = 1.8  10–6 M
x = [OH–] = 1.34  10–3 M
pOH = 2.87
pH = 11.13
12. Polyprotic Acids
A polyprotic acid has more than one removable proton. An example would be hydrogen sulfide:
H2S  H+ + HS–
Ka1 = 1.1  10–7
–
+
2–
HS  H + S
Ka2 = 1.2  10–13
Why are the equilibrium constants for the removal of the two protons different? This is because the first
proton is removed from a neutral molecule while the second is removed from an anion.
As an example, we will calculate the ion concentration in 0.100 M H2S solution.
[H+][HS-]
KA1 = [H S]
2
[𝑥][𝑥]
1.1 × 10−7 =
[0.100 − 𝑥]
After neglecting the x in the denominator, we solve to get:
x = [H +] = [HS -] = 1.0  10–4 M
Additional hydrogen ions will be formed in the second ionization. However, the H+ formed in later
steps is ALWAYS neglected. Let us now determine the sulfide concentration:
[H+][S-2]
KA2 =
[HS-]
[1.0 × 10−4 + 𝑦][𝑦]
1.2 × 10−13 =
[1.0 × 10−4 − 𝑦]
Notice that we used "y" instead of "x" in order to lessen confusion. After neglecting the additive y's, the
problem becomes enormously easy, and we get:
y = [S -2] = 1.2  10–13 M
Neglecting the y was clearly justified. Also note that "y" is equal to the H+ which comes from the
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second dissociation. As stated above, y is always small and can be ignored in calculating H+.
13. Hydrolysis
Hydrolysis occurs when the conjugate base of a weak acid or the conjugate acid of a weak base is
dissolved in water. To calculate the pH of a sodium acetate solution, one calculates the hydroxide
concentration caused by the hydrolysis of the acetate.
C2H3O–2 + H2O  HC2H3O2 + OH–
You will notice that in this reaction acetate functions as a base. Therefore to solve the problem you
must use Kb for the acetate ion — which is Kw divided by Ka. Thus for 0.15 M acetate you calculate:
𝐾𝑏 =
𝐾𝑤 1.0 × 10−14
=
= 5.55 × 10−10
𝐾𝑎
1.8 × 10−5
Setting x = [OH–] = [HC2H3O2]
[𝑂𝐻 − ][𝐻𝐶2 𝐻3 𝑂2 ]
𝑥𝑥
−10
𝐾𝑏 = 5.55 × 10
=
=
[𝐶2 𝐻3 𝑂2− ]
0.15 − 𝑥
Neglecting the x in the denominator gives:
x2 = 8.32  10–11
x = [OH–] = 9.12  10–6 M
pOH = 5.04 and pH = 8.96
In hydrolysis only the conjugate base (or conjugate acid) is present. If the conjugate base is
accompanied by the acid — if acetate is accompanied by acetic acid — the solution is a buffer. (See
below.) The following solutions should be treated as hydrolysis problems:
Sodium carbonate
Ammonium chloride
Equal moles of acetic acid and sodium hydroxide
A titration at its equivalence point
The following solutions should not be treated as hydrolysis problems:
Acetic acid
Ammonia
Unequal numbers of moles of acetic acid and sodium hydroxide
A titration anyplace except at its equivalence point
For qualitative questions — that is whether a given compound forms an acidic, basic or neutral solution
— you must know the following: The salt of a strong acid and a strong base is neutral; the salt of a
strong acid and a weak base is acidic; the salt of a weak acid and a strong base is basic; the salt of a
weak acid and a weak base is a nasty question which no one will ask you. Using this logic, you should
predict that ammonium nitrate is acidic, calcium chloride is neutral, potassium phosphate is basic and
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aluminum sulfate is acidic.
14. Buffers
A buffer is an aqueous solution which contains roughly equal concentrations of a weak acid and its
conjugate base or a weak base and its conjugate acid. A buffer resists pH change because no matter
whether H+ or OH– is added, there is something in the solution which will react with the added ion.
