This document explains how the concepts of chemical
equilibrium can be applied to the ionization of acids and bases in aqueous
solution. It is assumed that the
reader already has a fundamental understanding of chemical equilibrium.
If not, you need to first read my Web notes on “Chemical
Equilibrium”, in which the topic is developed from the ground up
There are several definitions for acids and bases.
The simplest, proposed by Svante Arrhenius in 1887, will satisfy our
needs here
An acid is a substance that produces H^{+} ions (or
equivalently, H_{3}O^{+} ions) when dissolved in water, and a
base is a substance that produces OH^{} ions when dissolved in water.
Let’s begin our look at acids using hydrochloric acid as
an example. The active ingredient
in hydrochloric acid is hydrogen chloride, which when pure, is a gas at room
temperature and normal atmospheric pressure.
When this gas dissolves in water, it produces the aqueous solution we
call “hydrochloric acid”.
H_{2}O
HCl(g) >
HCl(aq)
(1)
hydrochloric
acid
HCl(aq) >
H^{+}(aq) +
Cl^{}(aq)
(2)
Recognizing that hydrogen ions combine with water
molecules, we can write
H^{+}(aq)
+ H_{2}O(l)
> H_{3}O^{+}(aq)
(3)
If we add equations 2 and 3, the H^{+} cancels out
and we get
HCl(aq) +
H_{2}O(l) >
H_{3}O^{+}(aq)
+ Cl^{}(aq)
(4)
You will see both equations 2 and 4 used to represent the
ionization of HCl in water. They
tend to be used interchangably, so which one is used depends on which is more
convenient in a particular context. I
will normally opt for equation 4, since it is a more complete picture of what
happens, but there are times – such as working out net ionic equations for
acid / base neutralization reactions – when it is more convenient to represent
the ionized hydrogen simply as H^{+} rather than H_{3}O^{+}.
Whether we use equation 2 or 4, notice that the arrow
points only to the right. That is,
the equation is not represented as a reversible equilibrium reaction.
We say that HCl is a strong acid, and represent the ionization as a
oneway process. That is, when
molecular HCl dissolves in water, it begins to split into ions, and the process
continues until all the acid is ionized. We
don’t think of the ions as combining to form the molecular acid again.
Strong acids are 100% ionized in
water. In contrast, a weak acid
is only partially ionized in water. Acetic
acid is an example. The pure acid
is a liquid at room temperature and normal atmospheric pressure. It dissolves in water to form an aqueous acetic acid
solution.
H_{2}O
HC_{2}H_{3}O_{2}(l)
> HC_{2}H_{3}O_{2}(aq)
(5)
acetic acid
solution
The acetic acid ionizes in solution, but the ions can
recombine to form the molecular acid again.
HC_{2}H_{3}O_{2}(aq)
+ H_{2}O(l) <>
H_{3}O^{+}(aq)
+ C_{2}H_{3}O_{2}^{}(aq)
(6)
Because the reaction is reversible, only a fraction of the
acid will be ionized at any one point in time.
For this reason, we say that acetic acid is a weak acid.
A weak acid is only partially
ionized in aqueous solution.
Now let’s look at bases.
Sodium hydroxide is an example of a base. Pure sodium hydroxide is an ionic solid.
It dissolves in water to form a basic solution.
H_{2}O
NaOH(s) >
Na^{+}(aq) +
OH^{}(aq)
(7)
Like acids, bases can be characterized as either strong or
weak. Sodium hydroxide (NaOH) is an
example of a strong base. Notice
that the arrow points only to the right. When
the solid dissolves in water, it splits into ions, and these ions do not
recombine to form the solid again unless the solution is saturated.
A strong base is 100% ionized in
water. Now let’s consider a
weak base. Ammonia (NH_{3})
provides an example. Pure NH_{3} is a gas at room temperature and normal
atmospheric pressure. It dissolves
in water to form a basic solution known as aqueous ammonia or “ammonium
hydroxide”.
H_{2}O
NH_{3}(g) >
NH_{3}(aq)
(8)
NH_{3}(aq)
+ H_{2}O(l)
<> NH_{4}^{+}(aq)
+ OH^{}(aq)
(9
Notice the bidirectional arrow in equation 9. This
indicates that the reaction is reversible.
When the NH_{3} ionizes, the ions it generates are capable of
recombining to form the molecular NH_{3} again.
This means that not all of the NH_{3} will be ionized at any one
point in time. A
weak base is only partially ionized in aqueous solution.
You can see in equation 9 that an aqueous ammonia solution
will contain ammonium (NH_{4}^{+}) cations and hydroxide (OH^{})
anions. It is for this reason that
aqueous ammonia solutions are sometimes referred to as “ammonium hydroxide”.
But this name is something of a misnomer.
Ammonium hydroxide is not a stable compound that can be isolated.
If you have an aqueous solution of sodium chloride (table salt), you can
boil all the water away and recover the solid sodium chloride.
The same thing is not true of ammonia hydroxide, however.
You can not boil the water away to recover “ammonium hydroxide” from
the solution. The solution will
boil away as a mixture of NH_{3} and H_{2}O gases.
“Ammonium hydroxide” is known only in aqueous solution.
It has never been isolated.
Before we begin looking in detail at the equilibrium
chemistry of acidic and basic solutions, we need to look at the equilibrium of
water itself. Even in a sealed
container of pure water, there is an equilibrium going on.
You will sometimes see it written as
H_{2}O(l)
<> H^{+}(aq)
+ OH^{}(aq)
(10)
but recognizing that H^{+} will combine with an H_{2}O
molecule to form H_{3}O^{+}
H^{+}(aq)
+ H_{2}O(l)
> H_{3}O^{+}(aq)
(3)
we can add equations 10 and 3 to get
2H_{2}O(l)
<> H_{3}O^{+}(aq)
+ OH^{}(aq)
(11)
You may see equations 10 and 11 used interchangeably, but
in most cases, I will use equation 11. It
is a more accurate depiction of what is happening in solution.
Since equation 11 is an equilibrium equation (notice the
bidirectional arrow) it should have an equilibrium constant expression.
If we were to write a concentrationbased (K_{C}) equilibrium
constant expression as we learned to do in our first introduction to chemical
equilibrium, it would be
[H_{3}O^{+}]_{eq}[OH^{}]_{eq
}K_{C} =

