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Chemistry LibreTexts

Polyprotic Acids

Skills to Develop

  • Define a polyprotic acid.
  • Explain how a polyprotic behaves in its solution.
  • Write the equilibrium equations of ionization of polyprotic acids.
  • Calculate acidity constants, K1, K2, K3, and the overall K.
  • Calculate the concentrations of various species for a given set of data

Polyprotic acids contain more than one mole ionizable hydronium ions per mole of acids. They ionize to give more than one \(\ce{H+}\) ions per molecule. Possible forms of three polyprotic acids are given below after their dissociation into \(\ce{H+}\) ions.

\[\begin{align}
\ce{&H2S, &&HS- , &&S^2- &&}\\
\ce{&H2SO4, &&HSO4- , &&SO4^2- &&}\\
\ce{&H3PO4, &&H2PO4- , &&HPO4^2- , &&PO4^3-}
\end{align}\]

These acids ionize in several stages, giving out one proton at each stage. The acidity constants for these acids may be written as K1, K2, K3, ... Some polyprotic acids are given in Table \(\PageIndex{1}\) on the right here. Knowing their names and being familiar with their properties (ionization for example) is an asset for you. 

Figure \(\PageIndex{1}\): Some Polyprotic Acids
Formula Name
\(\ce{H2S}\) Hydrogen sulfide
\(\ce{H2SO4}\) Sulfuric acid
\(\ce{H2SO3}\) Sulfurous acid
\(\ce{H3PO4}\) Phosphoric acid
\(\ce{H2C2O4}\) Oxalic acid
\(\ce{H2CO3}\) Carbonic acid
\(\ce{H2C3H2O4}\) Malonic acid

Consider \(\ce{H2S}\),

\[\ce{H2S \rightleftharpoons H+ + HS-}\]


\[K_1 = \ce{\dfrac{[H+] [HS- ]}{[H2S]}}\]

and

\[\ce{HS- \rightleftharpoons H+ + S^2-}\]


\[K_2 = \ce{\dfrac{[H+] [S^2- ]}{[HS- ]}}\]

Obviously, for the overall ionization reaction,

\[\ce{H2S \rightleftharpoons 2 H+ + S^2-}\]


\[\begin{align}
K_{\ce{overall}} &= \ce{\dfrac{[H+]^2 [S^2- ]}{[H2S]}}\\
&= K_1 K_2
\end{align}\]

Confirm the above obvious result on a sheet of paper to satisfy yourself.

For polyprotic acids, the following is always true:

\[K_1 > K_2 > K_3 > \: ...\]

For most acids, K1/K2 = 1E5 or 100000, and K2/K3 = 1E5, but oxalic acid is different. For oxalic acid, K1 = 5.6E-2, and K2 = 5.4E-4. The two acidic groups are separated by a \(\ce{C-C}\) bond in oxalic acid.

Example \(\PageIndex{1}\): Oxalic Acid

Oxalic acid is an organic compound with the formula \(C_2H_2O_4\) and has to ionizable protons (white atoms on structure below). 

Calculate the overall equilibrium constant for oxalic acid.

\(\ce{H2C2O4 \rightleftharpoons 2 H+ + C2O4^2-}\)

\(K_1 = \textrm{5.6E-2}\)

\(K_2 = \textrm{5.4E-5}\)

SOLUTION

The calculation is straightforward:

\(\begin{align}
K_{\ce{overall}} &= K_1 K_2\\
      &= 3.0\textrm{E-}6
\end{align}\)

Example \(\PageIndex{2}\): Hydrogen Sulfide

A solution is acidified with \(\ce{HCl}\) so that its pH is 1.0, and is saturated with \(\ce{H2S}\) at 298 K. What is the sulfide \(\ce{S^2-}\) ion concentration in this solution? At 298 K, a saturated \(\ce{H2S}\) solution has \(\mathrm{[H_2S] = 0.10\: M}\) and

\(K_{\ce{overall}} = \textrm{1E-20}\)

for \(\ce{H2S}\).

SOLUTION

\(K_{\ce{overall}} = \ce{\dfrac{[H+]^2 [S^2- ]}{[H2S]}}\)

Thus,

\(\begin{align}
\ce{[S^2- ]} &= 1\textrm{E-}20 \, \dfrac{0.1}{(0.1)^2}\\
&= \textrm{1E-19 F}
\end{align}\)

Example \(\PageIndex{3}\): Sulfuric Acid

Sulfuric acid is a strong acid, and the \(\ce pK_{\large\textrm a_{\Large 2}}\) of \(\ce{HSO4-}\) is 1.92. What is the pH of a 0.100 M \(\ce{NaHSO4}\) solution?

