Chapter 18, Problem 98A

Interpretation Introduction

Interpretation:

The balance chemical equations for two successive ionization of carbonic acid in water are to be written. The conjugate-base pair is also to be identified.

Concept introduction:

The concept of conjugate acid-base pair is given by Arrhenius.

According to it,

Acid is a substance which donate hydrogen ion and Base is a substance which accept hydrogen ion.

When acid donates $H^{+}$, the species form is conjugate base and when base accepts $H^{+}$, the species form is conjugate acid.

Polyprotic acid is the acid which can give more than one proton in aqueous solution.

Acids are classified into monoprotic, diprotic and triprotic acid according to number of hydrogen ions it gives in aqueous solution.

Answer to Problem 98A

First ionization equation:

$H_{2}C{O}_3+H_{2}O\to H_3{O}^{+}+HC{O}_3{}^{-}$

Acid:

$H_{2}C{O}_3$

Base: $H_{2}O$

Conjugate base: $HC{O}_3{}^{-}$

Conjugate acid: $H_3{O}^{+}$

Second ionization equation:

$HC{O}_3{}^{-}+H_{2}O\to H_3{O}^{+}+C{O}_3{}^{2-}$

Acid:

$HC{O}_3{}^{-}$

Base: $H_{2}O$

Conjugate base: $C{O}_3{}^{2-}$

Conjugate acid: $H_3{O}^{+}$

Explanation of Solution

Carbonic acid is diprotic. It has two acidic protons. So it ionizes as:

First ionization equation:

$H_{2}C{O}_3+H_{2}O\to H_3{O}^{+}+HC{O}_3{}^{-}$

Second ionization equation:

$HC{O}_3{}^{-}+H_{2}O\to H_3{O}^{+}+C{O}_3{}^{2-}$

By Bronsted-Lowry concept:

In first equation,

Acid:

$H_{2}C{O}_3$

Base: $H_{2}O$

Conjugate base: $HC{O}_3{}^{-}$

Conjugate acid: $H_3{O}^{+}$

In second equation,

Acid:

$HC{O}_3{}^{-}$

Base: $H_{2}O$

Conjugate base: $C{O}_3{}^{2-}$

Conjugate acid: $H_3{O}^{+}$

Conclusion

Carbonic acid is diprotic. It has two acidic protons. So it ionizes in two steps.