(12)
[H_{2}O]_{eq}^{2}
but recently, we learned that solids and liquids are
omitted from the equilibrium constant expression.
As we have seen, the mathematical argument for this is that solids and
liquids have a constant concentration. If
[H_{2}O]_{eq} is constant, then so is [H_{2}O]_{eq}^{2}.
Multiplying both sides of equation 12 by [H_{2}O]_{eq}^{2}
gives
[H_{2}O]_{eq}^{2}
^{.} K_{C} =
[H_{3}O^{+}]_{eq}[OH^{}]_{eq}
(13)
The quantity [H_{2}O]_{eq}^{2}
cancels out on the right hand side of equation 12, so it does not appear on the
right hand side of equation 13. Looking
at the left hand side of equation 13, we see two constants multiplied together:
the square of the concentration of water and the concentrationbased (K_{C})
equilibrium constant. Since a
constant multiplied by a constant is just another constant, we rename the
product [H_{2}O]_{eq}^{2} ^{.} K_{C} to
be K_{w}, the water ionization constant.
So
K_{w} =
[H_{2}O]_{eq}^{2} ^{.} K_{C}
(14)
and substituting this into equation 13 gives
K_{w} =
[H_{3}O^{+}]_{eq}[OH^{}]_{eq}
(15)
Equation 15 is true at all temperatures (K_{w} is
always equal to the product of the hydronium and hydroxide ion concentrations)
but the numerical value of this constant varies with temperature.
At 25 ^{o}C, the numerical value of K_{w} is 1.0 x 10^{14}.
In General Chemistry, we will almost always carry out our calculations at
25 ^{o}C, so we can use this value of K_{w}.
The very small value of K_{w} indicates that the equilibrium very
much favors the left hand side of equation 11.
In a sample of pure water, almost all the water exists as electrically
neutral water molecules, and only small amounts of H_{3}O^{+}
and OH^{} are present. Just
how small? We can find out by doing an ICE table calculation.
We imagine that we start with a sample of water in which absolutely none
of the water molecules are ionized. That
is, the initial concentrations of H_{3}O^{+} and OH^{}
are zero. The reaction shown in
equation 11 then occurs in the forward direction (to the right) to whatever
extent is necessary to reach equilibrium. The
H_{3}O^{+} and OH^{} concentrations increase by equal
amounts (due to their onetoone reaction stoichiometry) so they have equal
concentrations at equilibrium. The
equilibrium concentrations are equal to the amount of the increase, since the
starting concentrations were zero. The
ICE table looks like this:
2H_{2}O(l) <> H_{3}O^{+}(aq) + OH^{}(aq) 


[H_{2}O]_{ } 
[H_{3}O^{+}] 
[OH^{}] 
Initial 
 
0 
0 
Change 
 
+X 
+X 
Equilibrium 
 
X 
X 
From the ICE table, we see that
[H_{3}O^{+}]_{eq}
= X
and
[OH^{}]_{eq} =
X
(16)
Substituting this into equation 15, we have
K_{w} =
[H_{3}O^{+}]_{eq}[OH^{}]_{eq}
= (X) (X)
= X^{2}
(17)
We have seen that the numerical value of K_{w} at
25 ^{o}C is 1.0 x 10^{14}.
Therefore, at 25 ^{o}C, we have
X^{2} =
1.0 x 10^{14}
(18)
Which solved for X, gives
X =
1.0 x 10^{7} mol/L
(19)
Although we don't carry the units through the equilibrium
constant expression, we know that this is a concentrationbased equilibrium
constant, so the answer should come out in molar units.
We now know the ion concentrations in pure water at 25 ^{o}C:
[H_{3}O^{+}]_{eq}
= [OH^{}]_{eq}
= 1.0 x 10^{7}
mol/L
(20)
This concentration in floating point notation is 0.0000001
mol/L, which is indeed small. This
is what we expected based on the very small value of the equilibrium constant.
The water exists almost entirely as electrically neutral molecules, with
extremely small amounts of H_{3}O^{+} and OH^{} ions,
as predicted.
The ions H_{3}O^{+} and OH^{} will
always be present in any aqueous solution.
Since we associate H_{3}O^{+} ions with acids and OH^{}
ions with bases, there is sometimes a student misconception that "acidic
solutions don’t contain OH^{} ions" and "basic solutions
don't contain H_{3}O^{+} ions". Both of these assertions are false! Both of these ions are present in any aqueous
solution, but in an acidic solution, there will be more H_{3}O^{+}
than OH^{}, and in a basic solution, there will be more OH^{}
than H_{3}O^{+}. If
the two ion concentrations are equal, then the solution is neutral in an acid /
base sense. Pure water is neutral
in an acid / base sense, because as equation 11 shows, water – on its own –
can only form the H_{3}O^{+} and OH^{} ions in equal
molar amounts. It is only when
other substances are added to water that the concentration of one of these ions
can be elevated above the other.
The degree of acidity or basisity in a solution is
frequently expressed by a quantity known as pH. The pH scale is defined such that a value of 7 is neutral for
aqueous solutions at 25 ^{o}C. Values
lower than 7 are acidic, while values higher than 7 are basic.
The pH scale can be derived from the equilibrium constant expression for
water seen in equation 15.
K_{w} =
[H_{3}O^{+}]_{eq}[OH^{}]_{eq}
(15)
We start by taking the common logarithm of both sides of
equation 15:
log K_{w} =
log ( [H_{3}O^{+}]_{eq}[OH^{}]_{eq}
)
(21)
Next, we recognize that the log of a product is the sum of
the logs. With this in mind, we can
separate the [H_{3}O^{+}]_{eq} and [OH^{}]_{eq}
terms on the right hand side of equation 21.
log K_{w}
= log [H_{3}O^{+}]_{eq}
+ log [OH^{}]_{eq}
(22)
Now multiply both sides of equation 22 by 1.
log K_{w}
= log [H_{3}O^{+}]_{eq}
+ ( log [OH^{}]_{eq}
)
(23)
Now we need to learn some new notation.
A lowercase p in front of something will mean to take the negative log of
that something. That is, for any variable X,
pX =
log X
(24)
When equation 24 is substituted into equation 23, the
result is
pK_{w} =
pH + pOH
(25)
where
pH =
log [H_{3}O^{+}]_{eq}
and
pOH =
log [OH^{}]
(26)
The H in pH is reminiscent of H^{+} which you may
recall, is sometimes used in place of H_{3}O^{+}.
The relationship seen in equation 25 is valid at all
temperatures, but the numerical value of pK_{w} will change with
temperature. Recall that at 25 ^{o}C,
K_{w} = 1.0 x 10^{14}. This
means that
pK_{w} =
log (1.0 x 10^{14})
= 14.00
at 25 ^{o}C
(27)
We now have enough theory to begin doing some calculations.
So here are some example problems to show what calculations can be done
with acidic and basic solutions.
Example 1:
HCl is a strong acid.
Calculate the pH, pOH, [H_{3}O^{+}] and [OH^{}]
in a 0.010 M aqueous HCl solution at 25 ^{o}C.
Discussion
and Answers:
The
concentration given for an acidic (or basic) solution just refers to how the
solution was prepared. Some
or all of the acid (or base) will have ionized at equilibrium, so the actual
concentration of molecular acid (or base) may not be that which is given as the
"concentration" of the acid (or base).
In the case of HCl, since it is a strong acid, we regard it as 100%
ionized. This means we will assume
that there is no molecular HCl present in the solution.
All of it will have ionized. The
designation "0.010 M" simply means that enough HCl was added to the
water to provide 0.010 moles of HCl for every liter of solution.
We can construct the following "ICE" table to describe what
happens.
HCl(aq)
+ H_{2}O(l)
> H_{3}O^{+}(aq)
+ Cl^{}(aq) 


