Diprotic and Triprotic Acids and Bases

The acid equilibrium problems discussed so far have focused on a family of compounds
known as **monoprotic acids**. Each of these acids has a single H^{+} ion, or
proton, it can donate when it acts as a Brnsted acid. Hydrochloric acid (HCl), acetic
acid (CH_{3}CO_{2}H or HOAc), nitric acid (HNO_{3}), and benzoic
acid (C_{6}H_{5}CO_{2}H) are all monoprotic acids.

Several important acids can be classified as **polyprotic acids**, which can lose
more than one H^{+} ion when they act as Brnsted acids. **Diprotic acids**,
such as sulfuric acid (H_{2}SO_{4}), carbonic acid (H_{2}CO_{3}),
hydrogen sulfide (H_{2}S), chromic acid (H_{2}CrO_{4}), and oxalic
acid (H_{2}C_{2}O_{4}) have two acidic hydrogen atoms. **Triprotic
acids**, such as phosphoric acid (H_{3}PO_{4}) and citric acid (C_{6}H_{8}O_{7}),
have three.

There is usually a large difference in the ease with which these acids lose the first
and second (or second and third) protons. When sulfuric acid is classified as a strong
acid, students often assume that it loses both of its protons when it reacts with water.
That isn't a legitimate assumption. Sulfuric acid is a strong acid because *K _{a}*
for the loss of the first proton is much larger than 1. We therefore assume that
essentially all the H

_{2}SO

_{4}molecules in an aqueous solution lose the first proton to form the HSO

_{4}

^{-}, or hydrogen sulfate, ion.

H_{2}SO_{4}(aq) + H_{2}O(l) H_{3}O^{+}(aq) + HSO_{4}^{-}(aq) |
K = 1 x 10_{a1}^{3} |

But *K _{a}* for the loss of the second proton is only 10

^{-2}and only 10% of the H

_{2}SO

_{4}molecules in a 1

*M*solution lose a second proton.

HSO_{4}^{-}(aq) + H_{2}O(l) H_{3}O^{+}(aq)
+ SO_{4}^{2-}(aq) |
K = 1.2 x 10_{a2}^{-2} |

H_{2}SO_{4} only loses both H^{+} ions when it reacts with a
base, such as ammonia.

The table below gives values of *K _{a}* for some common polyprotic acids.
The large difference between the values of

*K*for the sequential loss of protons by a polyprotic acid is important because it means we can assume that these acids dissociate one step at a timean assumption known as

_{a}**stepwise dissociation**.

*Acid-Dissociation Equilibrium Constants for Common Polyprotic Acids*

Acid |
K_{a1} |
K_{a2} |
K_{a3} |
|||

sulfuric acid (H_{2}SO_{4}) |
1.0 x 10^{3} |
1.2 x 10^{-2} |
||||

chromic acid (H_{2}CrO_{4}) |
9.6 | 3.2 x 10^{-7} |
||||

oxalic acid (H_{2}C_{2}O_{4}) |
5.4 x 10^{-2} |
5.4 x 10^{-5} |
||||

sulfurous acid (H_{2}SO_{3}) |
1.7 x 10^{-2} |
6.4 x 10^{-8} |
||||

phosphoric acid (H_{3}PO_{4}) |
7.1 x 10^{-3} |
6.3 x 10^{-8} |
4.2 x 10^{-13} |
|||

glycine (C_{2}H_{6}NO_{2}) |
4.5 x 10^{-3} |
2.5 x 10^{-10} |
||||

citric acid (C_{6}H_{8}O_{7}) |
7.5 x 10^{-4} |
1.7 x 10^{-5} |
4.0 x 10^{-7} |
|||

carbonic acid (H_{2}CO_{3}) |
4.5 x 10^{-7} |
4.7 x 10^{-11} |
||||

hydrogen sulfide (H_{2}S) |
1.0 x 10^{-7} |
1.3 x 10^{-13} |

Let's look at the consequence of the assumption that polyprotic acids lose protons one
step at a time by examining the chemistry of a saturated solution of H_{2}S in
water.

Hydrogen sulfide is the foul-smelling gas that gives rotten eggs their unpleasant odor.
It is an excellent source of the S^{2-} ion, however, and is therefore commonly
used in introductory chemistry laboratories. H_{2}S is a weak acid that
dissociates in steps. Some of the H_{2}S molecules lose a proton in the first step
to form the HS^{-}, or hydrogen sulfide, ion.

First step: | H_{2}S(aq) + H_{2}O(l)
H_{3}O^{+}(aq) + HS^{-}(aq) |

A small fraction of the HS^{-} ions formed in this reaction then go on to lose
another H^{+} ion in a second step.

