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15.6: Polyprotic Acids

  • Page ID
    45066
  • Skills to Develop

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

    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:

    \[\ce{HCl}(aq)+\ce{H2O}(l)⟶\ce{H3O+}(aq)+\ce{Cl-}(aq)\]

    \[\ce{HNO3}(aq)+\ce{H2O}(l)⟶\ce{H3O+}(aq)+\ce{NO3-}(aq)\]

    \[\ce{HCN}(aq)+\ce{H2O}(l)⟶\ce{H3O+}(aq)+\ce{CN-}(aq)\]

    Even though it contains four hydrogen atoms, acetic acid, 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

    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:

    \[\textrm{First ionization: }\ce{H2SO4}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HSO4-}(aq)\hspace{20px}K_{\ce a1}=\mathrm{more\: than\: 10^2;\: complete\: dissociation}\\
    \textrm{Second ionization: }\ce{HSO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{SO4^2−}(aq)\hspace{20px}K_{\ce a2}=1.2 \times 10^{−2}\]

    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. The hydrogen sulfate ion (\(HSO_4^−\)) is both the conjugate base of \(H_2SO_4\) and the conjugate acid of \(SO_4^{2−}\). Just like water, \(HSO_4^−\) can therefore act as either an acid or a base, depending on whether the other reactant is a stronger acid or a stronger base. Conversely, the sulfate ion (\(SO_4^{2−}\)) is a polyprotic base that is capable of accepting two protons in a stepwise manner:

    \[SO^{2−}_{4 (aq)} + H_2O_{(aq)} \ce{ <=>>} HSO^{−}_{4(aq)}+OH_{(aq)}^-\]

    \[HSO^{−}_{4 (aq)} + H_2O_{(aq)} \ce{ <=>>} H_2SO_{4(aq)}+OH_{(aq)}^-\]

    Like any other conjugate acid–base pair, the strengths of the conjugate acids and bases are related by \(pK_a\) + \(pK_b\) = pKw. Consider, for example, the \(HSO_4^−/ SO_4^{2−}\) conjugate acid–base pair. Yhe \(pK_a\) of \(HSO_4^−\) is 1.99. Hence the \(pK_b\) of \(SO_4^{2−}\) is 14.00 − 1.99 = 12.01. Thus sulfate is a rather weak base, whereas \(OH^−\) is a strong base, so the equilibrium lies to the left. The \(HSO_4^−\) ion is also a very weak base (\(pK_a\) of \(H_2SO_4\) = 2.0, \(pK_b\) of \(HSO_4^− = 14 − (−2.0) = 16\)), which is consistent with what we expect for the conjugate base of a strong acid. Once again, equilibrium favors the formation of the weaker acid–base pair.

    Note

    For a polyprotic acid, acid strength decreases and the \(pK_a\) increases with the sequential loss of each proton.

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

    \[\textrm{First ionization:}\\
    \ce{H2CO3}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO3-}(aq)
    \hspace{20px}K_{\ce{H2CO3}}=\ce{\dfrac{[H3O+][HCO3- ]}{[H2CO3]}}=4.3×10^{−7}\]

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

    \[\textrm{Second ionization:}\\
    \ce{HCO3-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CO3^2-}(aq) \hspace{20px}
    K_{\ce{HCO3-}}=\ce{\dfrac{[H3O+][CO3^2- ]}{[HCO3- ]}}=4.7×10^{−11}\]

    \(K_{\ce{H2CO3}}\) is larger than \(K_{\ce{HCO3-}}\) by a factor of 104, so H2CO3 is the dominant producer of hydronium ion in the solution. This means that little of the \(\ce{HCO3-}\) formed by the ionization of H2CO3 ionizes to give hydronium ions (and carbonate ions), and the concentrations of H3O+ and \(\ce{HCO3-}\) are practically equal in a pure aqueous solution of H2CO3.

    Example 16.6.1

    Predict whether the equilibrium for each reaction lies to the left or the right as written.

