Question

a. Draw the conjugate acid of each base: ${\mathbf{NH}}_{\mathbf{3}}\mathbf{,}\mathbf{}{\mathbf{Cl}}^{\mathbf{-}}\mathbf{,}\mathbf{}{\left({\mathrm{CH}}_3\right)}_{\mathbf{2}}\mathbf{C}\mathbf{=}\mathbf{O}$

b.Draw the conjugate base of each acid: HBr,${\mathbf{HSO}}_{\mathbf{4}}^{\mathbf{-}}\mathbf{,}\mathbf{}{\mathbf{CH}}_{\mathbf{3}}\mathbf{OH}$

Short answer

a. The conjugate acid of ${\mathrm{NH}}_3$is ${\mathrm{NH}}_4^{+}$.

The conjugate acid of ${\mathrm{Cl}}^{-}$is HCl.

The conjugate acid of ${\left({\mathrm{CH}}_3\right)}_2\mathrm{C}=\mathrm{O}$is role="math" localid="1648446092756" ${\left({\mathrm{CH}}_3\right)}_2\mathrm{C}=\stackrel{+}{\underset{..}{\mathrm{O}}}\mathrm{H}$.

b. The conjugate base of HBr is ${\mathrm{Br}}^{-}$.

The conjugate base of ${\mathrm{HSO}}_4^{-}$is ${\mathrm{SO}}_4^{2-}$.

The conjugate base of ${\mathrm{CH}}_3\mathrm{OH}$is ${\mathrm{CH}}_3{\mathrm{O}}^{-}$.

Step-by-step solution

Conjugate acid

A conjugate acid is created when the Lowry base obtains a proton.

For example, ${\mathrm{H}}_3{\mathrm{O}}^{+}$is the conjugate acid of ${\mathrm{H}}_2\mathrm{O}$.

Conjugate base

A conjugate base is created when the Brønsted Lowry acid deprives itself of a proton.

For example, ${\mathrm{NO}}_3^{-}$ is the conjugate base of ${\mathrm{HNO}}_3$.

Determination of conjugate acid of each base

a.${\mathrm{NH}}_3$receives a proton and forms ${\mathrm{NH}}_4^{+}$. So, the conjugate acid of ${\mathrm{NH}}_3$is ${\mathrm{NH}}_4^{+}$.

${\mathrm{Cl}}^{-}$receives a proton and forms HCl. So, the conjugate acid of ${\mathrm{Cl}}^{-}$is HCl.

${\left({\mathrm{CH}}_3\right)}_2\mathrm{C}=\mathrm{O}$receives a proton and forms ${\left({\mathrm{CH}}_3\right)}_2\mathrm{C}=\underset{..}{\overset{+}{\mathrm{O}}}\mathrm{H}$.So, the conjugate acid of ${\left({\mathrm{CH}}_3\right)}_2\mathrm{C}=\mathrm{O}$is ${\left({\mathrm{CH}}_3\right)}_2\mathrm{C}=\underset{..}{\overset{+}{\mathrm{O}}}\mathrm{H}$.

Determination of conjugate base of each acid

b. HBr loses a proton and forms ${\mathrm{Br}}^{-}$.Hence, the conjugate base of HBr is ${\mathrm{Br}}^{-}$.

${\mathrm{HSO}}_4^{-}$loses a proton and forms ${\mathrm{SO}}_4^{2-}$.Hence, the conjugate base of ${\mathrm{HSO}}_4^{-}$is ${\mathrm{SO}}_4^{2-}$.

${\mathrm{CH}}_3\mathrm{OH}$loses a proton and forms ${\mathrm{CH}}_3{\mathrm{O}}^{-}$. Hence, the conjugate base of ${\mathrm{CH}}_3\mathrm{OH}$is ${\mathrm{CH}}_3{\mathrm{O}}^{-}$.