Second step: | HS^{-}(aq) + H_{2}O(l)
H_{3}O^{+}(aq) + S^{2-}(aq) |

Since there are two steps in this reaction, we can write two equilibrium constant expressions.

Although each of these equations contains three terms, there are only four unknowns[H_{3}O^{+}],
[H_{2}S], [HS^{-}], and [S^{2-}] because
the [H_{3}O^{+}] and [HS^{-}] terms appear in both equations. The
[H_{3}O^{+}] term represents the total H_{3}O^{+} ion
concentration from both steps and therefore must have the same value in both equations.
Similarly, the [HS^{-}] term, which represents the balance between the HS^{-}
ions formed in the first step and the HS^{-} ions consumed in the second step,
must have the same value for both equations.

Four equations are needed to solve for four unknowns. We already have two equations:
the *K _{a1}* and

*K*expressions. We are going to have to find either two more equations or a pair of assumptions that can generate two equations. We can base one assumption on the fact that the value of

_{a2}*K*for this acid is almost a million times larger than the value of

_{a1}*K*.

_{a2}*K _{a1}* >>

*K*

_{a2}This means that only a small fraction of the HS^{-} ions formed in the first
step go on to dissociate in the second step. If this is true, most of the H_{3}O^{+}
ions in this solution come from the dissociation of H_{2}S, and most of the HS^{-}
ions formed in this reaction PSS remain in solution. As a result, we can assume that the H_{3}O^{+}
and HS^{-} ion concentrations are more or less equal.

First assumption: | [H_{3}O^{+}][HS^{-}] |

We need one more equation, and therefore one more assumption. Note that H_{2}S
is a weak acid (*K _{a1}* = 1.0 x 10

^{-7},

*K*= 1.3 x 10

_{a2}^{-13}). Thus, we can assume that most of the H

_{2}S that dissolves in water will still be present when the solution reaches equilibrium. In other words, we can assume that the equilibrium concentration of H

_{2}S is approximately equal to the initial concentration.

Second assumption: | [H_{2}S]
C_{H2S} |

We now have four equations in four unknowns.

[H_{3}O^{+}] [HS^{-}]

[H_{2}S]
*C*_{H2S}

Since there is always a unique solution to four equations in four unknowns, we are now
ready to calculate the H_{3}O^{+}, H_{2}S, HS^{-}, and S^{2-}
concentrations at equilibrium in a saturated solution of H_{2}S in water. All we
need to know is that a saturated solution of H_{2}S in water has an initial
concentration of about 0.10 *M*.

Because *K _{a1}* is so much larger than

*K*for this acid, we can work with the equilibrium expression for the first step without worrying about the second step for the moment. We therefore start with the expression for

_{a2}*K*for this acid.

_{a1}We then invoke one of our assumptions.

[H_{2}S] *C*_{H2S}
0.10 *M*

Substituting this approximation into the *K _{a1}* expression gives the
following equation.

We then invoke the other assumption.

[H_{3}O^{+}]
[HS^{-}]*C*

Substituting this approximation into the *K _{a1}* expression gives the
following result.

We now solve this approximate equation for *C.*

*C*
1.0 x 10^{-4}

If our two assumptions are valid, we are three-fourths of the way to our goal. We know
the H_{2}S, H_{3}O^{+}, and HS^{-} concentrations.

[H_{2}S]
0.10 *M*

[H_{3}O^{+}]
[HS^{-}]
1.0 x 10^{-4} *M*

Having extracted the values of three unknowns from the first equilibrium expression, we turn to the second equilibrium expression.

Substituting the known values of the H_{3}O^{+} and HS^{-} ion
concentrations into this expression gives the following equation.

Because the equilibrium concentrations of the H_{3}O^{+} and HS^{-}
ions are more or less the same, the S^{2-} ion concentration at equilibrium is
approximately equal to the value of *K _{a2}* for this acid.

[S^{2-}]
1.3 x 10^{-13} *M*

It is now time to check our assumptions. Is the dissociation of H_{2}S small
compared with the initial concentration? Yes. The HS^{-} and H_{3}O^{+}
ion concentrations obtained from this calculation are 1.0 x 10^{-4} *M*,
which is 0.1% of the initial concentration of H_{2}S. The following assumption is
therefore valid.