    • \(NH^+_{4(aq)}+PO^{3−}_{4(aq)} \rightleftharpoons NH_{3(aq)}+HPO^{2−}_{4(aq)}\)
    • \(CH_3CH_2CO_2H_{(aq)}+CN^−_{(aq)} \rightleftharpoons CH_3CH_2CO^−_{2(aq)}+HCN_{(aq)}\)

    Given: balanced chemical equation

    Asked for: equilibrium position

    Strategy:

    Identify the conjugate acid–base pairs in each reaction. Then refer to Table 16.6.1, Table 16.6.2, and Figure 16.6.2 to determine which is the stronger acid and base. Equilibrium always favors the formation of the weaker acid–base pair.

    Solution:

    The conjugate acid–base pairs are \(NH_4^+/NH_3\) and \(HPO_4^{2−}/PO_4^{3−}\). According to Table 16.6.1 and Table 16.6.2, \(NH_4^+\) is a stronger acid (\(pK_a = 9.25\)) than \(HPO_4^{2−}\) (pKa = 12.32), and \(PO_4^{3−}\) is a stronger base (\(pK_b = 1.68\)) than \(NH_3\) (\(pK_b = 4.75\)). The equilibrium will therefore lie to the right, favoring the formation of the weaker acid–base pair:

    \[ \underset{\text{stronger acid}}{NH^+_{4(aq)}} + \underset{\text{stronger base}}{PO^{3-}_{4(aq)}} \ce{<=>>} \underset{\text{weaker base}}{NH_{3(aq)}} +\underset{\text{weaker acid}} {HPO^{2-}_{4(aq)}} \]

    The conjugate acid–base pairs are \(CH_3CH_2CO_2H/CH_3CH_2CO_2^−\) and \(HCN/CN^−\). According to Table 16.6.1, HCN is a weak acid (pKa = 9.21) and \(CN^−\) is a moderately weak base (pKb = 4.79). Propionic acid (\(CH_3CH_2CO_2H\)) is not listed in Table 16.6.1, however. In a situation like this, the best approach is to look for a similar compound whose acid–base properties are listed. For example, propionic acid and acetic acid are identical except for the groups attached to the carbon atom of the carboxylic acid (\(\ce{−CH_2CH_3}\) versus \(\ce{−CH_3}\)), so we might expect the two compounds to have similar acid–base properties. In particular, we would expect the \(pK_a\) of propionic acid to be similar in magnitude to the \(pK_a\) of acetic acid. (In fact, the \(pK_a\) of propionic acid is 4.87, compared to 4.76 for acetic acid, which makes propionic acid a slightly weaker acid than acetic acid.) Thus propionic acid should be a significantly stronger acid than \(HCN\). Because the stronger acid forms the weaker conjugate base, we predict that cyanide will be a stronger base than propionate. The equilibrium will therefore lie to the right, favoring the formation of the weaker acid–base pair:

    \[ \underset{\text{stronger acid}}{CH_3CH_2CO_2H_{(aq)}} + \underset{\text{stronger base}}{CN^-_{(aq)}} \ce{<=>>} \underset{\text{weaker base}}{CH_3CH_2CO^-_{2(aq)}} +\underset{\text{weaker acid}} {HCN_{(aq)}} \]

    Exercise 16.6.2

    Predict whether the equilibrium for each reaction lies to the left or the right as written.

    1. \(H_2O_{(l)}+HS^−_{(aq)} \rightleftharpoons OH^−_{(aq)}+H_2S_{(aq)}\)
    2. \(HCO^−_{2(aq)}+HSO^−_{4(aq)} \rightleftharpoons HCO_2H_{(aq)}+SO^{2−}_{4(aq)}\)

    Answer:

    1. left
    2. left

    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 14.5.2: 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 \(\ce{[H3O+]}\), \(\ce{[HCO3- ]}\), and \(\ce{[CO3^2- ]}\) in a saturated solution of CO2 with an initial [H2CO3] = 0.033 M?

