# Acid / Base Equilibria

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, H3O+ 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”.

H2O
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)   +   H2O(l)   ---------->   H3O+(aq)                                                     (3)

If we add equations 2 and 3, the H+ cancels out and we get

HCl(aq)   +   H2O(l)   ---------->   H3O+(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 H3O+.

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 one-way 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.

H2O
HC2H3O2(l)   ---------->   HC2H3O2(aq)                                                        (5)
acetic acid
solution

The acetic acid ionizes in solution, but the ions can recombine to form the molecular acid again.

HC2H3O2(aq)   +   H2O(l)   <---------->   H3O+(aq)   +   C2H3O2-(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.

H2O
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 (NH3) provides an example.  Pure NH3 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”.

H2O
NH3(g)   ---------->   NH3(aq)                                                                       (8)

NH3(aq)   +   H2O(l)   <---------->   NH4+(aq)   +   OH-(aq)                         (9

Notice the bidirectional arrow in equation 9.  This indicates that the reaction is reversible.  When the NH3 ionizes, the ions it generates are capable of recombining to form the molecular NH3 again.  This means that not all of the NH3 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 (NH4+) 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 NH3 and H2O 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

H2O(l)   <---------->   H+(aq)   +   OH-(aq)                                                      (10)

but recognizing that H+ will combine with an H2O molecule to form H3O+

H+(aq)   +   H2O(l)   ---------->   H3O+(aq)                                                       (3)

we can add equations 10 and 3 to get

2H2O(l)   <---------->   H3O+(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 bi-directional arrow) it should have an equilibrium constant expression.  If we were to write a concentration-based (KC) equilibrium constant expression as we learned to do in our first introduction to chemical equilibrium, it would be

[H3O+]eq[OH-]eq
KC   =   --------------------                                                                               (12)
[H2O]eq2

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 [H2O]eq is constant, then so is [H2O]eq2.  Multiplying both sides of equation 12 by [H2O]eq2 gives

[H2O]eq2  .  KC   =   [H3O+]eq[OH-]eq                                                                (13)

The quantity [H2O]eq2 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 concentration-based (KC) equilibrium constant.  Since a constant multiplied by a constant is just another constant, we rename the product [H2O]eq2 . KC to be Kw, the water ionization constant.  So

Kw   =   [H2O]eq2 . KC                                                                                        (14)

and substituting this into equation 13 gives

Kw   =   [H3O+]eq[OH-]eq                                                                                 (15)

Equation 15 is true at all temperatures (Kw 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 oC, the numerical value of Kw is 1.0 x 10-14.  In General Chemistry, we will almost always carry out our calculations at 25 oC, so we can use this value of Kw.  The very small value of Kw 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 H3O+ 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 H3O+ 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 H3O+ and OH- concentrations increase by equal amounts (due to their one-to-one 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:

 2H2O(l)   <---------->   H3O+(aq)   +   OH-(aq) [H2O] [H3O+] [OH-] Initial --- 0 0 Change --- +X +X Equilibrium --- X X

From the ICE table, we see that

[H3O+]eq  =  X              and                   [OH-]eq  =  X                                       (16)

Substituting this into equation 15, we have

Kw  =  [H3O+]eq[OH-]eq  =  (X) (X)  =  X2                                                       (17)

We have seen that the numerical value of Kw at 25 oC is 1.0 x 10-14.  Therefore, at 25 oC, we have

X2   =   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 concentration-based equilibrium constant, so the answer should come out in molar units.  We now know the ion concentrations in pure water at 25 oC:

[H3O+]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 H3O+ and OH- ions, as predicted.

The ions H3O+ and OH- will always be present in any aqueous solution.  Since we associate H3O+ 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 H3O+ 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 H3O+ than OH-, and in a basic solution, there will be more OH- than H3O+.  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 H3O+ 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 oC.  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.