[H_{2}S]
*C*_{H2S}
0.10 *M*

Is the difference between the S^{2-} and HS^{-} ion concentrations
large enough to allow us to assume that essentially all of the H_{3}O^{+}
ions at equilibrium are formed in the first step and that essentially all of the HS^{-}
ions formed in this step remain in solution? Yes. The S^{2-} ion concentration
obtained from this calculation is 10^{9} times smaller than the HS^{-} ion
concentration. Thus, our other assumption is also valid.

[H_{3}O^{+}]
[HS^{-}]

We can therefore summarize the concentrations of the various components of this equilibrium as follows.

[H_{2}S]
0.10 *M*

[H_{3}O^{+}]
[HS^{-}]
1.0 x 10^{-4} *M*

[S^{2-}]
1.3 x 10^{-13} *M*

The techniques we have used with diprotic acids can be extended to diprotic bases. The
only challenge is calculating the values of *K _{b}* for the base.

Example: Let's calculate the H_{2}CO_{3}, HCO_{3}^{-},
CO_{3}^{2-}, and OH^{-} concentrations at equilibrium in a
solution that is initially 0.10 *M* in Na_{2}CO_{3}. (H_{2}CO_{3}:
*K _{a1}* = 4.5 x 10

^{-7};

*K*= 4.7 x 10

_{a2}^{-11})

Because it is a salt, sodium carbonate dissociates into its ions when it dissolves in water.

H_{2}O |
|||||

Na_{2}CO_{3}(aq) |
2 Na^{+}(aq) |
+ | CO_{3}^{2-}(aq) |

The carbonate ion then acts as a base toward water, picking up a pair of protons (one
at a time) to form the bicarbonate ion, HCO_{3}^{-} ion, and then
eventually carbonic acid, H_{2}CO_{3}.

CO_{3}^{2-}(aq) + H_{2}O(l) HCO_{3}^{-}(aq)
+ OH^{-}(aq) |
K = ?_{b1} |
||

HCO_{3}^{-}(aq) + H_{2}O(l) H_{2}CO_{3}(aq)
+ OH^{-}(aq) |
K = ?_{b2} |

The first step in solving this problem involves determining the values of *K _{b1}*
and

*K*for the carbonate ion. We start by comparing the

_{b2 }*K*expressions for the carbonate ion with the

_{b}*K*expressions for carbonic acid.

_{a}The expressions for *K _{b1}* and

*K*have something in commonthey both depend on the concentrations of the HCO

_{a2}_{3}

^{-}and CO

_{3}

^{2-}ions. The expressions for

*K*and

_{b2}*K*also have something in commonthey both depend on the HCO

_{a1}_{3}

^{-}and H

_{2}CO

_{3}concentrations. We can therefore calculate

*K*from

_{b1}*K*and

_{a2}*K*from

_{b2}*K*.

_{a1}We start by multiplying the top and bottom of the *K _{a1}* expression by
the OH

^{-}ion concentration to introduce the [OH

^{-}] term.

We then group terms in this equation as follows.

The first term in this equation is the inverse of the *K _{b2}* expression,
and the second term is the

*K*expression.

_{w}Rearranging this equation gives the following result.

*K _{a1}K_{b2}* =

*K*

_{w}Similarly, we can multiply the top and bottom of the *K _{a2}* expression
by the OH

^{-}ion concentration.

Collecting terms gives the following equation.

The first term in this equation is the inverse of *K _{b1}*, and the second
term is

*K*.

_{w}This equation can therefore be rearranged as follows.

*K _{a2}K_{b1}* =

*K*

_{w}We can now calculate the values of *K _{b1}* and

*K*for the carbonate ion.

_{b2}We are finally ready to do the calculations. We start with the *K _{b1}*
expression because the CO

_{3}

^{2-}ion is the strongest base in this solution and therefore the best source of the OH

^{-}ion.

The difference between *K _{b1}* and

*K*for the carbonate ion is large enough to suggest that most of the OH

_{b2}^{-}ions come from this step and most of the HCO

_{3}

^{-}ions formed in this reaction remain in solution.

[OH^{-}]
[HCO_{3}^{-}]
*C*

The value of *K _{b1}* is small enough to assume that

*C*is small compared with the initial concentration of the carbonate ion. If this is true, the concentration of the CO

_{3}

^{2-}ion at equilibrium will be roughly equal to the initial concentration of Na

_{2}CO

_{3}.

[CO_{3}^{2-}] *C*_{Na2CO3}

Substituting this information into the *K _{b1}* expression gives the
following result.

This approximate equation can now be solved for *C*.

*C
*
0.0046 *M*

We then use this value of *C*
to calculate the equilibrium concentrations of the OH^{-}, HCO_{3}^{-},
and CO_{3}^{2-} ions.

