4.9.5: Polyprotic Acids
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:
\[\tag{ First ionization} \ce{H_2 SO_4 (aq) + H2O (l) \rightleftharpoons H_3O^{+}(aq) + HSO_4^{-}(aq)} \]
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.
“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 \(\ce{[H2CO3]} = 0.033 M\); \(\ce{[H3O^{+}]} = 1.2 \times 10^{−4}\); \(\ce{[HCO3^{-}]}=1.2 \times 10^{−4}\,M\);
\(\ce{[CO32^{−}]}=4.7 \times 10^{−11}\,M\).
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
-
\([\ce{H2S}] = 0.1\, M\)
\([\ce{H3O^{+}}] = [\ce{HS^{-}}] = 0.000094\, M\)
\([\ce{S^{2-}}] = 1 \times 10^{−19}\, M\)
Triprotic Acids
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.
\[\ce{H_2 O (l) + CO_3^{2-}(aq) \rightleftharpoons HCO_3^{-}(aq) + OH^{-}(aq)} \quad K_{ b 1}=2.1 \times 10^{-4} \nonumber \]
\[\ce{H_2 O (l) + HCO_3^{-}(aq) \rightleftharpoons H_2 CO_3(aq) + OH^{-}(aq)} \quad K_{ b 2}=2.3 \times 10^{-8} \nonumber \]
Dependence of Degree of Protonation on pH for Polyprotic acids
Many natural bodies of water and living organisms contain high concentrations of salts that maintain (buffer) the proton concentration (pH) at particular values. This means that the protonation state of low concentration polyprotic acids is determined by the pH and not concentration of the particular acid. An important example of this is the amino acids that our bodies use to build proteins. They are all either diprotic or triprotic acids. We will consider arginine (Arg) which is triprotic. The four forms are Abbreviated \(\ce{H3Arg^{2+}}\), \(\ce{H2Arg+}\), \(\ce{HArg}\) and \(\ce{Arg-}\). The equilibria are:
\[\ce{H3Arg^{2+}(aq) <=> H2Arg+(aq) + H+(aq)}\quad pK_{a1} = 2.03\label{RXN:arg1}\]
\[\ce{H2Arg+(aq)<=> HArg(aq) + H+(aq)}\quad pK_{a2} = 9.00\label{RXN:arg2}\]
\[\ce{HArg(aq)<=> Arg-(aq) + H+(aq)}\quad pK_{a3} = 12.10\label{RXN:arg3}\]
We can solve the equilibrium constant expressions for each equilibrium for the concentration of the deprotonated species as follows:
\[K_{a1} = \frac{[H^+][H_{2}Arg^+]}{[H_{3}Arg^{2+}]} \implies [H_{2}Arg^+] = K_{a1}\frac{[H_{3}Arg^{2+}]}{[H^+]}\label{H2Arg}\]
\[K_{a2} = \frac{[H^+][HArg]}{[H_{2}Arg^{+}]} \implies [HArg] = K_{a2}\frac{[H_{2}Arg^{+}]}{[H^+]} = K_{a1}K_{a2}\frac{[H_{3}Arg^{2+}]}{[H^+]^2}\label{HArg}\]
\[K_{a3} = \frac{[H^+][Arg^-]}{[HArg]} \implies [Arg^-] = K_{a3}\frac{[HArg]}{[H^+]} = K_{a1}K_{a2}K_{a3}\frac{[H_{3}Arg^{2+}]}{[H^+]^3}\label{Arg}\]
Where the last expression in each of \(\ref{H2Arg}\), \(\ref{HArg}\) and \(\ref{Arg}\) results from substituting the final expression from the previous equation in for the numerator. To calculate the fraction in each protonation state we need the total of all the arginine species:
\[Arg_{tot} = [H_{3}Arg^{2+}]\left(1 + \frac{K_{a1}}{[H^+]} + \frac{K_{a1}K_{a2}}{[H^+]^2} + \frac{K_{a1}K_{a2}K_{a3}}{[H^+]^3}\right) \label{Argtot}\]
All the terms in in the parentheses in equation \(\ref{Argtot}\) are constant for a given pH, so we can calculate a value at each pH and to make it easier we can symbolize this value as A, allowing us to rewrite equation \(\ref{Argtot}\) as:
\[Arg_{tot} = [H_{3}Arg^{2+}]A \label{Argtot2}\]
This allows us to write the fraction of each species as:
\[f(H_{3}Arg^{2+}) = \frac{ [H_{3}Arg^{2+}]}{[H_{3}Arg^{2+}]A} = \frac{1}{A}\label{fH3Arg}\]
\[f(H_{2}Arg^+) = \frac{ K_{a1}[H_{3}Arg^{2+}]}{[H^+][H_{3}Arg^{2+}]A} = \frac{ K_{a1}}{[H^+]A}\label{fH2Arg}\]
\[f(HArg) = \frac{ K_{a1}K_{a2}[H_{3}Arg^{2+}]}{[H^+]^2[H_{3}Arg^{2+}]A} = \frac{ K_{a1}K_{a2}}{[H^+]^2A}\label{fHArg}\]
\[f(Arg)^- = \frac{ K_{a1}K_{a2}K_{a3}[H_{3}Arg^{2+}]}{[H^+]^3[H_{3}Arg^{2+}]A} = \frac{ K_{a1}K_{a2}K_{a3}}{[H^+]^3A}\label{fArg}\]
These equations allow you to determine at what [H + ] or pH each conjugate pair are in equal concentration. For example setting equation \(\ref{fH3Arg}\) equal to equation \(\ref{fH2Arg}\) and solving for [H + ] yields:
\[ \frac{1}{\cancel{A}} = \frac{ K_{a1}}{[H^+]\cancel{A}} \implies [H^+] = K_{a1} \implies pH = pK_{a1} \label{conjeq}\]
Similar algebra shows that each conjugate pair make up equal fractions of the mixture each time the pH = pK a . Calculation of the actual fraction show that they will each be close to 50% of the species in solution. See the plot of the calculated values (figure \(\PageIndex{1}\)).
Figure \(\PageIndex{1}\): Fraction of each of the protonation states of arginine versus pH.