[HCl] 
[H_{2}O] 
[H_{3}O^{+}] 
[Cl^{}] 
Initial 
0.010 
 
~0 
0 
Change 
0.010 
 
+0.010 
+0.010 
Equilibrium 
0 
 
~0.010 
0.010 
Because
the HCl is a strong acid, we assume all of it ionizes.
We can add 0.010 moles of HCl to enough water to make a liter a solution
and imagine that just for a moment, all the HCl remains unionized.
At that moment, there will be no Cl^{} ions present, since the
only source of Cl^{} is the ionization of HCl.
Notice that the initial concentration of H_{3}O^{+} is
given as ~0 (approximately 0) rather than simply 0 (exactly zero).
That's because in an aqueous environment, there will always be at
least a trace amount of both H_{3}O^{+} and OH^{}
present. These ion concentrations
never go to zero in a aqueous environment.
The self ionization of water (water's equilibrium with itself) guarantees
that H_{3}O^{+} and OH^{} will always be present.
However, the amount of H_{3}O^{+} that will be generated
by the ionization of the HCl will be so much larger than that which is already
there that we can ignore the small contribution from the selfionization of
water. We can treat it as if the
HCl ionization is the only source of H_{3}O^{+}.
Because
of the onetoone stoichiometry, the ionization of 0.010 moles of HCl (per
liter) will produce 0.010 moles of H_{3}O^{+} (per liter) and
0.010 moles of Cl^{} (per liter).
We will assume that at the end of the reaction, no molecular HCl exists
in the solution.
Notice
the dashes in the [H_{2}O] column.
We do not need to worry about entering anything for H_{2}O
because its concentration is essentially constant.
That was the basis for omitting liquids and solids from equilibrium
constant expressions.
We now
have one of the quantities asked for in this problem.
Our "ICE" table shows that [H_{3}O^{+}]_{eq}
= 0.010 mol/L.
We can
then find the [OH^{}] from equation 15:
K_{w} =
[H_{3}O^{+}]_{eq}[OH^{}]_{eq}
(15)
We can
solve this for [OH^{}]_{eq} and use the known values of K_{W}
(1.0 x 10^{14}) and [H_{3}O^{+}]_{eq} to
calculate [OH^{}]_{eq}.
K_{w}
1.0 x 10^{14}
[OH^{}]_{eq} =  =
 = 1.0
x 10^{12} mol/L
[H_{3}O^{+}]_{eq}
0.010
The
equilibrium calculation is carried out without units, but the units are restored
in the final answer. This is the
usual procedure, as explained previously.
To get pH
and pOH, we can use equation 26:
pH =
log [H_{3}O^{+}]_{eq}
and
pOH =
log [OH^{}]
(26)
pH = log
(0.010)
pOH = log (1.0 x 10^{12})
pH = 2.00
pOH =
12.00
As a
check on the work, notice that pH and pOH add up to 14.00:
pH
+ pOH
= 2.00 +
12.00 =
14.00
Based on
equations 25 and 27:
pK_{w} =
pH + pOH
(25)
pK_{w} =
log (1.0 x 10^{14})
= 14.00
at 25 ^{o}C
(27)
the above
total is what we expect.
Example 2:
HC_{2}H_{3}O_{2} is a weak acid.
Calculate the pH, pOH, [H_{3}O^{+}] and [OH^{}]
in a 0.010 M aqueous HC_{2}H_{3}O_{2} solution at 25 ^{o}C.
Note that the ionization constant for this acid is 1.8 x 10^{5}.
Discussion
and Answers:
As
before, the designation "0.010 M" refers to how the acid solution was
prepared. Enough HC_{2}H_{3}O_{2}
was added to water to provide 0.010 moles of HC_{2}H_{3}O_{2}
for each liter of solution. At
equilibrium, some of the HC_{2}H_{3}O_{2} will have
ionized, so the molecular acid will not actually have a concentration of 0.010
M. In the previous example, there
was no molecular acid left, because it was a strong acid. But here, we have a weak acid, so the concentration of
molecular acid that remains in the solution at equilibrium will still be close
to its original 0.010 M value.
Like any
acid, HC_{2}H_{3}O_{2} will react with water,
transferring a proton (H^{+}) to water to form H_{3}O^{+}.
The acid anion that remains will be the acetate ion, C_{2}H_{3}O_{2}^{}.
The equilibrium reaction is
HC_{2}H_{3}O_{2}(aq)
+ H_{2}O(l)
<> H_{3}O^{+}(aq)
+ C_{2}H_{3}O_{2}^{}(aq)
If we
were to write an equilibrium constant expression for this reaction using the
format we first learned about, the result would be
[H_{3}O^{+}]_{eq}
[C_{2}H_{3}O_{2}^{}]_{eq}
K_{C} =

