3.5: Polyprotic Acids

Learning Objectives

• Extend previously introduced equilibrium concepts to acids and bases that may donate or accept more than one proton

We can classify acids by the number of protons per molecule that they can give up in a reaction. Acids such as \(\ce{HCl}\), \(\ce{HNO3}\), and \(\ce{HCN}\) that contain one ionizable hydrogen atom in each molecule are called monoprotic acids. Their reactions with water are:

\[\ce{HCl}(aq)+\ce{H2O}(l)⟶\ce{H3O+}(aq)+\ce{Cl-}(aq)\]

\[\ce{HNO3}(aq)+\ce{H2O}(l)⟶\ce{H3O+}(aq)+\ce{NO3-}(aq)\]

\[\ce{HCN}(aq)+\ce{H2O}(l)⟶\ce{H3O+}(aq)+\ce{CN-}(aq)\]

Even though it contains four hydrogen atoms, acetic acid, \(\ce{CH3CO2H}\), is also monoprotic because only the hydrogen atom from the carboxyl group (\(\ce{-COOH}\)) reacts with bases:

Similarly, monoprotic bases are bases that will accept a single proton.

Diprotic Acids

Diprotic acids contain two ionizable hydrogen atoms per molecule; ionization of such acids occurs in two steps. The first ionization always takes place to a greater extent than the second ionization. For example, sulfuric acid, a strong acid, ionizes as follows:

• The first ionization is

\[ \ce{H2SO4}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HSO4-}(aq)\]

with \(K_{\ce a1} > 10^2;\: {complete\: dissociation}\).

• The second ionization is

\[ \ce{HSO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{SO4^{2−}}(aq)\]

with \( K_{\ce a2}=1.2×10^{−2}\).

This stepwise ionization process occurs for all polyprotic acids. When we make a solution of a weak diprotic acid, we get a solution that contains a mixture of acids. Carbonic acid, \(\ce{H2CO3}\), is an example of a weak diprotic acid. The first ionization of carbonic acid yields hydronium ions and bicarbonate ions in small amounts.

• First Ionization

\[\ce{H2CO3}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO3-}(aq)\]

with

\[K_{\ce{H2CO3}}=\ce{\dfrac{[H3O+][HCO3- ]}{[H2CO3]}}=4.3×10^{−7}\]

The bicarbonate ion can also act as an acid. It ionizes and forms hydronium ions and carbonate ions in even smaller quantities.

• Second Ionization

\[\ce{HCO3-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CO3^2-}(aq)\]

with

\[ K_{\ce{HCO3-}}=\ce{\dfrac{[H3O+][CO3^2- ]}{[HCO3- ]}}=4.7×10^{−11}\]

\(K_{\ce{H2CO3}}\) is larger than \(K_{\ce{HCO3-}}\) by a factor of 10⁴, so H₂CO₃ is the dominant producer of hydronium ion in the solution. This means that little of the \(\ce{HCO3-}\) formed by the ionization of H₂CO₃ ionizes to give hydronium ions (and carbonate ions), and the concentrations of H₃O⁺ and \(\ce{HCO3-}\) are practically equal in a pure aqueous solution of H₂CO₃.

If the first ionization constant of a weak diprotic acid is larger than the second by a factor of at least 20, it is appropriate to treat the first ionization separately and calculate concentrations resulting from it before calculating concentrations of species resulting from subsequent ionization. This can simplify our work considerably because we can determine the concentration of H₃O⁺ and the conjugate base from the first ionization, then determine the concentration of the conjugate base of the second ionization in a solution with concentrations determined by the first ionization.

