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

Expert verified
The stepwise $K_{a}$ reactions for citric acid (H3C6H5O7) are as follows: 1. \(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{-}(aq)\) 2. \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{-}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{HC}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{2-}(aq) \) 3. \(\mathrm{HC}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{2-}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{3-}(aq) \)

Step by step solution

01

First Dissociation Reaction

In the first dissociation reaction, citric acid donates its first acidic proton (H+). The equation for this step can be written as follows: \[\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{-}(aq)\]
02

Second Dissociation Reaction

In the second dissociation reaction, the resulting ion from the first step, H2C6H5O7-, donates its second acidic proton (H+). The equation for this step can be written as follows: \[ \mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{-}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{HC}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{2-}(aq) \]
03

Third Dissociation Reaction

In the third dissociation reaction, the resulting ion from the second step, HC6H5O7²-, donates its third and final acidic proton (H+). The equation for this step can be written as follows: \[ \mathrm{HC}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{2-}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{3-}(aq) \]

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