[HC_{2}H_{3}O_{2}]_{eq}
[H_{2}O]_{eq}
But we
now know that solids and liquids are omitted from equilibrium constant
expressions. As we have seen, the
justification is that their concentrations are essentially constant, so they
belong with the things that are constant, rather than the things that can vary.
If we multiply both sides of the above equilibrium constant expression by
[H_{2}O]_{eq} we get
[H_{3}O^{+}]_{eq} [C_{2}H_{3}O_{2}^{}]_{eq}
K_{C} ^{.} [H_{2}O]_{eq}
= 
[HC_{2}H_{3}O_{2}]_{eq}
Since a
constant multiplied by a constant is just another constant, we can replace K_{C}
^{.} [H_{2}O]_{eq} with K_{a}.
[H_{3}O^{+}]_{eq}
[C_{2}H_{3}O_{2}^{}]_{eq}
K_{a} = 
[HC_{2}H_{3}O_{2}]_{eq}
K_{a}
is known as the acid ionization constant. It
is the constant given in this problem to have a value of 1.8 x 10^{5}.
We are
now ready to set up an "ICE" table to describe this equilibrium
reaction. In the previous problem,
our ICE table had only numbers – no unknowns.
That's because we were dealing with a strong acid, so we knew the extent
of ionization would be 100%. But
now, we are dealing with a weak acid, so we do not know the extent of
ionization. Our ICE table will
include a variable (we will call it X) for which the value is not initially
known. We will use the equilibrium
constant expression to solve for this unknown X.
This will give us the value of [H_{3}O^{+}]_{eq}
and we can then obtain the other quantities we are interested in using the same
methods as before.
HC_{2}H_{3}O_{2}
+ H_{2}O(l)
<> H_{3}O^{+}(aq)
+ C_{2}H_{3}O_{2}^{}(aq) 


[HC_{2}H_{3}O_{2}] 
[H_{2}O] 
[H_{3}O^{+}] 
[C_{2}H_{3}O_{2}^{}] 
Initial 
0.010 
 
~0 
0 
Change 
X 
 
+X 
+X 
Equilibrium 
0.010X 
 
~X 
X 
Notice
that in the ICE table, we assume a reduction in the concentration of HC_{2}H_{3}O_{2}
by some unknown amount X. We enter
a –X as the change for HC_{2}H_{3}O_{2} to indicate
the loss of molecular acid due to its ionization.
This leaves a concentration of 0.010X in solution at equilibrium.
Since H_{2}O
is a liquid, we don't need to enter anything in the column for H_{2}O.
Solids and liquids are omitted from equilibrium constant expressions.
Notice
that the initial concentration of H_{3}O^{+} is entered as ~0
(approximately zero) rather than 0 (exactly zero). There will be a small amount of H_{3}O^{+} in
solution even before the acid ionizes.
Recall that aqueous solutions always contain H_{3}O^{+}
and OH^{} due to the selfionization of water.
However, the concentration of H_{3}O^{+} from the
selfionization of water is far smaller than that which will be provided by the
ionization of HC_{2}H_{3}O_{2}.
Therefore, we can regard all the equilibrium H_{3}O^{+}
as having come from the ionization of HC_{2}H_{3}O_{2}.
The C_{2}H_{3}O_{2}^{}
has an initial concentration of exactly zero, since there is no C_{2}H_{3}O_{2}^{}
before the HC_{2}H_{3}O_{2} ionizes.
Notice
that in the change row, we see –X, +X and +X.
This is obtained from stoichiometry.
For every mole of HC_{2}H_{3}O_{2} that ionizes,
we get a mole of H_{3}O^{+} and a mole of C_{2}H_{3}O_{2}^{}.
So if the concentration of HC_{2}H_{3}O_{2}
"decreases by X amount" (change of –X), then the concentrations of H_{3}O^{+}
and C_{2}H_{3}O_{2}^{} must both "increase
by X amount" (+X and +X).
The
bottom row of the ICE table gives the equilibrium concentrations; these must
satisfy the equilibrium constant expression:
[H_{3}O^{+}]_{eq}
[C_{2}H_{3}O_{2}^{}]_{eq}
(X) (X)
K_{a} = 
= 
[HC_{2}H_{3}O_{2}]_{eq}
(0.010X)
We know
the value for K_{a} is 1.8 x 10^{5} so we can solve for X:
(X) (X)
X^{2}
 =
 =
1.8 x 10^{5}
0.010X
0.010X
Because
the above equation contains both an X^{2} term and an X term, it must be
solved using the quadratic formula. In
many cases, it is possible to make an approximation that simplifies the equation
and avoids the use of the quadratic formula.
In what follows, the equation is first solved exactly using the quadratic
formula, and then we explore how the equation can be simplified.
It is beneficial to know how to work the problem "the long way
around", because you may occasionally encounter acid / base equilibrium
problems that can not be simplified.
We take
the equation
X^{2}
 =
1.8 x 10^{5}
0.010X
and
multiply both sides by 0.010X to get
X^{2}
= 1.8 x 10^{5}
(0.010X) =
1.8 x 10^{7} – (1.8 X 10^{5})X
and then
transpose the linear (Xcontaining) term and the constant to the left hand side
to get
X^{2}
+ (1.8 x 10^{5})X
 1.8 x 10^{7}
= 0
This has
the form of a standard quadratic equation:
aX^{2}
+ bX
+ c =
0
which has
the solution
b
± √(b^{2}4ac)
X =

