14.5: Polyprotic Acids

Learning Objectives

By the end of this section, you will be able to:

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

Acids are classified 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:

$$\begin{align*} \ce{HCl (aq) + H_2 O (l)} & \ce{\longrightarrow H3O^{+}(aq) + Cl^{-}(aq) } \\[4pt] \ce{HNO_3(aq) + H_2 O (l)} & \ce{\longrightarrow H3O^{+}(aq) + NO_3^{-}(aq) } \\[4pt] \ce{HCN (aq) + H_2 O (l)} & \ce{\rightleftharpoons H3O^{+}(aq) + CN^{-}(aq) } \end{align*} \nonumber$$

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 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:

$$\tag{ First ionization} \ce{H_2 SO_4 (aq) + H2O (l) \rightleftharpoons H_3O^{+}(aq) + HSO_4^{-}(aq)}$$

with $K_{ a 1} > 10^2$.

$$\tag{ Second ionization} \ce{HSO_4^{-}(aq) + H_2 O (l) \rightleftharpoons H3O^{+}(aq) + SO_4^{2-}(aq) }$$

with $K_{ a 2}=1.2 \times 10^{-2}$.

This stepwise ionization process occurs for all polyprotic 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.

$$\tag{First ionization} \ce{H_2 CO_3(aq) + H_2 O (l) \rightleftharpoons H_3O^{+}(aq) + HCO_3^{-}(aq) }$$

with

$$K_{ H_2 CO_3}=\frac{\left[\ce{H3O^{+}} \right]\left[ \ce{HCO_3^{-}} \right]}{\left[ \ce{H_2CO_3} \right]}=4.3 \times 10^{-7} \nonumber$$

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

$$\tag{Second ionization} \ce{ HCO_3^{-}(aq) + H_2 O (l) \rightleftharpoons H3O^{+}(aq) + CO_3^{2-}(aq) }$$

with

$$K_{\ce{HCO_3^{-}}}=\frac{\left[ \ce{H_3O^{+}}\right]\left[ \ce{CO_3^{2-}}\right]}{\left[ \ce{HCO_3^{-}}\right]}=4.7 \times 10^{-11} \nonumber$$

$K_{\ce{H_2CO_3}}$ is larger than $K_{\ce{HCO_3^{-}}}$ by a factor of $10^{4}$, so $\ce{H2CO3}$ is the dominant producer of hydronium ion in the solution. This means that little of the formed by the ionization of $\ce{H2CO3}$ ionizes to give hydronium ions (and carbonate ions), and the concentrations of $\ce{H3O^{+}}$ and are practically equal in a pure aqueous solution of $\ce{H2CO3}$.

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 approach is demonstrated in the following example and exercise.

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

“Carbonated water” contains a palatable amount of dissolved carbon dioxide. The solution is acidic because $\ce{CO2}$ reacts with water to form carbonic acid, $\ce{H2CO3}$. What are and in a saturated solution of $\ce{CO2}$ with an initial $\ce{[H2CO3] = 0.033 \,M}$?

$$\ce{H_2 CO_3(aq) + H_2 O (l) \rightleftharpoons H3O^{+}(aq) + HCO_3^{-}(aq)} \quad K_{ a 1}=4.3 \times 10^{-7} \nonumber$$

$$\ce{HCO_3^{-}(aq) + H_2 O (l) \rightleftharpoons H3O^{+}(aq) + CO_3^{2-}(aq)} \quad K_{ a 2}=4.7 \times 10^{-11} \nonumber$$

Solution

As indicated by the ionization constants, $\ce{H2CO3}$ is a much stronger acid than so the stepwise ionization reactions may be treated separately.

The first ionization reaction is

$$\ce{H_2 CO_3(aq) + H_2 O (l) \rightleftharpoons H3O^{+}(aq) + HCO_3^{-}(aq)} \quad K_{ a 1}=4.3 \times 10^{-7} \nonumber$$

Using provided information, an ICE table for this first step is prepared:

Substituting the equilibrium concentrations into the equilibrium equation gives

$$K_{ H_2 CO_3}=\frac{\left[ H3O^{+}\right]\left[ HCO_3^{-}\right]}{\left[ H_2 CO_3\right]}=\frac{(x)(x)}{0.033-x}=4.3 \times 10^{-7} \nonumber$$