Kw   =   [H3O+]eq[OH-]eq                                                                                 (15)

We start by taking the common logarithm of both sides of equation 15:

log Kw  =   log (  [H3O+]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 [H3O+eq and [OH-]eq terms on the right hand side of equation 21.

log Kw   =   log [H3O+]eq   +   log [OH-]eq                                                        (22)

Now multiply both sides of equation 22 by -1.

-log Kw   =   -log [H3O+]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

pKw   =   pH   +   pOH                                                                                     (25)

where

pH   =   -log [H3O+]eq          and           pOH   =   -log [OH-]                           (26)

The H in pH is reminiscent of H+ which you may recall, is sometimes used in place of H3O+.

The relationship seen in equation 25 is valid at all temperatures, but the numerical value of pKw will change with temperature.  Recall that at 25 oC, Kw = 1.0 x 10-14.  This means that

pKw   =   -log (1.0 x 10-14)   =   14.00   at 25 oC                                            (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, [H3O+] and [OH-] in a 0.010 M aqueous HCl solution at 25 oC.

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)  +  H2O(l)  ----->  H3O+(aq)  +  Cl-(aq) [HCl] [H2O] [H3O+] [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 H3O+ 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 H3O+ 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 H3O+ and OH- will always be present.  However, the amount of H3O+ 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 self-ionization of water.  We can treat it as if the HCl ionization is the only source of H3O+.

Because of the one-to-one stoichiometry, the ionization of 0.010 moles of HCl (per liter) will produce 0.010 moles of H3O+ (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 [H2O] column.  We do not need to worry about entering anything for H2O 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 [H3O+]eq = 0.010 mol/L.

We can then find the [OH-] from equation 15:

Kw   =   [H3O+]eq[OH-]eq                                                                                 (15)

We can solve this for [OH-]eq and use the known values of KW (1.0 x 10-14) and [H3O+]eq to calculate [OH-]eq.

Kw          1.0 x 10-14
[OH-]eq =   ------   =   -----------   =   1.0 x 10-12 mol/L
[H3O+]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 [H3O+]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:

pKw   =   pH   +   pOH                                                                                 (25)

pKw   =   -log (1.0 x 10-14)   =   14.00   at 25 oC                                          (27)

the above total is what we expect.

Example 2:

HC2H3O2 is a weak acid.  Calculate the pH, pOH, [H3O+] and [OH-] in a 0.010 M aqueous HC2H3O2 solution at 25 oC.  Note that the ionization constant for this acid is 1.8 x 10-5.

As before, the designation "0.010 M" refers to how the acid solution was prepared.  Enough HC2H3O2 was added to water to provide 0.010 moles of HC2H3O2 for each liter of solution.  At equilibrium, some of the HC2H3O2 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, HC2H3O2 will react with water, transferring a proton (H+) to water to form H3O+.  The acid anion that remains will be the acetate ion, C2H3O2-.  The equilibrium reaction is

HC2H3O2(aq)  +  H2O(l)  <----->  H3O+(aq)  +  C2H3O2-(aq)

If we were to write an equilibrium constant expression for this reaction using the format we first learned about, the result would be

[H3O+]eq [C2H3O2-]eq
KC  =  --------------------
[HC2H3O2]eq [H2O]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 [H2O]eq we get

[H3O+]eq [C2H3O2-]eq
KC . [H2O]eq  =    -------------------
[HC2H3O2]eq

Since a constant multiplied by a constant is just another constant, we can replace KC . [H2O]eq with Ka.

[H3O+]eq [C2H3O2-]eq
Ka =  -------------------
[HC2H3O2]eq

Ka 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 [H3O+]eq and we can then obtain the other quantities we are interested in using the same methods as before.

 HC2H3O2  +  H2O(l)  <----->  H3O+(aq)  +  C2H3O2-(aq) [HC2H3O2] [H2O] [H3O+] [C2H3O2-] Initial 0.010 --- ~0 0 Change -X --- +X +X Equilibrium 0.010-X --- ~X X

Notice that in the ICE table, we assume a reduction in the concentration of HC2H3O2 by some unknown amount X.  We enter a –X as the change for HC2H3O2 to indicate the loss of molecular acid due to its ionization.  This leaves a concentration of 0.010-X in solution at equilibrium.