2a
In our
equation, a = 1, b = 1.8 x 10^{5} and c = 1.8 x 10^{7}.
X must be
positive, because it is the equilibrium concentration of H_{3}O^{+},
and concentrations can never be negative. Since
b is positive, b must be negative, so it is clear that we must take the
positive value of the square root. Otherwise,
we will get a negative value for X, which is not allowed.
1.8 x 10^{5} +
√( (1.8 x 10^{5})^{2}4(1)(1.8 x 10^{7}) )
X =

2(1)
1.8 x 10^{5} +
√( 3.24 x 10^{10} + 7.2 x 10^{7} )
X =

2
1.8 x 10^{5} +
√( 7.20324 x 10^{7} )
X =

2
1.8 x 10^{5} +
8.487190348 x 10^{4}
X =

2
8.307190348 x 10^{4}
X =
 =
4.153595174 x 10^{4}
2
[H_{3}O^{+}]_{eq}
= X ≈ 4.2 x 10^{4} mol/L
Now that
[H_{3}O^{+}]_{eq} is known, the other quantities can be
calculated in the same way as before:
We can
find the [OH^{}] from equation 15:
K_{w} =
[H_{3}O^{+}]_{eq}[OH^{}]_{eq}
(15)
We can
solve this for [OH^{}]_{eq} and use the known values of K_{W}
(1.0 x 10^{14}) and [H_{3}O^{+}]_{eq} to
calculate [OH^{}]_{eq}.
K_{w}
1.0
x 10^{14}
[OH^{}]_{eq} =  = 
= 2.4 x 10^{11}
mol/L
[H_{3}O^{+}]_{eq}
4.2 x 10^{4}
The
equilibrium calculation is carried out without units, but the units are restored
in the final answer. This is the
usual procedure, as explained previously.
To get pH
and pOH, we can use equation 26:
pH = log [H_{3}O^{+}]_{eq} and pOH = log [OH^{}] (26)
pH = log
(4.2 x 10^{4}) pOH =
log (2.4 x 10^{11})
pH = 3.38
pOH = 10.62
As a
check on the work, notice that pH and pOH add up to 14.00:
pH
+ pOH
= 3.38 +
10.62 =
14.00
Based on
equations 25 and 27:
pK_{w} =
pH + pOH
(25)
pK_{w} =
log (1.0 x 10^{14})
= 14.00
at 25 ^{o}C
(27)
the above
total is what we expect.
We solved
this problem using the quadratic formula. However,
we could have made an approximation that would have greatly simplified the work.
Now that you have seen it "the long way around", let's explore
an approximation we can make and see how much easier it make things.
Recall
that the equilibrium reaction was
HC_{2}H_{3}O_{2}(aq)
+ H_{2}O(l)
<> H_{3}O^{+}(aq)
+ C_{2}H_{3}O_{2}^{}(aq)
for which
the acid ionization constant (K_{a}) had a value of
1.8 x 10^{5}.
After setting up the ICE table for this reaction, we found that the
equation we had to solve was
X^{2}
 =
1.8 x 10^{5}
0.010X
Since
this reaction has a fairly small equilibrium constant, we expect the left hand
side to be favored over the right hand side.
This means that the equilibrium concentrations of H_{3}O^{+}
and C_{2}H_{3}O_{2}^{} (which are equal to X)
are likely to be small in comparison to the concentration of HC_{2}H_{3}O_{2}
originally present. We propose that
X is much smaller than 0.010. That
is,
X
<< 0.010 (The
symbol << is read "much much less than")
If this
is true, then subtracting X from 0.010 does not change it very much.
When you subtract a very small quantity from a much larger quantity, the
larger quantity is not changed very much. So
we propose that
0.010 –
X ≈ 0.010
This
allows our original equation
X^{2}
 =
1.8 x 10^{5}
0.010X
to be
rewritten
X^{2}
 =
1.8 x 10^{5}
0.010
All we
have done is neglected the subtraction of X in the denominator, but that small
change has made a big difference in the ease of solution.
The equation no longer has both an X^{2} term and an X term.
It now has only the X^{2} term.
All we have to do is transpose the 0.010 to the right hand side of the
equation and take the square root of both sides!
X^{2}
= 0.010 (1.8 x 10^{5})
= 1.8 x 10^{7}
X
= √(1.8 x 10^{7})
= 4.242640687 x 10^{4}
≈ 4.2 x 10^{4}
Recall
that the "exact" value of X was 4.153595174 x 10^{4} which
rounded off to 2 significant figures, is 4.2 x 10^{4}, just like the
approximated value of X. Acid (and
base) ionization constants are usually only known to 2 significant figures, and
that's normally the best precision we can expect from our calculated X values.
The "exact" and approximate X's are identical within the
uncertainty of the data.
If you
make the approximation discussed here, it is always a good idea to check at the
end of the calculation to be sure the approximation was justified.
If you discover the approximation was not justified, you need to go back
and solve the problem again using the quadratic formula.
In this case, we know the approximation is justified because we
originally solved the problem using the quadratic formula and therefore, have
the "exact" value of X to compare with the approximate one.
But of course, if you're going to have to solve the quadratic formula to
confirm your approximation was valid, you might as well not make the
approximation in the first place. After
all, the purpose of the approximation was to avoid having to solve the
quadratic formula. So we need
another way to check the validity of our result.
Recall
that the approximation relied on the assumption that X was small in comparison
with 0.010. That is, that
0.010 –
X ≈
0.010
All we
have to do is substitute the approximate value of X into the above equation and
see if the assumption is valid. Our
approximate value of X was 4.2 x 10^{4}.
0.010 –
4.2 x 10^{4} =
9.58 x 10^{3} =
0.00958 ≈
0.010
That is,
rounded off to 3 decimal places (the precision of the original acid
concentration) the value is still 0.010.
In
general chemistry, you will almost always be able to make this approximation.
The smaller the acid (or base) ionization constant, the better it works.
The smaller the ionization constant is, the more the reaction favors the
left hand side (where the molecular acid or base is) and the less ionization
there is. It also works better when
the acid or base is more concentrated. With
a higher initial concentration of the molecular acid or base, the ionization
makes up a smaller percentage of the total amount of acid or base.
If you are working with a very dilute solution of an acid or base with an
ionization constant on the order of 10^{1} or 10^{2}, this
approximation may cause problems. You
will probably have to use the quadratic formula to get an accurate answer.
But the typical general chemistry exam or quiz problem will give you
acids (or bases) with ionization constants of 10^{5} or smaller, and
concentrations of 0.1 M or larger. These
should work fine with the approximation discussed here.
Example 3:
Hydroxylamine, NH_{2}OH, is a weak base with an
ionization constant (K_{b}) of 9.1 x 10^{‑9}.
Calculate the pH, pOH, [H_{3}O^{+}] and [OH^{}]
in a 0.100 M aqueous NH_{2}OH solution at 25 ^{o}C
Discussion
and Answers
A base
will react with water to remove an H^{+}, which combines with the
molecule of the base to form a cation. This
leaves the H_{2}O molecule as an OH^{} ion. You are already familiar with the basic properties of NH_{3}.
It extracts an H^{+} from H_{2}O to become NH_{4}^{+},
while leaving the H_{2}O as OH^{}.
NH_{3}(aq)
+ H_{2}O(l)
<> NH_{4}^{+}(aq)
+ OH^{}(aq)
You can
think of hydroxylamine as a derivative of NH_{3}, in which one of the
hydrogens is replaced with an OH group. Its
reaction in water is similar to that of NH_{3}.
The reaction is
NH_{2}OH(aq)
+ H_{2}O(l)
<> NH_{3}OH^{+}(aq)
+ OH^{}(aq)
If we
were to write an equilibrium constant expression for this reaction in the form
we first learned about, the result would be
[NH_{3}OH^{+}]_{eq}
[OH^{}]_{eq}
K_{C} =