Assuming $x \ll 0.033$ and solving the simplified equation yields

$$x=1.2 \times 10^{-4} \nonumber$$

The ICE table defined x as equal to the bicarbonate ion molarity and the hydronium ion molarity:

$$\left[ H_2 CO_3\right]=0.033 M \nonumber$$

$$\left[ H3O^{+}\right]=\left[ HCO_3^{-}\right]=1.2 \times 10^{-4} M \nonumber$$

Using the bicarbonate ion concentration computed above, the second ionization is subjected to a similar equilibrium calculation:

$$\ce{HCO_3^{-}(aq) + H_2 O (l) \rightleftharpoons H3O^{+}(aq) + CO_3^{2-}(aq)} \nonumber$$

$$K_{ HCO_{3^{-}}}=\frac{\left[ H3O^{+}\right]\left[ CO_3^{2-}\right]}{\left[ HCO_3^{-}\right]}=\frac{\left(1.2 \times 10^{-4}\right)\left[ CO_3^{2-}\right]}{1.2 \times 10^{-4}} \nonumber$$

$$\left[ CO_3^{2-}\right]=\frac{\left(4.7 \times 10^{-11}\right)\left(1.2 \times 10^{-4}\right)}{1.2 \times 10^{-4}}=4.7 \times 10^{-11} M \nonumber$$

To summarize: at equilibrium $\ce{[H2CO3]} = 0.033 M$; $\ce{[H3O^{+}]} = 1.2 \times 10^{−4}$; $\ce{[HCO3^{-}]}=1.2 \times 10^{−4}\,M$;
$\ce{[CO32^{−}]}=4.7 \times 10^{−11}\,M$.

Exercise $\PageIndex{1}$

The concentration of $\ce{H2S}$ 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:

$$\ce{H_2 S (aq) + H_2 O (l) \rightleftharpoons H3O^{+}(aq) + HS^{-}(aq)} \quad K_{ a 1}=8.9 \times 10^{-8} \nonumber$$

$$\ce{HS^{-}(aq) + H_2 O (l) \rightleftharpoons H3O^{+}(aq) + S^{2-}(aq)} \quad K_{ a 2}=1.0 \times 10^{-19} \nonumber$$

Answer

$[\ce{H2S}] = 0.1\, M$

$[\ce{H3O^{+}}] = [\ce{HS^{-}}] = 0.000094\, M$

$[\ce{S^{2-}}] = 1 \times 10^{−19}\, M$

A triprotic acid is an acid that has three ionizable H atoms. Phosphoric acid is one example:

$$\tag{First ionization} \ce{H3PO_4(aq) + H2O (l) \rightleftharpoons H3O^{+}(aq) + H_2 PO_4^{-}(aq)} \quad K_{ a 1}=7.5 \times 10^{-3}$$

$$\tag{Second ionization} \ce{H2PO4^{-}(aq) + H2O (l) \rightleftharpoons H3O^{+}(aq) + HPO_4^{2-}(aq)} \quad K_{ a 2}=6.2 \times 10^{-8}$$

$$\tag{Third ionization} \ce{HPO4^{2-}(aq) + H2O (l) \rightleftharpoons H3O^{+}(aq) + PO_4^{3-}(aq)} \quad K_{ a 3}=4.2 \times 10^{-13}$$

As for the diprotic acid examples, each successive ionization reaction is less extensive than the former, reflected in decreasing values for the stepwise acid ionization constants. 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 $\ce{H3PO4}$ complicated. However, because the successive ionization constants differ by a factor of $10^{5}$ to $10^{6}$, large differences exist in the small changes in concentration accompanying the ionization reactions. This allows the use of math-simplifying assumptions and processes, as demonstrated in the examples above.

Polyprotic bases are capable of accepting more than one hydrogen ion. The carbonate ion is an example of a diprotic base, because it can accept two protons, as shown below. Similar to the case for polyprotic acids, note the ionization constants decrease with ionization step. Likewise, equilibrium calculations involving polyprotic bases follow the same approaches as those for polyprotic acids.

$$\ce{H_2 O (l) + CO_3^{2-}(aq) \rightleftharpoons HCO_3^{-}(aq) + OH^{-}(aq)} \quad K_{ b 1}=2.1 \times 10^{-4} \nonumber$$

$$\ce{H_2 O (l) + HCO_3^{-}(aq) \rightleftharpoons H_2 CO_3(aq) + OH^{-}(aq)} \quad K_{ b 2}=2.3 \times 10^{-8} \nonumber$$