Since H2O is a liquid, we don't need to enter anything in the column for H2O.  Solids and liquids are omitted from equilibrium constant expressions.

Notice that the initial concentration of H3O+ is entered as ~0 (approximately zero) rather than 0 (exactly zero).  There will be a small amount of H3O+ in solution even before the acid ionizes.  Recall that aqueous solutions always contain H3O+ and OH- due to the self-ionization of water.  However, the concentration of H3O+ from the self-ionization of water is far smaller than that which will be provided by the ionization of HC2H3O2.  Therefore, we can regard all the equilibrium H3O+ as having come from the ionization of HC2H3O2.

The C2H3O2- has an initial concentration of exactly zero, since there is no C2H3O2- before the HC2H3O2 ionizes.

Notice that in the change row, we see –X, +X and +X.  This is obtained from stoichiometry.  For every mole of HC2H3O2 that ionizes, we get a mole of H3O+ and a mole of C2H3O2-.  So if the concentration of HC2H3O2 "decreases by X amount" (change of –X), then the concentrations of H3O+ and C2H3O2- 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:

[H3O+]eq [C2H3O2-]eq        (X) (X)
Ka =  -------------------  =  -------------
[HC2H3O2]eq           (0.010-X)

We know the value for Ka is 1.8 x 10-5 so we can solve for X:

(X) (X)            X2
--------   =   ---------   =   1.8 x 10-5
0.010-X        0.010-X

Because the above equation contains both an X2 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

X2
--------   =   1.8 x 10-5
0.010-X

and multiply both sides by 0.010-X to get

X2  =  1.8 x 10-5 (0.010-X)  =  1.8 x 10-7 – (1.8 X 10-5)X

and then transpose the linear (X-containing) term and the constant to the left hand side to get

X2  +  (1.8 x 10-5)X  -  1.8 x 10-7  =  0

This has the form of a standard quadratic equation:

aX2  +  bX  +  c  =  0

which has the solution

-b  ± √(b2-4ac)
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 H3O+, 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

[H3O+]eq = X ≈ 4.2 x 10-4 mol/L

Now that [H3O+]eq is known, the other quantities can be calculated in the same way as before:

We can find the [OH-] from equation 15:

Kw   =   [H3O+]eq[OH-]eq                                                                                 (15)

We can solve this for [OH-]eq and use the known values of KW (1.0 x 10-14) and [H3O+]eq to calculate [OH-]eq.

Kw          1.0 x 10-14
[OH-]eq =   ------   =   -----------   =   2.4 x 10-11 mol/L
[H3O+]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 [H3O+]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:

pKw   =   pH   +   pOH                                                                                 (25)

pKw   =   -log (1.0 x 10-14)   =   14.00   at 25 oC                                           (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

HC2H3O2(aq)  +  H2O(l)  <----->  H3O+(aq)  +  C2H3O2-(aq)

for which the acid ionization constant (Ka) 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

X2
--------   =   1.8 x 10-5
0.010-X

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 H3O+ and C2H3O2- (which are equal to X) are likely to be small in comparison to the concentration of HC2H3O2 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

X2
--------   =   1.8 x 10-5
0.010-X

to be rewritten

X2
------   =   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 X2 term and an X term.  It now has only the X2 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!

X2  =  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, NH2OH, is a weak base with an ionization constant (Kb) of 9.1 x 10‑9.  Calculate the pH, pOH, [H3O+] and [OH-] in a 0.100 M aqueous NH2OH solution at 25 oC

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 H2O molecule as an OH- ion.  You are already familiar with the basic properties of NH3.  It extracts an H+ from H2O to become NH4+, while leaving the H2O as OH-.

