6.2: Polyprotic Acids
- Extend previously introduced equilibrium concepts to acids and bases that may donate or accept more than one proton
- Predict whether amphoteric species will act as acids or bases in solution and calculate the pH of a solution
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. 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. This means that little of the \(\ce{HCO3-}\) formed by the ionization of H 2 CO 3 ionizes to give hydronium ions (and carbonate ions), and the concentrations of H 3 O + and \(\ce{HCO3-}\) are practically equal in a pure aqueous solution of H 2 CO 3 .
The sequential Ka values for a polyprotic acid are labeled K a1 , K a2 , K a3 , etc., and refer respectively to the acid ionization constant for the proton that ionizes first, the proton that ionizes next, and so on.
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 H 3 O + 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.
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. Thus there are two parts in the solution of this problem:
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).
\[\ce{H2CO3}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO3-}(aq) \hspace{20px} K_{\ce a1}=4.3×10^{−7} \nonumber\]
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:
\[\ce{[H2CO3]}=0.033\:M \nonumber\]
\[\ce{[H3O+]}=\ce{[HCO3- ]}=1.2×10^{−4}\:M \nonumber\]
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 | \(\ce{HCO3-}(aq)\) | \(\ce{H2O}(l)\) | \( \ce{H3O+}(aq) \) | \(\ce{CO3^2-}(aq)\) |
|---|---|---|---|---|
| I nitial (M) | \(1.2×10^{−4}\:M\) | - | \(1.2×10^{−4}\:M\) | \(0\) |
| C hange (M) | \(- y\) | - | \(+y\) | \(+y\) |
| E quilibrium (M) | \(1.2×10^{−4}\:M - y\) | - | \(1.2×10^{−4}\:M + y\) | \( y\) |
\[ \begin{align*} K_{\ce{HCO3-}}&=\ce{\dfrac{[H3O+][CO3^2- ]}{[HCO3- ]}} \\[4pt] &=\dfrac{(1.2×10^{−4}\:M + y) (y)}{(1.2×10^{−4}\:M - y)} \end{align*}\]
We can assume \(y \ll 1.2×10^{−4}\:M \) so
\[K_{\ce{HCO3-}} = 4.7×10^{−11} \approx \dfrac{(1.2×10^{−4}\:M ) (y)}{(1.2×10^{−4}\:M)} \nonumber\]
\[[\ce{CO3^2-}]=y \approx 4.7×10^{−11} \nonumber\]
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 10 5 to 10 6 . This set of three dissociation reactions may appear to make calculations of equilibrium concentrations in a solution of H 3 PO 4 complicated. However, because the successive ionization constants differ by a factor of 10 5 to 10 6 , 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
\[\ce{H2O}(l)+\ce{HCO3-}(aq)⇌\ce{H2CO3}(aq)+\ce{OH-}(aq)\]
Polyprotic Acids Contain Amphoteric Species
Because polyprotic acids contain more than one ionizable proton, the conjugate bases of polyprotic acids are amphoteric species - species that can either accept or donate a proton. When an amphoteric species exists in solution, it has the ability to act as either an acid or a base. For example, bicarbonate participates in the following equilibria:
\[ \ce{HCO3^{-}(aq)} + \ce{H2O(l)} \rightleftharpoons \ce{H3O^+(aq)} + \ce{CO3^{2-}(aq)} \]
\[ \ce{HCO3^{-}(aq)} + \ce{H2O(l)} \rightleftharpoons \ce{H2CO3(aq)} + \ce{OH^{-}(aq)} \:\:\:K_{b}=2.3\times10^{-8} \]
Which of these equilibrium reactions dominates the pH of the solution depends on the relative magnitudes of the K a and K b values.
Typically we are only given the K a values for the ionizable protons of the polyprotic acid. We can use the K a for the ionization of the proton in question to determine the K b for the base reaction.