[CO_{3}^{2-}] = 0.10 - *C*
0.095
*M*

[OH

^{-}] [HCO_{3}^{-}]C0.0046M

We now turn to the *K _{b2}* expression.

Substituting what we know about the OH^{-} and HCO_{3}^{-} ion
concentrations into this equation gives the following result.

According to this equation, the H_{2}CO_{3} concentration at
equilibrium is approximately equal to *K _{b2}* for the carbonate ion.

[H_{2}CO_{3}]
2.2 x 10^{-8} *M*

Summarizing the results of our calculations allows us to test the assumptions made generating these results.

[CO

_{3}^{2-}] 0.095M

[OH^{-}]
[HCO_{3}^{-}]
4.6 x 10^{-3} *M*

[H_{2}CO_{3}] 2.2
x 10^{-8} *M*

All of our assumptions are valid. The extent of the reaction between the CO_{3}^{2-}
ion and water to give the HCO_{3}^{-} ion is less than 5% of the initial
concentration of Na_{2}CO_{3}. Furthermore, most of the OH^{-} ion
comes from the first step, and most of the HCO_{3}^{-} ion formed in this
step remains in solution.

Our techniques for working diprotic acid or diprotic base equilibrium problems can be
applied to triprotic acids and bases as well. To illustrate this, let's calculate the H_{3}O^{+},
H_{3}PO_{4}, H_{2}PO_{4}^{-}, HPO_{4}^{2-},
and PO_{4}^{3-} concentrations at equilibrium in a 0.10 *M* H_{3}PO_{4}
solution, for which *K _{a1}* = 7.1 x 10

^{-3},

*K*= 6.3 x 10

_{a2}^{-8}, and

*K*= 4.2 x 10

_{a3}^{-13}.

Let's assume that this acid dissociates by steps and analyze the first stepthe most extensive reaction.

We now assume that the difference between *K _{a1}* and

*K*is large enough that most of the H

_{a2}_{3}O

^{+}ions come from this first step and most of the H

_{2}PO

_{4}

^{-}ions formed in this step remain in solution.

[H_{3}O^{+}]
[H_{2}PO_{4}^{-}]*C*

Substituting this assumption into the *K _{a1}* expression gives the
following equation.

The assumption that is small compared with the initial concentration of the acid fails in this problem. But we don't really need this assumption because we can use the quadratic formula or successive approximations to solve the equation. Either way, we obtain the same answer.

*C*
= 0.023 *M*

We can then use this value of *C*
to obtain the following information.

[H_{3}PO_{4}]
0.10 - *C*
0.077 *M*

[H_{3}O^{+}]
[H_{2}PO_{4}^{-}]
0.023 *M*

We now turn to the second strongest acid in this solution.

Substituting what we know about the H_{3}O^{+} and H_{2}PO_{4}^{-}
ion concentrations into this expression gives the following equation.

If our assumptions so far are correct, the HPO_{4}^{2-} ion
concentration in this solution is equal to *K _{a2}*.

[HPO_{4}^{2-}]
6.3 x 10^{-8}

We have only one more equation, the equilibrium expression for the weakest acid in the solution.

Substituting what we know about the concentrations of the H_{3}O^{+}
and HPO_{4}^{2-} ions into this expression gives the following equation.

This equation can be solved for the phosphate ion concentration at equilibrium.

[PO_{4}^{3-}]1.2
x 10^{-18} *M*

Summarizing the results of the calculations helps us check the assumptions made along the way.

[H_{3}PO_{4}]
0.077 *M*

[H_{3}O^{+}]
[H_{2}PO_{4}^{-}]
0.023 *M*

[HPO_{4}^{2-}]
6.3 x 10^{-8} *M*

[PO_{4}^{3-}]
1.2 x 10^{-18} *M*

The only approximation used in working this problem was the assumption that the acid
dissociates one step at a time. Is the difference between the concentrations of the H_{2}PO_{4}^{-}
and HPO_{4}^{2-} ions large enough to justify the assumption that
essentially all of the H_{3}O^{+} ions come from the first step? Yes. Is
it large enough to justify the assumption that essentially all of the H_{2}PO_{4}^{-}
formed in the first step remains in solution? Yes.

You may never encounter an example of a polyprotic acid for which
the difference between successive values of *K _{a}* are too small to allow us
to assume stepwise dissociation. This assumption works even when we might expect it to
fail.

Practice Problem 7:Calculate the H ^{-4}, K = 1.7 x 10_{a2}^{-5}, K
= 4.0 x 10_{a3}^{-7}) |