NH3(aq)  +  H2O(l)  <----->  NH4+(aq)  +  OH-(aq)

You can think of hydroxylamine as a derivative of NH3, in which one of the hydrogens is replaced with an OH group.  Its reaction in water is similar to that of NH3.  The reaction is

NH2OH(aq)  +  H2O(l)  <----->  NH3OH+(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

[NH3OH+]eq [OH-]eq
KC  =  -----------------
[NH2OH]eq [H2O]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 [H2O]eq to get

[NH3OH+]eq [OH-]eq
[H2O]eq . KC  =  ------------------
[NH2OH]eq

Then, since a constant multiplied by a constant is just another constant, we replace [H2O]eq . KC with Kb, the base ionization constant.  We then have

[NH3OH+]eq [OH-]eq
Kb  =  ------------------
[NH2OH]eq

We were given a value of 9.1 x 10-9 for Kb at 25 oC.  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 Kb to solve for X.

 NH2OH(aq)  +  H2O(l)  <----->  NH3OH+(aq)  +  OH-(aq) [NH2OH] [H2O] [NH3OH+] [OH-] Initial 0.100 --- 0 ~0 Change -X --- +X +X Equilibrium 0.100-X --- X ~X

The designation that this is "a 0.100 M NH2OH solution" refers to how the solution was prepared.  Enough NH2OH was added to provide 0.100 moles of NH2OH for each liter of solution.  At equilibrium, some of the NH2OH will have ionized, so in principle, the concentration of molecular (unionized) NH2OH in the solution at equilibrium will be less than 0.100 M.  However, the small value of Kb 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 (NH2OH) concentration will still be 0.100 M.

The usual principles were followed in filling out the ICE table.  The concentration of NH2OH decreases by some small unknown amount, so a –X is written for the change in NH2OH concentration.  This leaves a concentration of 0.100-X for NH2OH at equilibrium.  Because of the one-to-one stoichiometry, the NH3OH+ and OH- must increase by the same amount that the NH2OH decreased.  So if we write a –X for NH2OH, we must write a +X for NH3OH+ and OH-.  The initial concentration of NH3OH+ is exactly zero, because the only source of NH3OH+ is the ionization of NH2OH.  There will be no NH3OH+ 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 H3O+ and OH- from the self-ionization of water.  The concentrations of NH3OH+ and OH- increase by X, giving equilibrium concentrations of exactly X for NH3OH+ 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 NH2OH reaction sufficiently outweighs the contribution from the self-ionization 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 Kb.

[NH3OH+]eq [OH-]eq       (X) (X)
Kb  =  ------------------  =  ------------  =  9.1 x 10-9
[NH2OH]eq           0.100-X

Multiplying the two X's on top, we have

X2
--------  =  9.1 x 10-9
0.100-X

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 Ka on the order of 10-5 and a concentration of only 0.010 M.  Here, we have a base with a Kb 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

X2
--------  =  9.1 x 10-9
0.100-X

can be rewritten as

X2
--------  =  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 X2 term and an X term, and therefore, required the quadratic formula to solve it.  The modified equation has only the X2 term.  All we have to do is multiply both sides by 0.100 and then take the square root of both sides.

X2  =  (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 [H3O+]eq but [OH-]eq.  This sometimes causes problems for students because they automatically take X to be the same thing as [H3O+]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,

Kw   =   [H3O+]eq[OH-]eq                                                                                 (15)

to calculate the [H3O+]eq.

Solving for [H3O+]eq, we have

Kw           1.0 x 10-14
[H3O+]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 [H3O+]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:

pKw   =   pH   +   pOH                                                                                    (25)

pKw   =   -log (1.0 x 10-14)   =   14.00   at 25 oC                                            (27)

the above total is what we expect.

Example 4:

Sodium acetate, NaC2H3O2, hydrolyzes in water to form a solution that is not acid / base neutral.  Calculate the pH, pOH, [H3O+] and [OH-] in a 0.250 M NaC2H3O2 solution.

Pure sodium acetate is a an ionic solid at room temperature.  Like all water soluble ionic compounds, it dissociates when it dissolves in water:

NaC2H3O2(s)  ---------->  Na+(aq)  +  C2H3O2-(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+  +  H2O(l)  ---------->  N.R.

