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7.16: Polyprotic Acids

  • Page ID
    221522
  • Learning Objectives

    • 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 \(\ce{HCl}\), \(\ce{HNO3}\), and \(\ce{HCN}\) that contain one ionizable hydrogen atom in each molecule are called monoprotic acids. Their reactions with water are:

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

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

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

    Even though it contains four hydrogen atoms, acetic acid, \(\ce{CH3CO2H}\), 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

    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:

    • The first ionization is

    \[ \ce{H2SO4}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HSO4-}(aq)\]

    with \(K_{\ce a1} > 10^2;\: {complete\: dissociation}\).

    • The second ionization is

    \[ \ce{HSO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{SO4^{2−}}(aq)\]

    with \( K_{\ce a2}=1.2×10^{−2}\).

    This stepwise ionization process occurs for all polyprotic acids, meaning that one H+ is lost at the time. This process contrasts with the ionization of strong bases such as Ca(OH)2 in which all the OH- ions are lost in a single dissociation reaction:

    Ca(OH)2(aq) → Ca2+(aq) + 2 OH-(aq)

    This is because hydroxides are ionic compounds, while acids are covalent compounds.

    When we make a solution of a weak diprotic acid, we get a solution that contains a mixture of 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.

    • First Ionization

    \[\ce{H2CO3}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO3-}(aq)\]

    with

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

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

    • Second Ionization

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

    with

    \[ K_{\ce{HCO3-}}=\ce{\dfrac{[H3O+][CO3^2- ]}{[HCO3- ]}}=4.7×10^{−11}\]

    \(K_{\ce{H2CO3}}\) is larger than \(K_{\ce{HCO3-}}\) by a factor of 104, so H2CO3 is the dominant producer of hydronium ion in the solution. 

     

    Example \(\PageIndex{1}\): 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?

    \[\ce{H2CO3}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO3-}(aq) \hspace{20px} K_{\ce a1}=4.3×10^{−7} \label{step1} \tag{equilibrium step 1}\]

    \[\ce{HCO3-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CO3^2-}(aq) \hspace{20px} K_{\ce a2}=4.7×10^{−11} \label{step2} \tag{equilibrium step 2}\]

    Solution

    As indicated by the ionization constants, H2CO3 is a much stronger acid than \(\ce{HCO3-}\), so \(\ce{H2CO3}\) is the dominant producer of hydronium ion in solution. 

    1. Using the customary four steps, we determine the concentration of H3O+ and \(\ce{HCO3-}\) produced by ionization of H2CO3.

     

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

     

     

    An abbreviated table of changes and concentrations shows:

    ICE Table \(\ce{H2CO3}(aq)\) \(\ce{H2O}(l)\) \( \ce{H3O+}(aq) \) \(\ce{HCO3-}(aq)\)
    Initial (M) \(0.033 \:M\) - \(0\) \(0\)
    Change (M) \(- x\) - \(+x\) \(+x\)
    Equilibrium (M) \(0.033 \:M - x\) - \(x\) \( x\)

    Substituting the equilibrium concentrations into the equilibrium constant gives us:

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

    Solving the preceding equation making our standard assumptions gives:

    \[x=1.2×10^{−4} \nonumber\]

    Thus:

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

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

    pH=-log[H3O+]=3.92

     

    Triprotic Acids

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

    • The first ionization is

    \[\ce{H3PO4}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{H2PO4-}(aq) \]

    with \(K_{\ce a1}=7.5×10^{−3} \).

    • The second ionization is

    \[\ce{H2PO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HPO4^2-}(aq)\]

    with \( K_{\ce a2}=6.2×10^{−8} \).

    • The third ionization is

    \[\ce{HPO4^2-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{PO4^3-}(aq) \]

    with \( K_{\ce a3}=4.2×10^{−13} \).

    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. Again, using the first dissociation constant is a good approximation to calculate the pH for a solution of phosphoric acid.

    Contributors and Attributions

    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 significant, it is a good estimation to only consider the first dissociation constant to calculate pH

     

    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

    Contributors and Attributions