Chemistry » Acid-Base Equilibria » Polyprotic Acids

Polyprotic Acids

Polyprotic Acids

We can classify acids by the number of protons per molecule that they can give up in a reaction. Acids such as HCl, HNO3, and HCN that contain one ionizable hydrogen atom in each molecule are called monoprotic acids. Their reactions with water are:

\(\begin{array}{c}\text{HCl}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{Cl}}^{\text{−}}\left(aq\right)\\ {\text{HNO}}_{3}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{NO}}_{3}{}^{\text{−}}\left(aq\right)\\ \text{HCN}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{CN}}^{\text{−}}\left(aq\right)\end{array}\)

Even though it contains four hydrogen atoms, acetic acid, CH3CO2H, is also monoprotic because only the hydrogen atom from the carboxyl group (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:

\(\begin{array}{}\\ \\ \text{First ionization:}\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}{\text{SO}}_{4}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)⇌{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{HSO}}_{4}{}^{\text{−}}\left(aq\right)\phantom{\rule{5em}{0ex}}{K}_{\text{a}1}=\text{more than}\phantom{\rule{0.2em}{0ex}}{10}^{\text{2}}\text{; complete dissociation}\\ \text{Second ionization:}\phantom{\rule{0.2em}{0ex}}{\text{HSO}}_{4}{}^{\text{−}}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)⇌{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{SO}}_{4}{}^{2-}\left(aq\right)\phantom{\rule{5em}{0ex}}{K}_{\text{a}2}=1.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}\end{array}\)

This stepwise ionization 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, H2CO3, is an example of a weak diprotic acid. The first ionization of carbonic acid yields hydronium ions and bicarbonate ions in small amounts.

\(\begin{array}{l}\text{First ionization:}\\ {\text{H}}_{2}{\text{CO}}_{3}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)⇌{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{HCO}}_{3}{}^{\text{−}}\left(aq\right)\phantom{\rule{5em}{0ex}}{K}_{{\text{H}}_{2}{\text{CO}}_{3}}=\phantom{\rule{0.2em}{0ex}}\cfrac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{HCO}}_{3}{}^{\text{−}}\right]}{\left[{\text{H}}_{2}{\text{CO}}_{3}\right]}\phantom{\rule{0.2em}{0ex}}=4.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}\end{array}\)

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

\(\begin{array}{l}\text{Second ionization:}\\ {\text{HCO}}_{3}{}^{\text{−}}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)⇌{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{CO}}_{3}{}^{2-}\left(aq\right)\phantom{\rule{5em}{0ex}}{K}_{{\text{HCO}}_{3}{}^{\text{−}}}=\phantom{\rule{0.2em}{0ex}}\cfrac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{CO}}_{3}{}^{2-}\right]}{\left[{\text{HCO}}_{3}{}^{\text{−}}\right]}\phantom{\rule{0.2em}{0ex}}=5.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-11}\end{array}\)

\({K}_{{\text{H}}_{2}{\text{CO}}_{3}}\) is larger than \({K}_{{\text{HCO}}_{3}{}^{\text{−}}}\) by a factor of 104, so H2CO3 is the dominant producer of hydronium ion in the solution. This means that little of the \({\text{HCO}}_{3}{}^{\text{−}}\) formed by the ionization of H2CO3 ionizes to give hydronium ions (and carbonate ions), and the concentrations of H3O+ and \({\text{HCO}}_{3}{}^{\text{−}}\) are practically equal in a pure aqueous solution of 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 can simplify our work considerably because we can determine the concentration of H3O+ 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.

Example

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 CO2 reacts with water to form carbonic acid, H2CO3. What are \(\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right],\)\(\left[{\text{HCO}}_{3}{}^{\text{−}}\right],\) and \(\left[{\text{CO}}_{3}{}^{2-}\right]\) in a saturated solution of CO2 with an initial [H2CO3] = 0.033 M?

\({\text{H}}_{2}{\text{CO}}_{3}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)⇌{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{HCO}}_{3}{}^{\text{−}}\left(aq\right)\phantom{\rule{5em}{0ex}}{K}_{\text{a}1}=4.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}\)

\({\text{HCO}}_{3}{}^{\text{−}}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)⇌{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{CO}}_{3}{}^{2-}\left(aq\right)\phantom{\rule{5em}{0ex}}{K}_{\text{a2}}=5.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-11}\)

Solution

As indicated by the ionization constants, H2CO3 is a much stronger acid than \({\text{HCO}}_{3}{}^{\text{−}},\) so H2CO3 is the dominant producer of hydronium ion in solution. Thus there are two parts in the solution of this problem: (1) Using the customary four steps, we determine the concentration of H3O+ and \({\text{HCO}}_{3}{}^{\text{−}}\) produced by ionization of H2CO3. (2) Then we determine the concentration of \({\text{CO}}_{3}{}^{2-}\) in a solution with the concentration of H3O+ and \({\text{HCO}}_{3}{}^{\text{−}}\) determined in (1). To summarize:

Four tan rectangles are shown that are connected with right pointing arrows. The first is labeled “left bracket H subscript 2 C O subscript 3 right bracket.” The second is labeled “left bracket H subscript 3 O superscript plus right bracket and left bracket H C O subscript 3 superscript negative right bracket from H subscript 2 C O subscript 3.” The third is labeled “left bracket C O subscript 3 superscript 2 negative right bracket from H C O subscript 3 superscript negative.”