The acid equilibrium reaction here represents the ionization of the second proton from \(\ce{H2CO3} \), thus the acid ionization constant is K a2 .
\[ \ce{HCO3^{-}(aq)} + \ce{H2O(l)} \rightleftharpoons \ce{H3O^+(aq)} + \ce{CO3^{2-}(aq)} \:\:\:K_{a2}=4.7\times10^{-11} \]
The base equilibrium reaction here represents re-association of the first proton from \(\ce{H2CO3} \). To calculate the K b for this reaction, we will use K a1 :
\[ \ce{H2CO3(aq)} + \ce{H2O(l)} \rightleftharpoons \ce{H3O^+(aq)} + \ce{HCO3^{-}(aq)} \:\:\:K_{a1}=4.3\times10^{-7} \]
\[ K_{b} = \dfrac{K_w}{K_{a1}} = \dfrac{1\times10^{-14}}{4.3\times10^{-7}} = 2.3\times10^{-8} \]
In this case, K b (\(2.3\times10^{-8} \)) is much larger than K a (\(4.7\times10^{-11} \)) so the bicarbonate will act as a base in solution.
What is the pH of a solution that contains 0.20 M \(\ce{NaHSO3} \)?
Solution
In this solution, the Na+ is pH neutral. The \(\ce{HSO3^{-}} \) can act as either an acid or a base. The acid equilibrium reaction and Ka value are:
\[ \ce{HSO3^{-}(aq)} + \ce{H2O(l)} \rightleftharpoons \ce{H3O^+(aq)} + \ce{SO3^{2-}(aq)} \:\:\:K_{a2}=6.4\times10^{-8} \]
The base equilibrium reaction is:
\[ \ce{HSO3^{-}(aq)} + \ce{H2O(l)} \rightleftharpoons \ce{H2SO3(aq)} + \ce{OH^{-}(aq)} \]
To determine the K b , we can use the K a for sulfurous acid, \(K_{a1} = 1.7\times10^{-2} \).
\[K_{b} = \dfrac{K_w}{K_{a1}} = \dfrac{1\times10^{-14}}{1.7\times10^{-2}} = 5.9\times10^{-13} \]
The acid equilibrium reaction will dominate the pH of the solution.
| \(\ce{HSO3^{-}(aq)} +\) | \(\ce{H2O(l)} \rightleftharpoons\) | \(\ce{H3O^+(aq)} +\) | \(\ce{SO3^{2-} (aq)}\) | |
| Initial | 0.20 M | excess | ~ 0 | 0 |
| Change | -x | -x | +x | +x |
| Equilibrium | (0.20 - x) | excess | x | x |
Solving for \(x\),
\[K_a = \dfrac{\ce{[H3O^+][SO3^{2-}]}}{\ce{[HSO3^{-}]}} =6.4\times10^{-8} \]
\[K_a = \dfrac{\ce{(x)(x)}}{\ce{(0.20-x)}} =6.4\times10^{-8} \]
\[x = \ce{[H3O^+]} = 1.1\times10^{-4} \]
\[pH=-log( \ce{[H3O^+]}) = -log(1.1\times10^{-4}) = 3.95 \]
What is the pH of a solution that contains 0.100 M K 2 HPO 4 ? (K a values are provided above.)
- Answer
-
The \(\ce{K^+}\) ion is pH neutral, so we need only to focus on the equilibrium reactions for \(\ce{HPO4^{2-}} \).
The \(\ce{HPO4^{2-}} \) ion can act as an acid as follows:
\[\ce{HPO4^{2-}}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{PO4^3-}(aq)\:\:\: K_{a3} = 4.2\times10^{-13} \]
The \(\ce{HPO4^{2-}} \) ion can act as a base as follows:
\[\ce{HPO4^{2-}}(aq)+\ce{H2O}(l)⇌\ce{H2PO4^{-}}(aq)+\ce{OH^-}(aq) \]
The \(K_b\) for this reaction is calculated using \(K_{a2}\):
\[K_b = \dfrac{1\times10^{-14}}{6.2\times10^{-8}} = 1.6\times10^{-7} \]
In this case, K b is much larger than K a and the species predominately acts as a base.
pH = 10.10
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