SOLUTION

The salt is completely ionized in its solution.

\(\ce{NaHSO4 \rightarrow Na+ + HSO4-}\)

The anion further ionizes. If \(\ce{[H+]} = x\), then the equilibrium concentrations of various species are:

\(\begin{array}{cccccl}
\ce{HSO4- &\rightleftharpoons &H+ &+ &SO4^2-} &\hspace{20px} K_{\large\textrm a_{\Large 2}} = 10^{-1.92} = 0.0120\\
0.100 - x   &&x,       &&x &
\end{array}\)

\(K_{\large\textrm a_{\Large 2}} = \dfrac{x^2}{0.100-x} = 0.0120\)

\(\begin{align}
\ce{[H+]}&= x\\
    &= \dfrac{-0.120 + (0.012^2 + 4\times0.00120)^{1/2}}{2}\\
    &= \textrm{0.0292 M}
\end{align}\)

Thus,

\(\mathrm{pH = - \log 0.0292 = 1.54}\)

Exercise \(\PageIndex{3A}\)

Is the \(\ce{NaHSO4}\) salt solution acidic?

Although no concentration is stated, such a solution is acidic because of the acidity of \(\ce{HSO4-}\).

Exercise \(\PageIndex{3B}\)

Given that \(\ce{H2SO4}\), \(\ce p K_{\large\textrm a_{\Large 2}} = 1.92\)

For \(\ce{H3SO4}\), \(\ce p K_{\large\textrm a_{\Large 1}} = 2.12\); \(\ce p K_{\large\textrm a_{\Large 2}} = 7.21\); \(\ce p K_{\large\textrm a_{\Large 3}} = 12.67\)

Which of the following solutions are acidic, basic, or neutral?

  • \(\ce{Na2SO4}\)
  • \(\ce{NaH2PO4}\)
  • \(\ce{Na2HPO4}\)
  • \(\ce{Na3PO4}\)
  • \(\ce{NaNO3}\)

Work out the answer please; some of these will appear on the examinations. If the concentration of a salt solution is given, you may be required to evaluate the pH or pOH of the solution.

Example \(\PageIndex{1}\): \(\ce{NaHSO4}\)

What is the pH of a solution containing 0.500 M \(\ce{NaHSO4}\) and 0.300 M \(\ce{Na2SO4}\)?

SOLUTION

The 0.500 M solution of \(\ce{NaHSO4}\) supplies 0.500 M \(\ce{HSO4-}\) as an acid, and similarly, the solution also contains 0.300 M \(\ce{SO4^2-}\).

Using the equation:

\(\begin{align}
\ce{pH} &= \mathrm{p\mathit K_{\large a} - \log \dfrac{[salt]}{[acid]}}\\
   &= 1.92 + \log \left(\dfrac{0.300}{0.500}\right)\\
   &= 1.70
\end{align}\)

Questions

  1. The two acid ionization constants for sulfurous acid are 1.2E-2 and 6.6E-8 respectively. Calculate the overall equilibrium constant for

    \(\ce{H2SO3 \rightleftharpoons 2 H+ + SO3^2-}\)

  2. What is the pH of a 1.0 M \(\ce{H2SO3}\) solution?
    (K1 = 1.2E-2, and K2 = 6.6E-8)

  3. If the pH of a 1.0 M \(\ce{H2SO3}\) solution is 1.0, what is the sulfite ion concentration? (K1 = 1.2E-2, and K2 = 6.6E-8)

Solutions

  1. Answer 7.9E-10
    Consider...

    \(K_{\ce{overall}} = K_1 \times K_2\)

  2. Answer 0.98

    Consider...
    Only K1 matters in this calculation. Using the quadratic formula yields a pH of 0.98. When approximation is used, you'll get a pH of 0.96.

  3. Answer 7.1E-8
    Consider...

    \(\begin{align}
    K_{\ce{overall}} = 7.9\textrm{E-}10 &= \ce{\dfrac{[H+]^2 [SO3^2- ]}{[H2SO3]}}\\
    &= \dfrac{0.12 \ce{[SO3^2- ]}}{0.9}
    \end{align}\)

    \(\ce{[SO3^2- ]} =\: ??\)

    This is not an easy question to answer!

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