Consider a buffer made from the weak acid HF and its conjugate base F–. If an acid is added, it reacts
with the fluoride:
H+ + F–  HF
If a base is added, it will react with the hydrofluoric acid:
OH– + HF  H2O + F–
The pH of a buffer is easy to calculate. Consider a buffer based on HF, which has a KA = 7.2  10–4. In
a solution where concentrations of both the weak acid and its conjugate base are equal to 0.10 M, the
pH can be calculated from the mass action expression.
[𝐻 + ][𝐹 − ] [𝐻 + ](0.10)
𝐾𝑎 = 7.2 × 10−4 =
=
[𝐻𝐹]
(0.10)
[H+] = 7.2  10–4 M
pH = 3.14
If H+ is added to the buffer, it all reacts with F– to form HF. (This is important!) So if we add (without
dilution) 0.01 M H+, we end up with 0.11 M HF and 0.90 M F–. Therefore:
[𝐻 + ][𝐹 − ] [𝐻 + ](0.090)
−4
𝐾𝑎 = 7.2 × 10 =
=
Remember that adding a strong
[𝐻𝐹]
(1.10)
base to a mixture of HA and A+
–4
[H ] = 8.8  10 M
both decreases the HA and
increases the A-.
pH = 3.05
Now let us try another buffer calculation, based on hypochlorous acid, HOCl, for which
Ka = 3.5  10–8.
Suppose we start with 1 liter of 0.50 M HOCl and add,
HOCl
OH–
OCl–
without diluting the solution, 0.20 moles of KOH. This
is typical of how buffer problems start; so it is
Initial
.50
.20
0
important you understand it. The strong base (KOH)
-.20
-.20
+.20
reacts completely with the weak acid (HOCl) to form Change
–
H2O and OCl .
Equilibrium
.30
0
.20
HOCl + OH–  OCl– + H2O
This can be summarized in the ICE chart on the right:
The H+ is formed by "bounce back" and its concentration is determined by the equilibrium expression:
[H+][OCl-]
KA = [HOCl]
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3.5 × 10−8 =
[𝐻 + ](0.20)
(0.30)
[H+] = 5.25  10–8 M
pH = 7.28
Now consider a buffer made by mixing 70.0 mL of 0.100 M ammonia with 30.0 mL of 0.100 M HCl.
We will assume the volumes are additive. After dilution, the concentrations are:
[NH3] = 0.070 M
[HCl] = 0.030 M
The HCl (actually the H+ from the HCl) reacts with the NH3 as shown below:
H+ + NH3  NH4+
This gives:
[NH4+] = 0.030 M
[NH3] = 0.040 M
Plugging this in to the mass action expression gives us:
[𝑁𝐻4+ ][𝑂𝐻 − ]
𝐾𝑏 =
[𝑁𝐻3 ]
(0.30)[𝑂𝐻 − ]
1.8 × 10−5 =
[0.40]
[OH–] = 2.4  10–5 M
pOH = 4.62
pH = 9.38
Many textbooks explain buffers in terms of the Henderson-Hasselbalch Equation:
[base]
pH = pKA + log [acid]
This is equivalent to what we have been doing, and many students find Henderson-Hasselbalch makes
simple problems even simpler. However, it makes hard problems even more difficult. So we are not
using it here.
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15. Indicators
How do people determine the pH of a solution? An electrochemical device called a pH meter can be
used. However, these are delicate and expensive and they need frequent calibration. A simpler method
involves the use of an organic acid or base which has different colors in its acidic and basic forms. Such
a material, an example of which is phenolphthalein, is called an indicator.
Phenolphthalein is colorless in its acid form (Figure 1) but
becomes pink when it loses a proton. The equilibrium constant
for the loss of a proton by this indicator (HIn  H+ + In–) is
Ka = 1  10–9 — which gives a pKA of 9. The pKA of an acid is
significant since it equals the pH at which half the material is in
the acid form and half is in the basic form. Thus the pK roughly
equals the pH at which color change occurs. Of course one can see
the pink color of the basic form long before the indicator is half
converted. Consider what happens at pH 8.