Example \(\PageIndex{1}\): Ionization of a Diprotic Acid

When we buy soda water (carbonated water), we are buying a solution of carbon dioxide in water. The solution is acidic because CO₂ reacts with water to form carbonic acid, H₂CO₃. What are \(\ce{[H3O+]}\), \(\ce{[HCO3- ]}\), and \(\ce{[CO3^2- ]}\) in a saturated solution of CO₂ with an initial [H₂CO₃] = 0.033 M?

\[\ce{H2CO3}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO3-}(aq) \hspace{20px} K_{\ce a1}=4.3×10^{−7} \label{step1} \tag{equilibrium step 1}\]

\[\ce{HCO3-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CO3^2-}(aq) \hspace{20px} K_{\ce a2}=4.7×10^{−11} \label{step2} \tag{equilibrium step 2}\]

Solution

As indicated by the ionization constants, H₂CO₃ is a much stronger acid than \(\ce{HCO3-}\), so \(\ce{H2CO3}\) is the dominant producer of hydronium ion in solution. Thus there are two parts in the solution of this problem:

1. Using the customary four steps, we determine the concentration of H₃O⁺ and \(\ce{HCO3-}\) produced by ionization of H₂CO₃.
2. Then we determine the concentration of \(\ce{CO3^2-}\) in a solution with the concentration of H₃O⁺ and \(\ce{HCO3-}\) determined in (1).

To summarize:

1. First Ionization: Determine the concentrations of \(\ce{H3O+}\) and \(\ce{HCO3-}\).

Since \ref{step1} is has a much bigger \(K_{a1}=4.3×10^{−7}\) than \(K_{a2}=4.7×10^{−11}\) for \ref{step2}, we can safely ignore the second ionization step and focus only on the first step (but address it in next part of problem).

\[\ce{H2CO3}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO3-}(aq) \hspace{20px} K_{\ce a1}=4.3×10^{−7} \nonumber\]

As for the ionization of any other weak acid:

An abbreviated table of changes and concentrations shows:

ICE Table	\(\ce{H2CO3}(aq)\)	\(\ce{H2O}(l)\)	\( \ce{H3O+}(aq) \)	\(\ce{HCO3-}(aq)\)
Initial (M)	\(0.033 \:M\)	-	\(0\)	\(0\)
Change (M)	\(- x\)	-	\(+x\)	\(+x\)
Equilibrium (M)	\(0.033 \:M - x\)	-	\(x\)	\( x\)

Substituting the equilibrium concentrations into the equilibrium constant gives us:

\[K_{\ce{H2CO3}}=\ce{\dfrac{[H3O+][HCO3- ]}{[H2CO3]}}=\dfrac{(x)(x)}{0.033−x}=4.3×10^{−7} \nonumber\]

Solving the preceding equation making our standard assumptions gives:

\[x=1.2×10^{−4} \nonumber\]

Thus:

\[\ce{[H2CO3]}=0.033\:M \nonumber\]

\[\ce{[H3O+]}=\ce{[HCO3- ]}=1.2×10^{−4}\:M \nonumber\]

2. Second Ionization: Determine the concentration of \(CO_3^{2-}\) in a solution at equilibrium.

Since the \ref{step1} is has a much bigger \(K_a\) than \ref{step2}, we can the equilibrium conditions calculated from first part of example as the initial conditions for an ICER Table for the \ref{step2}:

\[\ce{HCO3-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CO3^2-}(aq) \nonumber\]

ICE Table	\(\ce{HCO3-}(aq)\)	\(\ce{H2O}(l)\)	\( \ce{H3O+}(aq) \)	\(\ce{CO3^2-}(aq)\)
Initial (M)	\(1.2×10^{−4}\:M\)	-	\(1.2×10^{−4}\:M\)	\(0\)
Change (M)	\(- y\)	-	\(+y\)	\(+y\)
Equilibrium (M)	\(1.2×10^{−4}\:M - y\)	-	\(1.2×10^{−4}\:M + y\)	\( y\)

\[ \begin{align*} K_{\ce{HCO3-}}&=\ce{\dfrac{[H3O+][CO3^2- ]}{[HCO3- ]}} \\[4pt] &=\dfrac{(1.2×10^{−4}\:M + y) (y)}{(1.2×10^{−4}\:M - y)} \end{align*}\]

To avoid solving a quadratic equation, we can assume \(y \ll 1.2×10^{−4}\:M \) so

\[K_{\ce{HCO3-}} = 4.7×10^{−11} \approx \dfrac{(1.2×10^{−4}\:M ) (y)}{(1.2×10^{−4}\:M)} \nonumber\]

Rearranging to solve for \(y\)

\[y \approx \dfrac{ (4.7×10^{−11})(1.2×10^{−4}\:M )}{ 1.2×10^{−4}\:M} \nonumber\]

\[[\ce{CO3^2-}]=y \approx 4.7×10^{−11} \nonumber\]

To summarize:

In part 1 of this example, we found that the \(\ce{H2CO3}\) in a 0.033-M solution ionizes slightly and at equilibrium \([\ce{H2CO3}] = 0.033\, M\), \([\ce{H3O^{+}}] = 1.2 × 10^{−4}\), and \(\ce{[HCO3- ]}=1.2×10^{−4}\:M\). In part 2, we determined that \(\ce{[CO3^2- ]}=5.6×10^{−11}\:M\).