  1. Determine the concentrations of\({H}_{\mathit{3}}{O}^{\mathit{\text{+}}}\)and\(HC{O}_{\mathit{3}}{}^{\mathit{\text{−}}}.\)

    \({\text{H}}_{2}{\text{CO}}_{3}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)⇌{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{HCO}}_{3}{}^{\text{−}}\left(aq\right)\phantom{\rule{5em}{0ex}}{K}_{\text{a}1}=4.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}\)

    As for the ionization of any other weak acid:

    Four tan rectangles are shown that are connected with right pointing arrows. The first is labeled “Determine the direction of change.” The second is labeled “Determine x and the equilibrium concentrations.” The third is labeled “Solve for x and the equilibrium concentrations.” The fourth is labeled “Check the math.”

    An abbreviated table of changes and concentrations shows:

    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 constant ( 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, x, x. The third column has the following: 0, x, x.

    Substituting the equilibrium concentrations into the equilibrium gives us:

    \({K}_{{\text{H}}_{2}{\text{CO}}_{3}}=\cfrac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{HCO}}_{3}{}^{\text{−}}\right]}{\left[{\text{H}}_{2}{\text{CO}}_{3}\right]}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{\left(x\right)\left(x\right)}{0.033-x}\phantom{\rule{0.2em}{0ex}}=4.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}\)

    Solving the preceding equation making our standard assumptions gives:

    \(x=1.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\)

    Thus:

    \(\left[{\text{H}}_{2}{\text{CO}}_{3}\right]=0.033\phantom{\rule{0.2em}{0ex}}M\)

    \(\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]=\left[{\text{HCO}}_{3}{}^{\text{−}}\right]=1.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\phantom{\rule{0.2em}{0ex}}M\)

  2. Determine the concentration of \(C{O}_{\mathit{3}}{}^{2-}\) in a solution at equilibrium with\(\left[{H}_{\mathit{3}}{O}^{\mathit{\text{+}}}\right]\)and\(\left[HC{O}_{\mathit{3}}{}^{\mathit{\text{−}}}\right]\)both equal to 1.2\(\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\)10−4 M.

    \({\text{HCO}}_{3}{}^{\text{−}}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)⇌{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{CO}}_{3}{}^{2-}\left(aq\right)\)

    \({K}_{{\text{HCO}}_{3}{}^{\text{−}}}=\cfrac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{CO}}_{3}{}^{2-}\right]}{\left[{\text{HCO}}_{3}{}^{\text{−}}\right]}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{\left(1.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\right)\left[{\text{CO}}_{3}{}^{2-}\right]}{1.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}}\)

    \(\left[{\text{CO}}_{3}{}^{2-}\right]=\phantom{\rule{0.2em}{0ex}}\cfrac{\left(5.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-11}\right)\left(1.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\right)}{1.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}}\phantom{\rule{0.2em}{0ex}}=5.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-11}\phantom{\rule{0.2em}{0ex}}M\)

To summarize: In part 1 of this example, we found that the H2CO3 in a 0.033-M solution ionizes slightly and at equilibrium [H2CO3] = 0.033 M; \(\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\) = 1.2 \(\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\) 10−4; and \(\left[{\text{HCO}}_{3}{}^{\text{−}}\right]=1.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\phantom{\rule{0.2em}{0ex}}M.\) In part 2, we determined that \(\left[{\text{CO}}_{3}{}^{2-}\right]=5.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-11}\phantom{\rule{0.2em}{0ex}}M.\)

A triprotic acid is an acid that has three dissociable protons that undergo stepwise ionization: Phosphoric acid is a typical example:

\(\begin{array}{}\\ \text{First ionization:}\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{PO}}_{4}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)⇌{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{H}}_{2}{\text{PO}}_{4}{}^{\text{−}}\left(aq\right)\phantom{\rule{5em}{0ex}}{K}_{\text{a}1}=7.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\\ \text{Second ionization:}\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}{\text{PO}}_{4}{}^{\text{−}}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)⇌{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{HPO}}_{4}{}^{2-}\left(aq\right)\phantom{\rule{5em}{0ex}}{K}_{\text{a}2}=6.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-8}\\ \text{Third ionization:}\phantom{\rule{0.2em}{0ex}}{\text{HPO}}_{4}{}^{2-}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)⇌{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{PO}}_{4}{}^{3-}\left(aq\right)\phantom{\rule{5em}{0ex}}{K}_{\text{a}3}=4.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-13}\end{array}\)

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 105 to 106.

This set of three dissociation reactions may appear to make calculations of equilibrium concentrations in a solution of H3PO4 complicated. However, because the successive ionization constants differ by a factor of 105 to 106, 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:

\({\text{H}}_{2}\text{O}\left(l\right)+{\text{CO}}_{3}{}^{2-}\left(aq\right)⇌{\text{HCO}}_{3}{}^{\text{−}}\left(aq\right)+{\text{OH}}^{\text{−}}\left(aq\right)\phantom{\rule{1.5em}{0ex}}\text{and}\phantom{\rule{1.5em}{0ex}}{\text{H}}_{2}\text{O}\left(l\right)+{\text{HCO}}_{3}{}^{\text{−}}\left(aq\right)⇌{\text{H}}_{2}{\text{CO}}_{3}\left(aq\right)+{\text{OH}}^{\text{−}}\left(aq\right)\)

[Attributions and Licenses]


This is a lesson from the tutorial, Acid-Base Equilibria and you are encouraged to log in or register, so that you can track your progress.

Log In

Share Thoughts