[𝐻 + ][𝐼𝑛− ]
𝐾𝑎 =
[𝐻𝐼𝑛]
10−8 [𝐼𝑛− ]
−9
10 =
[𝐻𝐼𝑛]
[𝐼𝑛− ]
0.1 =
[𝐻𝐼𝑛]
Dropping the pH by one unit causes
the fraction of indicator in the basic
form to decrease from 50% to about
9%. However, a tinge of pink in an
otherwise colorless solution can
probably be seen at even less than
9%. For most indicators the color
changes over a range of about one
pH unit on either side of the pK.
Fortunately, this is narrow enough
for most purposes.
Figure 1. The acidic form
of phenolphthalein
Indicator
pH Range pKa Acid Form Base Form
Methyl violet
0.0 - 1.6
0.8 yellow
blue
Methyl yellow
2.9 - 4.0
3.3 red
yellow
Methyl orange
3.1 - 4.4
4.2 red
yellow
Methyl red
4.2 - 6.2
5.0 red
yellow
Chlorophenol red
4.8 - 6.4
6.0 yellow
red
Bromothymol blue 6.0 - 7.6
7.1 yellow
blue
Phenol red
6.4 - 8.0
7.4 yellow
red
Cresol purple
7.4 - 9.0
8.3 yellow
purple
Thymol blue
8.0 - 9.6
8.9 yellow
blue
Phenolphthalein
8.0
9.8
9.0
colorless
red
Figure 2. Properties of Acid-Base Indicators at 25°C
The strips of pH paper which most of
you have used for measuring pH are based on indicators. Typically a mixture of three indicators, which
change color at different pH values, is used to give a solution which changes color continuously over a
wide pH range. Litmus is also an indicator, one which turns red in acidic solution and blue in basic
solution. (Remember bbblue for bbbasic.)
Indicators are widely used to determine the end point in a titration. What you need to know, how to
choose which indicator to use in a given titration, will be discussed in a later section.
16. Titrations
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Titrations are used to determine the amount of acid or base in a sample. If the sample to be analyzed is
an acid, a base solution of known concentration (the titrant) is added. The point at which the moles of
acid and the moles of base are equal is known as the "end point" or "equivalence point." Calculations
are based on the volume of titrant needed to reach the end point. Thus one can calculate the moles of
acid in a sample:
MolesA = MB VB
If the acid is in a solution of known volume, its molarity can be calculated from:
MA VA = MB VB
Suppose a sample of organic acid weighing 0.255 g was found to require 35.6 mL of 0.0110 M KOH to
reach the equivalence point. What is the molecular weight of the unknown acid? We know the
molecular weight = grams/moles and that at the end point molesA = molesB. So:
𝑚𝑎𝑠𝑠𝐴
𝑚𝑎𝑠𝑠𝐴
𝑚𝑎𝑠𝑠𝐴
0.255
𝑀. 𝑊. =
=
=
=
= 651 𝑔/𝑚𝑜𝑙
𝑚𝑜𝑙𝑒𝑠𝐴 𝑚𝑜𝑙𝑒𝑠𝐵
𝑀𝐵 𝑉𝐵
0.0110 × 0.0356
What is the molarity of an acid solution if 5.0 mL are titrated by 28.8 mL of 0.0110 M KOH?
MA VA = MB VB
MA  5.0 = 0.0110  28.8
MA = 0.0634 M
Since volume is stated in mL on both sides of the equation, the units cancel.
All this assumes a monoprotic acid. If the acid is diprotic, and assuming that both protons are titrated, it
is necessary to multiply the acid molarity by two. Thus:
2  MA VA = MB VB
17. Titration Curves
To understand what happens during a titration it is helpful to graph the way pH changes as the titration
proceeds. This graph is known as a titration curve.
First look at what happens when a strong acid is titrated with a
strong base (Figure 3). The pH starts low. As the titration
proceeds (as base is added) the pH increases slowly. Suddenly, as
the titration nears the equivalence point, the slope increases and the
pH changes more rapidly. The equivalence point is, in fact, the
point at which the titration curve has its greatest slope. This is why
titrations work so well. At the equivalence point, the very point you
want to determine, a small amount of titrant will cause a large
change in pH. A single drop could cause the pH to change from 4
to 10, for example. This is fortunate, since indicators change color
gradually over a range of several pH units. This titration is often
done using phenolphthalein (pKA = 9) as an indicator.