Exercise \(\PageIndex{2}\): Hydrogen Sulfide

The concentration of \(H_2S\) in a saturated aqueous solution at room temperature is approximately 0.1 M. Calculate \(\ce{[H3O+]}\), \(\ce{[HS^{−}]}\), and \(\ce{[S^{2−}]}\) in the solution:

\[\ce{H2S}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HS-}(aq) \hspace{20px} K_{\ce a1}=8.9×10^{−8} \nonumber\]

\[\ce{HS-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{S^2-}(aq) \hspace{20px} K_{\ce a2}=1.0×10^{−19} \nonumber\]

Answer

\([\ce{H2S}] = 0.1 M\), \(\ce{[H3O+]} = [HS^{−}] = 0.0001\, M\), \([S^{2−}] = 1 × 10^{−19}\, M\)

We note that the concentration of the sulfide ion is the same as K_a2. This is due to the fact that each subsequent dissociation occurs to a lesser degree (as acid gets weaker).

Triprotic Acids

A triprotic acid is an acid that has three dissociable protons that undergo stepwise ionization: Phosphoric acid is a typical example:

• The first ionization is

\[\ce{H3PO4}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{H2PO4-}(aq) \]

with \(K_{\ce a1}=7.5×10^{−3} \).

• The second ionization is

\[\ce{H2PO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HPO4^2-}(aq)\]

with \( K_{\ce a2}=6.2×10^{−8} \).

• The third ionization is

\[\ce{HPO4^2-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{PO4^3-}(aq) \]

with \( K_{\ce a3}=4.2×10^{−13} \).

As with the diprotic acids, the differences in the ionization constants of these reactions tell us that in each successive step the degree of ionization is significantly weaker. This is a general characteristic of polyprotic acids and successive ionization constants often differ by a factor of about 10⁵ to 10⁶. This set of three dissociation reactions may appear to make calculations of equilibrium concentrations in a solution of H₃PO₄ complicated. However, because the successive ionization constants differ by a factor of 10⁵ to 10⁶, the calculations can be broken down into a series of parts similar to those for diprotic acids.

Polyprotic bases can accept more than one hydrogen ion in solution. The carbonate ion is an example of a diprotic base, since it can accept up to two protons. Solutions of alkali metal carbonates are quite alkaline, due to the reactions:

\[\ce{H2O}(l)+\ce{CO3^2-}(aq)⇌\ce{HCO3-}(aq)+\ce{OH-}(aq) \]

and

\[\ce{H2O}(l)+\ce{HCO3-}(aq)⇌\ce{H2CO3}(aq)+\ce{OH-}(aq)\]

Summary

An acid that contains more than one ionizable proton is a polyprotic acid. The protons of these acids ionize in steps. The differences in the acid ionization constants for the successive ionizations of the protons in a polyprotic acid usually vary by roughly five orders of magnitude. As long as the difference between the successive values of K_a of the acid is greater than about a factor of 20, it is appropriate to break down the calculations of the concentrations of the ions in solution into a series of steps.

Glossary

diprotic acid

acid containing two ionizable hydrogen atoms per molecule. A diprotic acid ionizes in two steps

diprotic base

base capable of accepting two protons. The protons are accepted in two steps

monoprotic acid

acid containing one ionizable hydrogen atom per molecule

stepwise ionization

process in which an acid is ionized by losing protons sequentially

triprotic acid

acid that contains three ionizable hydrogen atoms per molecule; ionization of triprotic acids occurs in three steps

Contributors and Attributions

• Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/85abf193-2bd...a7ac8df6@9.110).