However, the acetate ion, C2H3O2-, is the conjugate base of the acid acetic acid, HC2H3O2.  Since HC2H3O2 is a weak acid, its conjugate base has enough strength to react with water.  It reacts as a typical base, abstracting an H+ from H2O to leave OH-.

C2H3O2-(aq)  +  H2O(l)  <---------->  HC2H3O2(aq)  +  OH-(aq)

In order to calculate the equilibrium concentrations for this reaction, we need its base ionization constant, Kb.  However, if you look in a typical table of base ionization constants, you will probably find that the Kb for C2H3O2- is not listed.  It doesn’t need to be, because you can get it from the Ka for its conjugate acid, HC2H3O2 and the Kw for the ionization of water.  The ionization reactions for HC2H3O2 and H2O are

HC2H3O2(aq) + H2O(l) <-----> H3O+(aq) + C2H3O2-(aq)

[H3O+]eq[C2H3O2-]eq
Ka =   ----------------  =  1.8 x 10-5
[HC2H3O2]eq

2H2O(l)  <----->  H3O+(aq)  +  OH-(aq)

Kw  =  [H3O+]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 C2H3O2- ion.  Recall that when you reverse a reaction, you take the reciprocal of its equilibrium constant

H3O+(aq)  +  C2H3O2-(aq)  <----->  HC2H3O2(aq)  +  H2O(l)

1         [HC2H3O2]eq
----  =  -----------------  =  5.6 x 104
Ka      [H3O+]eq[C2H3O2-]eq

2H2O(l)  <----->  H3O+(aq)  +  OH-(aq)

Kw  =  [H3O+]eq[[OH-]eq  =  1.0 x 10-14

When the two reactions are added the H2O(l) on the right hand side of the first equation cancels with one of the 2 H2O(l) on the left hand side of the second equation, leaving a single H2O on the left hand side in the total.  The H3O+ 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

C2H3O2-(aq)  +  H2O(l)  <----->  HC2H3O2(aq)  +  OH-(aq)

1
--- . Kw  =  5.6 x 104 (1.0 x 10-14)  =  5.6 x 10-10  =  Kb
Ka

From this, you see that

1
Kb  = --- . Kw
Ka

Multiplying both sides of this by Ka you get

Ka . Kb  =  Kw

a more conveniently remembered form of equation for the relationship between the Ka of an acid and the Kb of its conjugate base (or Kb of a base and Ka 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

Ka . Kb = Kw

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.

 C2H3O2-(aq) +H2O(l <----->HC2H3O2(aq) +OH-(aq) C2H3O2- H2O HC2H3O2 OH- Initial 0.250 --- 0 ~0 Change -X --- +X +X Equilibrium 0.250-X === X ~X

The 0.250 in the above ICE table comes from the molar concentration given for the salt.  The salt’s formula, NaC2HO2, shows that one mole of the salt will produce one mole of Na+ ions and one mole of C2H3O2- 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 C2H3O2- for every liter of solution.  The Kb for this reaction has been found to be 5.6 x 10-10, so you can now write the following equation:

[HC2H3O2]eq[OH-]eq    (X)(X)           X2
----------------  =  --------  =  ---------  =  5.6 x 10-10
[C2H3O2-]eq        0.250-X       0.250-X

Because of the small value of Kb, it seems reasonable that the extent of ionization will be small.  It should be safe to assume that X << 0.250, in which case 0.250-X ≈ 0.250.  With this assumption, the equation becomes

X2
------  =  5.6 x 10-10
0.250

Multiplying both sides by 0.250 gives

X2  =  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,

Kw   =   [H3O+]eq[OH-]eq                                                                                 (15)

to calculate the [H3O+]eq.  Solving for [H3O+]eq, we have

Kw          1.0 x 10-14
[H3O+]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 [H3O+]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:

pKw   =   pH   +   pOH                                                                                    (25)

pKw   =   -log (1.0 x 10-14)   =   14.00   at 25 oC                                            (27)

the above total is what we expect.