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Figure 3. The titration of a
strong acid with a strong base.
The titration of a weak acid with a strong base gives a similar curve
(Figure 4), but there are differences. The initial pH is higher, since
the acid is weaker, and the pH at the equivalence point will be higher
as well. What this means is that the vertical portion of the titration
curve, the segment near the inflection point, is smaller than with a
strong acid. This means that determining the end point is more
difficult with a weak acid than with a strong acid, and choice of
indicator is more critical.
Figure 4. The titration of a
This is a good time to address a point which some students find
weak acid with a strong base.
confusing. The simple comparison between strong and weak
acids given in the previous paragraph works only if the concentrations
are the same. A concentrated solution of a weak acid can be more acidic than a dilute solution of strong
acid.
You must know what species are present at various points on the titration curve. Consider the titration
of an HF solution with a solution of KOH. At first the solution contains only HF. As the reaction
proceeds, the HF reacts with OH– to form H2O and F–. Thus, the principal species are K+, HF and F–.
At the equivalence point, all the HF has been converted to F– and the principal species present are K+
and F–. What happens when the volume of titrant is half of that needed to reach the equivalence point?
(This is called the half equivalence point). Here the concentrations of HF and F– are equal. If [HF] =
[F-], we get the following:
[H+][F-]
KA = [HF] and KA = [H+]
Therefore the pH is equal to the pK. This is not a calculation which a practicing chemist does often, but
it occurs frequently on the AP test. Watch for the half-equivalence point!
To calculate the pH at various points on the curve you should remember that the initial solution is a
weak acid, a calculation which you know well. The solution at the half-equivalence point (or at any
other point part way through the titration) is a buffer. Finally, at the equivalence point all the weak acid
and strong base have reacted. All the HF has been converted to F– and therefore the [H+] and [OH–]
concentrations are based on the reaction:
F– + H2O  HF + OH–
The titration of a weak base with a strong acid is simply the reverse of
this. It starts at a high pH (but not as high a pH as it would if it were a
strong base) and has its equivalence point below 7. Similarly the titration
of a strong base with a strong acid gives a curve which is the same as the
titration of a strong acid with a strong base. They both have their
equivalence points at 7, but in one the pH increases and in the other it
decreases.
Students worry about what happens when you titrate a weak acid with a
weak base. Don't! The experimenter gets to choose the titrant and no one
would choose to titrate with a weak base.
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Figure 5. Titration Curve
for a diprotic acid.
The titration of a polyprotic acid is more complex. Consider a diprotic acid with: KA1 = 10–3 and Ka2 =
10–10. Since the acid has two K's, the titration curve has two inflection points. When the K's of a
polyprotic acid are close, one deprotonation starts before the other has finished and determining the end
point is difficult.
18. Mixing Different Types of Acid/Base Problems
Acid/base problems are not difficult. Unfortunately students often don't know what category a problem
falls into and try to solve the wrong type. There are four types of acid/base problems. We talk about
acids, but everything here is equally applicable to bases.
Strong Acid -- nothing present except for a strong acid
Weak Acid -- nothing present except for the weak acid and its dissociation products. Finding the
pH of an acetic acid solution is a weak acid problem.
Buffer -- a weak acid and its conjugate base. Acetic acid and sodium acetate would make a buffer.
Acetic acid and sodium hydroxide would also make a buffer, because the hydroxide would
react with the acetic acid to form acetate. However hydrochloric acid and sodium chloride
would not form a buffer, because hydrochloric acid is strong.
Hydrolysis -- only the conjugate base of a weak acid. Thus, sodium acetate solution is a hydrolysis
problem. However, if you add an acid to this solution, it becomes a buffer.
Some examples:
1.0 M SODIUM FLUORIDE. Forms sodium ions and fluoride ions in solution. The fluoride ions react:
F– + H2O  HF + OH–, which is hydrolysis.
1.0 M SODIUM FLUORIDE + 0.5 M HYDROFLUORIC ACID. These dissolve to form F– and HF. They
don't react; they form a buffered equilibrium.