    \[\ce{H2CO3}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO3-}(aq) \hspace{20px} K_{\ce a1}=4.3×10^{−7}\]

    \[\ce{HCO3-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CO3^2-}(aq) \hspace{20px} K_{\ce a2}=4.7×10^{−11}\]

    Solution

    As indicated by the ionization constants, H2CO3 is a much stronger acid than \(\ce{HCO3-}\), 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 \(\ce{HCO3-}\) produced by ionization of H2CO3. (2) Then we determine the concentration of \(\ce{CO3^2-}\) in a solution with the concentration of H3O+ and \(\ce{HCO3-}\) 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.”

    Determine the concentrations of \(\ce{H3O+}\) and \(\ce{HCO3-}\).

    \[\ce{H2CO3}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO3-}(aq) \hspace{20px} K_{\ce a1}=4.3×10^{−7}\]

    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:

    \[K_{\ce{H2CO3}}=\ce{\dfrac{[H3O+][HCO3- ]}{[H2CO3]}}=\dfrac{(x)(x)}{0.033−x}=4.3×10^{−7}\]

    Solving the preceding equation making our standard assumptions gives:

    \[x=1.2×10^{−4}\]

    Thus:

    \[\ce{[H2CO3]}=0.033\:M\]

    \[\ce{[H3O+]}=\ce{[HCO3- ]}=1.2×10^{−4}\:M\]

    1. Determine the concentration of \(CO_3^{2-}\) in a solution at equilibrium with \([H_3O^+]\) and \([HCO_3^- ]\) both equal to 1.2 × 10−4 M.

    \[\ce{HCO3-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CO3^2-}(aq)\]

    \[K_{\ce{HCO3-}}=\ce{\dfrac{[H3O+][CO3^2- ]}{[HCO3- ]}}=\dfrac{(1.2×10^{−4})[\ce{CO3^2-}]}{1.2×10^{−4}}\]

    \[[\ce{CO3^2-}]=\dfrac{(4.7×10^{−11})(1.2×10^{−4})}{1.2×10^{−4}}=4.7×10^{−11}\:M\]

    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; \(\ce{[H3O+]}\) = 1.2 × 10−4; and \(\ce{[HCO3- ]}=1.2×10^{−4}\:M\). In part 2, we determined that \(\ce{[CO3^2- ]}=4.7×10^{−11}\:M\).

    Exercise 14.5.2

    The concentration of H2S in a saturated aqueous solution at room temperature is approximately 0.1 M. Calculate \(\ce{[H3O+]}\), [HS], and [S2−] in the solution:

    \[\ce{H2S}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HS-}(aq) \hspace{20px} K_{\ce a1}=1.0×10^{−7}\]

    \[\ce{HS-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{S^2-}(aq) \hspace{20px} K_{\ce a2}=1.0×10^{−19}\]

    Answer:

    [H2S] = 0.1 M; \(\ce{[H3O+]}\) = [HS] = 0.0001 M; [S2−] = 1 × 10−19 M

    We note that the concentration of the sulfide ion is the same as Ka2. 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:

    \[\textrm{First ionization: }\ce{H3PO4}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{H2PO4-}(aq) \hspace{20px} K_{\ce a1}=7.5×10^{−3}\\
    \textrm{Second ionization: }\ce{H2PO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HPO4^2-}(aq) \hspace{20px} K_{\ce a2}=6.3×10^{−8}\\
    \textrm{Third ionization: }\ce{HPO4^2-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{PO4^3-}(aq) \hspace{20px} K_{\ce a3}=3.6×10^{−13}\]

    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 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:

    \[\ce{H2O}(l)+\ce{CO3^2-}(aq)⇌\ce{HCO3-}(aq)+\ce{OH-}(aq)\]

    and

    \[ \ce{H2O}(l)+\ce{HCO3-}(aq)⇌\ce{H2CO3}(aq)+\ce{OH-}(aq)\]

    Key Concepts and 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 Ka 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

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