[NH_{2}OH]_{eq} [H_{2}O]_{eq}
But we
now know that liquids should not appear in the equilibrium constant expression.
Since the concentration of water is constant, we multiply both sides of
the equation by [H_{2}O]_{eq} to get
[NH_{3}OH^{+}]_{eq}
[OH^{}]_{eq}
[H_{2}O]_{eq} ^{.} K_{C}
= 
[NH_{2}OH]_{eq}
Then,
since a constant multiplied by a constant is just another constant, we replace
[H_{2}O]_{eq} ^{.} K_{C} with K_{b}, the
base ionization constant. We then
have
[NH_{3}OH^{+}]_{eq}
[OH^{}]_{eq}
K_{b} =

[NH_{2}OH]_{eq}
We were
given a value of 9.1 x 10^{9} for K_{b} at 25 ^{o}C.
We can set up an ICE table for this reaction and substitute the
equilibrium values from the ICE table into the above equation.
Since this is a weak base, the extent of reaction will not be initially
known, and our ICE table will have an unknown variable X in it.
We will use the equilibrium constant expression and the known value of K_{b}
to solve for X.
NH_{2}OH(aq)
+ H_{2}O(l)
<> NH_{3}OH^{+}(aq)
+ OH^{}(aq) 


[NH_{2}OH] 
[H_{2}O] 
[NH_{3}OH^{+}] 
[OH^{}] 
Initial 
0.100 
 
0 
~0 
Change 
X 
 
+X 
+X 
Equilibrium 
0.100X 
 
X 
~X 
The
designation that this is "a 0.100 M NH_{2}OH solution" refers
to how the solution was prepared. Enough
NH_{2}OH was added to provide 0.100 moles of NH_{2}OH for each
liter of solution. At equilibrium,
some of the NH_{2}OH will have ionized, so in principle, the
concentration of molecular (unionized) NH_{2}OH in the solution at
equilibrium will be less than 0.100 M. However,
the small value of K_{b} tells us that this equilibrium will greatly
favor the left hand side over the right hand side, so we expect the extent of
ionization to be very small. We
will see that to 3 decimal places, the molecular base (NH_{2}OH)
concentration will still be 0.100 M.
The usual
principles were followed in filling out the ICE table.
The concentration of NH_{2}OH decreases by some small unknown
amount, so a –X is written for the change in NH_{2}OH concentration.
This leaves a concentration of 0.100X for NH_{2}OH at
equilibrium. Because of the
onetoone stoichiometry, the NH_{3}OH^{+} and OH^{}
must increase by the same amount that the NH_{2}OH decreased.
So if we write a –X for NH_{2}OH, we must write a +X for NH_{3}OH^{+}
and OH^{}. The initial
concentration of NH_{3}OH^{+} is exactly zero, because the only
source of NH_{3}OH^{+} is the ionization of NH_{2}OH.
There will be no NH_{3}OH^{+} before the reaction takes
place. However, there is already
some OH^{} in solution even before the reaction takes place.
Recall that all aqueous solutions contain both H_{3}O^{+}
and OH^{} from the
selfionization of water. The
concentrations of NH_{3}OH^{+} and OH^{} increase by X,
giving equilibrium concentrations of exactly X for NH_{3}OH^{+}
and approximately X for OH^{}. The
OH^{} concentration is not exactly X at equilibrium because it was not
exactly zero initially. However, we
will assume that the contribution to OH^{} from the NH_{2}OH
reaction sufficiently outweighs the contribution from the selfionization of
water that we can ignore the contribution from water.
Therefore, in our calculations, we treat the OH^{} concentration
as if it were exactly X.
We now
substitute the equilibrium values from the ICE table into the equilibrium
constant expression and set it equal to the numerical value of the base
ionization constant K_{b}.
[NH_{3}OH^{+}]_{eq}
[OH^{}]_{eq}
(X) (X)
K_{b} =
 =
 = 9.1 x 10^{9}
[NH_{2}OH]_{eq}
0.100X
Multiplying
the two X's on top, we have
X^{2}
 =
9.1 x 10^{9}
0.100X
The exact
solution of this equation involves the quadratic formula, but we should be able
to make the simplifying approximation discussed earlier to avoid such a
complicated calculation. As you may
recall, we successfully applied this approximation to an acid having a K_{a}
on the order of 10^{5} and a concentration of only 0.010 M.
Here, we have a base with a K_{b} on the order of 10^{9}
and a concentration of 0.100 M. Because
the extent of ionization will be smaller (smaller equilibrium constant) and the
initial concentration is higher, the ionization should be even less significant
here than it was in the pervious problem. The
approximation worked in the previous problem, so it should work even better
here.
The
assumption will be that X is very small in comparison to 0.100, that is, that
0.100 –
X ≈
0.100
With this
in mind, the equation
X^{2}
 =
9.1 x 10^{9}
0.100X
can be
rewritten as
X^{2}
 =
9.1 x 10^{9}
0.100
where the
subtraction of X in the denominator has been neglected.
As we have seen, this seemingly small change in the equation makes a big
difference in the ease of solution. The
original equation contained both an X^{2} term and an X term, and
therefore, required the quadratic formula to solve it.
The modified equation has only the X^{2} term.
All we have to do is multiply both sides by 0.100 and then take the
square root of both sides.
X^{2}
= (0.100) (9.1 x 10^{9})
= 9.1 x 10^{10}
X
= √(9.1 x 10^{10})
= 3.0 x 10^{5}
Now we
check that the assumption 0.100 – X ≈ 0.100 is valid:
0.100 –
3.0 x 10^{5} =
9.997 x 10^{2} =
0.09997 ≈
0.100
Expressed
to 3 decimal places (the precision to which the concentration was reported) the