1.0 M HYDROFLUORIC ACID + 1.0 M SODIUM HYDROXIDE. They form HF and OH-, which react
according to HF + OH–  F– + H2O. Because these are present in equal concentrations, you
end up with 1.0 M fluoride which then hydrolyzes.
1.0 M SODIUM FLUORIDE + 1.0 M HYDROCHLORIC ACID. The solution contains 1.0 M F– and 1.0 M
H+. They react to form 1.0 M HF which then dissociates in a typical weak acid reaction.
0.5 M SODIUM FLUORIDE + 1.0 M HYDROCHLORIC ACID. These form H+ and F– ions. All the
fluoride is protonated by 0.5 M of the H+ and the remaining 0.5 M H+ is left to act as a strong
acid.
1.0 M HYDROFLUORIC ACID + 0.5 M SODIUM HYDROXIDE. These form HF and OH-. All of the
hydroxide reacts with 0.5 M HF to form 0.5 M F-. This mixture, which now contains 0.5 M
fluoride and 0.5 M HF, acts as a buffer.
1.0 M SODIUM FLUORIDE AND 0.5 M HYDROCHLORIC ACID. These dissolve to give a solution
containing H+ and F– ions. All the H+ from the hydrochloric acid is used to protonate 0.5 M
of the fluoride forming 0.5 M HF. This leaves a mixture containing 0.5 M HF and the
remaining 0.5 M F-, which is a buffer.
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19. Bronsted and Lewis Models
After all we have learned about acids and bases and their properties, it is strange to realize that the
definition of acids and bases which we have been using is only one of several. The definition which we
have used, that an acid supplies H+ ions and a base gives OH– ions, is the original definition as put forth
by the Swedish chemist Svante Arrhenius. It is, therefore, called the Arrhenius model or the Arrhenius
definition. It is very useful, but there are others.
The Bronsted Model, or more properly the Bronsted-Lowry Model, defines an acid as a proton donor
and a base as a proton acceptor. Consider the reaction of hydrofluoric acid with ammonia
HF + NH3  F– + NH4+
The hydrofluoric acid donates a proton to the ammonia, making the hydrofluoric acid an acid and the
ammonia a base. Fluoride, which results from the loss of a proton by hydrofluoric acid, is considered a
base since in the reverse reaction it gains a proton. It is said to be the "conjugate base" of hydrofluoric
acid. Similarly the ammonium ion, which loses a proton in the reverse reaction, is said to be the
"conjugate acid" of ammonia.
Altlhough this is less obvious, the dissolving of hydrogen chloride in water can also be considered an
acid base reaction.
HCl + H2O  H3O+ + Cl–
Here HCl is the acid, water is the base, hydronium ion (H3O+, essentially another way of writing H+) is
the conjugate acid and chloride is the conjugate base.
Water is a base. How strange! Now look at what happens when ammonia is added to water:
NH3 + H2O  NH4+ + OH–
Ammonia is a base, water is an acid, ammonium ion is the conjugate acid and hydroxide ion is the
conjugate base. So water is either an acid or base, depending on the circumstances.
In the Bronsted model acid/base reactions can be considered to be a competition between the base
and the conjugate as to which one wants the proton more.
Imagine a creature on some distant
planet where lakes are made of
ammonia instead of water. Acid/base
chemistry in ammonia would be based
on the reaction:
NH3 + NH3  NH4+ + NH2–
Chemistry students on this planet might
think it strange that acid/base
neutralization reactions could occur in
liquid water. Yet they could easily
understand aqueous chemistry in terms
of the Bronsted model.
Another way of defining acids and bases is seen in the Lewis model. Lewis acids are electron pair
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acceptors and Lewis bases are electron pair donors. This is consistent with the Arrhenius model, since a
proton is itself an excellent acceptor of electrons while hydroxide is a good electron donor. Thus a
neutralization reaction is water is still an acid base reaction.
