Q. 10.74 - An Introduction to General, Organic, and Biological Chemistry

Question

Step-by-Step Solution

Answer

Step 1:Introduction

Definition of Conjugate Acid Conjugate acids and bases are Bronsted-Lowry acid and base pairs that differ depending on which species gains or loses a proton. The species that gains a hydrogen (proton) when a base dissolves in water is the base's conjugate acid.

Step 2:Given data

A base, according to the Bronsted-Lowry theory, is a substance that can accept a proton.(hydrogen ion, $\left.{\mathrm{H}}^{+}\right)$.A base is converted to its conjugate acid when it accepts a proton from an acid. When a base accepts a proton from another molecule or ion, a conjugate acid is formed. For example, if${\mathrm{B}}^{-}$is a base, it becomes $\mathrm{BH}$ by accepting a proton, ${\mathrm{H}}^{+}$. Hence, $\mathrm{BH}$ is the conjugate acid of base${\mathrm{B}}^{-}$.

Step 3:Bronsted-Lowry theory,

A conjugate acid-base pair, according to the Bronsted-Lowry theory, is any pair of molecules or ions that can be interconverted by proton transfer. A base is converted to its conjugate acid when it accepts a proton from an acid. By donating a proton, a conjugate acid can be converted to a base.

Step 4:A conjugate acid,

A conjugate acid,${\mathrm{HS}}^{-}$(hydrogen sulfide ion),becomes converted to${\mathrm{S}}^{2-}$ (sulfide ion),by contributing a proton${\mathrm{H}}^{+}$. Hence, ${\mathrm{S}}^{2-}$ is the base of the conjugate acid ${\mathrm{HS}}^{-}$.

A hydronium ion, ${\mathrm{H}}_3{\mathrm{O}}^{+}$,becomes converted to ,${\mathrm{H}}_2\mathrm{O}$by contributing a proton, . Hence, ${\mathrm{H}}_2\mathrm{O}$ is the base of the conjugate acid ${\mathrm{H}}_3{\mathrm{O}}^{+}$.

Ammonia, ${\mathrm{NH}}_3$,  is transformed into its conjugate acid ${{\mathrm{NH}}_4}^{+}$, ammonium ion, by accepting a proton, ${\mathrm{H}}^{+}$.

Bicarbonate ion,${\mathrm{HCO}}^{3-}$, is transformed into its conjugate acid${\mathrm{H}}_2{\mathrm{CO}}_3$, carbonic acid, by taking a proton , ${\mathrm{H}}^{+}$.

Step 5:Solution