Skip to main content
Chemistry LibreTexts

14.5: Polyprotic Acids

  1. Last updated
  2. Save as PDF
  • Page ID
    414702

    • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    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:

    This image contains two equilibrium reactions. The first shows a C atom bonded to three H atoms and another C atom. The second C atom is double bonded to an O atom and also forms a single bond to another O atom. The second O atom is bonded to an H atom. There is a plus sign and then the molecular formula H subscript 2 O. An equilibrium arrow follows the H subscript 2 O. To the right of the arrow is H subscript 3 O superscript positive sign. There is a plus sign. The final structure shows a C atom bonded the three H atoms and another C atom. This second C atom is double bonded to an O atom and single bonded to another O atom. The entire structure is in brackets and a superscript negative sign appears outside the brackets. The second reaction shows C H subscript 3 C O O H ( a q ) plus H subscript 2 O ( l ) equilibrium arrow H subscript 3 O ( a q ) plus C H subscript 3 C O O superscript negative sign ( a q ).

    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:

    This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header of “H subscript 2 C O subscript 3 plus sign H subscript 2 O equilibrium arrow H subscript 3 O superscript positive sign plus sign H C O subscript 3 superscript negative sign.” Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.033, negative sign x, 0.033 minus sign x. The second column has the following: approximately 0, positive x, x. The third column has the following: 0, positive x, x.

    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.

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

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

    This page titled 14.5: Polyprotic Acids is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.