7.16: 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 ioniza tion process occurs for all polyprotic acids, meaning that one H + is lost at the time. This process contrasts with the ionization of strong bases such as Ca(OH) 2 in which all the OH- ions are lost in a single dissociation reaction:
Ca(OH) 2 (aq) → Ca 2+ (aq) + 2 OH - (aq)
This is because hydroxides are ionic compounds, while acids are covalent compounds.
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 10 4 , so H 2 CO 3 is the dominant producer of hydronium ion in the solution.
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 CO 2 reacts with water to form carbonic acid, H 2 CO 3 . What are \(\ce{[H3O+]}\) , \(\ce{[HCO3- ]}\), and \(\ce{[CO3^2- ]}\) in a saturated solution of CO 2 with an initial [H 2 CO 3 ] = 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, H 2 CO 3 is a much stronger acid than \(\ce{HCO3-}\), so \(\ce{H2CO3}\) is the dominant producer of hydronium ion in solution.
- Using the customary four steps, we determine the concentration of H 3 O + and \(\ce{HCO3-}\) produced by ionization of H 2 CO 3 .
1. First Ionization: Determine the concentrations of \(\ce{H3O+}\) and \(\ce{HCO3-}\).
An abbreviated table of changes and concentrations shows:
| ICE Table | \(\ce{H2CO3}(aq)\) | \(\ce{H2O}(l)\) | \( \ce{H3O+}(aq) \) | \(\ce{HCO3-}(aq)\) |
|---|---|---|---|---|
| I nitial (M) | \(0.033 \:M\) | - | \(0\) | \(0\) |
| C hange (M) | \(- x\) | - | \(+x\) | \(+x\) |
| E quilibrium (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:
\[\ce{[H2CO3]}=0.033\:M \nonumber\]
\[\ce{[H3O+]}=\ce{[HCO3- ]}=1.2×10^{−4}\:M \nonumber\]
pH=-log[H 3 O + ]=3.92
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 10 5 to 10 6 . Again, using the first dissociation constant is a good approximation to calculate the pH for a solution of phosphoric acid.
Contributors and Attributions
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 Ka of the acid is significant, it is a good estimation to only consider the first dissociation constant to calculate pH
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 ).