number rounds off to 0.100. The
approximation is indeed valid.
Note that
in this problem, X is not [H_{3}O^{+}]_{eq} but [OH^{}]_{eq}.
This sometimes causes problems for students because they automatically
take X to be the same thing as [H_{3}O^{+}]_{eq}.
In any equilibrium calculation, it is important that you look carefully
at your ICE table to see what X represents.
So now we
know that [OH^{}]_{eq} = 3.0 x 10^{5} mol/L.
We can
use equation 15,
K_{w} =
[H_{3}O^{+}]_{eq}[OH^{}]_{eq}
(15)
to
calculate the [H_{3}O^{+}]_{eq}.
Solving
for [H_{3}O^{+}]_{eq}, we have
K_{w}
1.0 x 10^{14}
[H_{3}O^{+}]_{eq} =
 =

= 3.3 x 10^{10}
mol/L
[OH^{}]_{eq}
3.0 x 10^{5}
The
equilibrium calculation is carried out without units, but the units are restored
in the final answer. This is the
usual procedure, as explained previously.
To get pH
and pOH, we can use equation 26:
pH =
log [H_{3}O^{+}]_{eq}
and
pOH =
log [OH^{}]
(26)
pH = log
(3.3 x 10^{10}) pOH = log
(3.0 x 10^{5})
pH = 9.48
pOH = 4.5
As a
check on the work, notice that pH and pOH add up to 14.00:
pH
+ pOH
= 9.48 +
4.52 =
14.00
Based on
equations 25 and 27:
pK_{w} =
pH + pOH
(25)
pK_{w} =
log (1.0 x 10^{14})
= 14.00
at 25 ^{o}C
(27)
the above
total is what we expect.
Example
4:
Sodium
acetate, NaC_{2}H_{3}O_{2}, hydrolyzes in water to form
a solution that is not acid / base neutral.
Calculate the pH, pOH, [H_{3}O^{+}] and [OH^{}]
in a 0.250 M NaC_{2}H_{3}O_{2} solution.
Discussion
and Answers
Pure
sodium acetate is a an ionic solid at room temperature.
Like all water soluble ionic compounds, it dissociates when it dissolves
in water:
NaC_{2}H_{3}O_{2}(s)
> Na^{+}(aq) +
C_{2}H_{3}O_{2}^{}(aq)
Now we
have to consider what reaction, if any, these ion will have with water.
The Na^{+} can be thought of as the conjugate acid of the base
NaOH. The stronger an acid or base
is, the weaker will be its conjugate. Since
NaOH is a strong base (100% ionized in water), its conjugate acid will be too
weak to react. Therefore, we do not expect Na^{+} to react with
water:
Na^{+}
+ H_{2}O(l)
> N.R.
However,
the acetate ion, C_{2}H_{3}O_{2}^{}, is the
conjugate base of the acid acetic acid, HC_{2}H_{3}O_{2}.
Since HC_{2}H_{3}O_{2} is a weak acid, its
conjugate base has enough strength to react with water.
It reacts as a typical base, abstracting an H^{+} from H_{2}O
to leave OH^{}.
C_{2}H_{3}O_{2}^{}(aq)
+ H_{2}O(l)
<> HC_{2}H_{3}O_{2}(aq)
+ OH^{}(aq)
In order
to calculate the equilibrium concentrations for this reaction, we need its base
ionization constant, K_{b}. However,
if you look in a typical table of base ionization constants, you will probably
find that the K_{b} for C_{2}H_{3}O_{2}^{}
is not listed. It doesn’t need to
be, because you can get it from the K_{a} for its conjugate acid, HC_{2}H_{3}O_{2}
and the K_{w} for the ionization of water. The ionization reactions for HC_{2}H_{3}O_{2}
and H_{2}O are
HC_{2}H_{3}O_{2}(aq)
+ H_{2}O(l) <> H_{3}O^{+}(aq) + C_{2}H_{3}O_{2}^{}(aq)
[H_{3}O^{+}]_{eq}[C_{2}H_{3}O_{2}^{}]_{eq}
K_{a} =  =
1.8 x 10^{5
} [HC_{2}H_{3}O_{2}]_{eq}^{
}
2H_{2}O(l)
<> H_{3}O^{+}(aq)
+ OH^{}(aq)
K_{w} = [H_{3}O^{+}]_{eq}[OH^{}]_{eq}
If you reverse the acid ionization reaction and add the water ionization reaction, you get the reaction for the hydrolysis (reaction with water) of C_{2}H_{3}O_{2}^{} ion. Recall that when you reverse a reaction, you take the reciprocal of its equilibrium constant
H_{3}O^{+}(aq)
+ C_{2}H_{3}O_{2}^{}(aq)
<> HC_{2}H_{3}O_{2}(aq)
+ H_{2}O(l)
1
[HC_{2}H_{3}O_{2}]_{eq}
 =
 =
5.6 x 10^{4}
K_{a} [H_{3}O^{+}]_{eq}[C_{2}H_{3}O_{2}^{}]_{eq}
2H_{2}O(l)
<> H_{3}O^{+}(aq)
+ OH^{}(aq)
K_{w} = [H_{3}O^{+}]_{eq}[[OH^{}]_{eq}
= 1.0 x 10^{14}
When the
two reactions are added the H_{2}O(l) on the right hand side of the
first equation cancels with one of the 2 H_{2}O(l) on the left hand side
of the second equation, leaving a single H_{2}O on the left hand side in
the total. The H_{3}O^{+}
appears on the left hand side of the first reaction and on the right hand side
of the second reaction, so it completely cancels out. Finally, recall that when reactions are added, their
equilibrium constants are multiplied. With
all these considerations, you come up with
C_{2}H_{3}O_{2}^{}(aq)
+ H_{2}O(l)
<> HC_{2}H_{3}O_{2}(aq)
+ OH^{}(aq)
1
 ^{.} K_{w}
= 5.6 x 10^{4} (1.0
x 10^{14}) = 5.6 x 10^{10}
= K_{b}
K_{a}
From
this, you see that
K_{b} =  ^{.} K_{w}
K_{a}
Multiplying
both sides of this by K_{a} you get
K_{a}
^{.} K_{b} =
K_{w}
a more
conveniently remembered form of equation for the relationship between the K_{a}
of an acid and the K_{b} of
its conjugate base (or
K_{b} of a base and K_{a} of its conjugate acid). The above analysis is just to show you where this equation
comes from. You don’t actually
have to carry out all of this chemical arithmetic each time you deal with a salt
hydrolysis problem. You just
remember the equation
K_{a} ^{.} K_{b} = K_{w}
and
calculate whichever equilibrium constant you don’t know from the one you do
know. From the work above, you can
now set up the ICE table and write an equation to be solved for X.
C_{2}H_{3}O_{2}^{}(aq)
+H_{2}O(l <>HC_{2}H_{3}O_{2}(aq)
+OH^{}(aq) 


