Q. 10.12 - An Introduction to General, Organic, and Biological Chemistry

Question

Write the formula for the conjugate acid for each of the following bases:

a.${\mathrm{SO}}_4^{2-}$

b. $C{N}^{-}$

c. $N{H}_3$

d. $Cl{O}_2^{-}$

Step-by-Step Solution

Answer

(Part a)    $$\begin{gathered} H^{+}+S{O}_4^{-2}\rightleftarrows HS{O}_4^{-} \\ \left(base\right) (conjugate acid) \end{gathered}$$

(Part b)  $$\begin{gathered} H^{+}+C{N}^{-}\rightleftarrows HCN \\ \left(base\right) (conjugate acid) \end{gathered}$$

(Part c)   $$\begin{gathered} H^{+}+N{H}_3\rightleftarrows N{H}_4^{+} \\ \left(base\right) (conjugate acid) \end{gathered}$$

(Part d)  $$\begin{gathered} H^{+}+Cl{O}_2^{-}\rightleftarrows HCl{O}_2 \\ \left(base\right) (conjugate acid) \end{gathered}$$

Step 1: Introduction (Part a)

The given are the bases $S{O}_4^{2-},C{N}^{-},N{H}_3,Cl{O}_2^{-}$. To find  the formula for the conjugate acid for each of the given bases

Step 2: explanation of (Part a)

Conjugate acid is formed from the base by a gain of $\mathrm{H}+$ ion. ${\mathrm{SO}}_4^{2-}$gain one proton and form $HS{O}_4^{-}$ , it is the conjugate acid for ${\mathrm{SO}}_4^{2-}$

$$\begin{gathered} H^{+}+S{O}_4^{-2}\rightleftarrows HS{O}_4^{-} \\ \left(base\right) (conjugate acid) \end{gathered}$$

Step 3: explanation of (Part b)

Conjugate acid is formed from the base by a gain of $\mathrm{H}+$ ion. ${\mathrm{CN}}^{-}$gain one proton and form $HCN$ , it is the conjugate acid for $C{N}^{-}$.

$$\begin{gathered} H^{+}+C{N}^{-}\rightleftarrows HCN \\ \left(base\right) (conjugate acid) \end{gathered}$$

Step 4: explanation of (Part c)

Conjugate acid is formed from the base by the gain of $\mathrm{H}+$ an ion.  $N{H}_3$ gain one proton and form $N{H}_4^{+}$ , it is the conjugate acid for $N{H}_3$.

$$\begin{gathered} H^{+}+N{H}_3\rightleftarrows N{H}_4^{+} \\ \left(base\right) (conjugate acid) \end{gathered}$$

Step 5: explanation of (Part d)

Conjugate acid is formed from the base by a gain of $\mathrm{H}+\mathrm{ion}.{\mathrm{ClO}}_2^{-}$gain one proton and forms $HCl{O}_2$, it is the conjugate acid for $Cl{O}_2^{-}$.

$$\begin{gathered} H^{+}+Cl{O}_2^{-}\rightleftarrows HCl{O}_2 \\ \left(base\right) (conjugate acid) \end{gathered}$$