Question

Write out the stepwise \(K_{\mathrm{a}}\) reactions for citric acid \(\left(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}\right)\) a triprotic acid.

Short answer

The stepwise ionization reactions of citric acid (H3C6H5O7) are as follows: 1. \[ \mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7} \rightleftharpoons \mathrm{H}^{+} + \mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{-} \] 2. \[ \mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{-} \rightleftharpoons \mathrm{H}^{+} + \mathrm{H}_{\mathrm{C}}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{2-} \] 3. \[ \mathrm{H}_{\mathrm{C}}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{2-} \rightleftharpoons \mathrm{H}^{+} + \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{3-} \]

Step-by-step solution

Write the first ionization reaction

The first ionization reaction is the process of citric acid losing one of its acidic protons (H+) and forming the first anion. The reaction is as follows: \[ \mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7} \rightleftharpoons \mathrm{H}^{+} + \mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{-} \]

Write the second ionization reaction

The second ionization reaction involves the first anion (formed in step 1) losing another acidic proton (H+) and forming the second anion. The reaction is as follows: \[ \mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{-} \rightleftharpoons \mathrm{H}^{+} + \mathrm{H}_{\mathrm{C}}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{2-} \]

Write the third ionization reaction

The third ionization reaction involves the second anion (formed in step 2) losing its last acidic proton (H+) and forming the third anion. The reaction is as follows: \[ \mathrm{H}_{\mathrm{C}}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{2-} \rightleftharpoons \mathrm{H}^{+} + \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{3-} \] Now we have completed the stepwise reactions for the ionization of citric acid.