H+ + OH–  H2O
However, other reactions which we do not think of as acid/base reactions are acid/base reactions
according to the Lewis model. Consider the
reaction of trimethylamine, which we know to
have an unshared pair of electrons and to thus be a
Lewis base. Suppose it reacts with BF3, a molecule
Figure 6. A Lewis acid – Lewis base reaction
which lacks an octet and is, therefore, capable of
acting as an electron acceptor. The molecule which results, BF3-N(CH3)3, is referred to as a "Lewis
acid/Lewis base adduct."
20. Relation of Structure to Acidity
A compound which contains a hydrogen atom will often give it up when dissolved, forming an acidic
solution. However many compounds, for example organic compounds other than carboxylic acids, do
not lose protons in water. The question of why some proton-containing compounds ionize easily
(strong acids), others ionize with difficulty (weak acids) and still others don't ionize at all (non-acids)
has to do with bond strength. Why bonds strengths vary as they do is complex.
The simplest example of a structure/acidity relationship
occurs in the hydrogen halides — HF, HCl, HBr and HI.
Since HF contains the most electronegative, and therefore
the most non-metallic, of the halogens, it seems that it should
form the most acidic hydride. However that's not the way it
works. Being the most electronegative element means it will
form the strongest bond with hydrogen, which means it will
form the weakest acid.
Acid
HF
HCl
HBr
HI
Bond
Strength
(kJ/mol)
565
427
363
295
Acid
Strength
in Water
Weak
Strong
Strong
Strong
The next trend is among the oxy-acids, for example HClO4,
HClO3, HClO2 and HClO. The trend is that the more oxygens there are, the stronger the acid. Thus the
trend in acid strength is: HClO4 > HClO3 > HClO2 > HClO. To understand this you must first realize
that the formulas written above give a misleading picture of the structures. In almost all oxy-acids the
hydrogen is bound to an oxygen and not to the central atom. Thus, HClO2 is actually H-O-Cl-O. (That's
why HCl, a strong acid, doesn't fit the trend. The hydrogen in HCl is bound directly to chlorine.)
Why do additional oxygen atoms make an oxy-acid stronger? The easiest way to view this (and there
are several legitimate ways) is to say that oxygen, being very electronegative, withdraws electron
density from the chlorine. Since losing a proton (a hydrogen cation) will leave behind a negative
charge, something which withdraws negative charge will make the ion more stable. The acidity of
other oxy-acids follows the same trend. Thus: H2SO4 > H2SO3 and HNO3 > HNO2.
Another trend you should know is that metal hydroxides are basic while non-metal hydroxides are
acidic. Thus ClOH is an acid (usually written as HOCl) and KOH is a base. Similarly metal oxides
generally react with water to form basic solutions, for example:
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CaO + H2O  Ca2+ + 2 OH–
Non-metal oxides, on the other hand, react with water to form acidic solutions, for example:
SO2 + H2O  H2SO3
Why is this? The answer is that an electronegative element, such as chlorine, forms a covalent bond
with oxygen. This bond, like other covalent bonds, does not break easily in water. Thus the only bond
which can break is the H-O bond. In metal hydroxides, however, the central atom is electropositive and
thus forms an ionic bond. The ionic bond between the metal ion and the hydroxide ion is strong, but
unlike the covalent bond it comes apart easily in a polar solvent such as water. (As you may recall,
polar water molecules surround the ions and stabilize the charges.)
The intermediate case, in which the central atom is neither strongly electronegative nor strongly
electropositive, leads to a situation where both the X-O bond and the O-H bond are capable of breaking.
This leads to a phenomenon known as "amphoterism," in which the oxide or hydroxide is capable of
acting as either an acid or a base. It is difficult to predict whether a transition metal oxide will be acidic,
basic or amphoteric. In general you will not have to do that. However, when we get to the reaction
section of this course, you will learn that aluminum and zinc form amphoteric oxides. But that's for
another day.
21. "How many acids" problems
One good way of testing your knowledge of acid-base chemistry is to ask how many compounds in a
list form acidic (or basic) solutions in water. Consider the following list of compounds.
CaBr2, NH4Cl, SO3, AlCl3, KCN
If you answered that "two form acidic solutions," you are wrong. The correct answer is "three,"
specifically NH4Cl, SO3, and AlCl3. Make sure you understand this.
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