C_{2}H_{3}O_{2}^{}

H_{2}O 
HC_{2}H_{3}O_{2}

OH^{}

Initial 
0.250 
 
0 
~0 
Change 
X 
 
+X 
+X 
Equilibrium 
0.250X 
=== 
X 
~X 
The 0.250
in the above ICE table comes from the molar concentration given for the salt.
The salt’s formula, NaC_{2}H_{3}O_{2}, shows
that one mole of the salt will produce one mole of Na^{+} ions and one
mole of C_{2}H_{3}O_{2}^{} ions.
So if there are 0.250 moles of salt for every liter of solution, then
when the salt dissociates, there will be 0.250 moles of Na^{+} and 0.250
moles of C_{2}H_{3}O_{2}^{} for every liter of
solution. The K_{b} for
this reaction has been found to be 5.6 x 10^{10}, so you can now write
the following equation:
[HC_{2}H_{3}O_{2}]_{eq}[OH^{}]_{eq}
(X)(X)
X^{2}
 =
 =
 =
5.6 x 10^{10}
[C_{2}H_{3}O_{2}^{}]_{eq}
0.250X 0.250X
Because
of the small value of K_{b}, it seems reasonable that the extent of
ionization will be small. It should
be safe to assume that X << 0.250, in which case 0.250X ≈ 0.250. With this assumption, the equation becomes
X^{2}
 =
5.6 x 10^{10}
0.250
Multiplying
both sides by 0.250 gives
X^{2}
= 1.4 x 10^{10}
and then
taking the square root of both sides,
X
= 1.2 x 10^{5}
From the
ICE table, you see that X is the [OH^{}]_{eq}. Therefore,
[OH^{}]_{eq}
= 1.2 x 10^{5}
We can
use equation 15,
K_{w} =
[H_{3}O^{+}]_{eq}[OH^{}]_{eq}
(15)
to
calculate the [H_{3}O^{+}]_{eq}.
Solving for [H_{3}O^{+}]_{eq}, we have
K_{w}
1.0
x 10^{14}
[H_{3}O^{+}]_{eq} =
 =
 =
8.3 x 10^{10} mol/L
[OH^{}]_{eq}
1.2 x 10^{5}
The
equilibrium calculation is carried out without units, but the units are restored
in the final answer. This is the
usual procedure, as explained previously.
To get pH
and pOH, we can use equation 26:
pH =
log [H_{3}O^{+}]_{eq}
and
pOH =
log [OH^{}]
(26)
pH = log
(8.3 x 10^{10}) pOH = log
(1.2 x 10^{5})
pH = 9.08
pOH = 4.92
As a
check on the work, notice that pH and pOH add up to 14.00:
pH
+ pOH
= 9.08 +
4.92 =
14.00
Based on
equations 25 and 27:
pK_{w} =
pH + pOH
(25)
pK_{w} =
log (1.0 x 10^{14})
= 14.00
at 25 ^{o}C
(27)
the above
total is what we expect.