Q. 10.9 - An Introduction to General, Organic, and Biological Chemistry

Question

Write the formula for the conjugate base for each of the following acids:

a. $\mathrm{HF}$

b. ${\mathrm{H}}_2\mathrm{O}$

c. ${\mathrm{H}}_2{\mathrm{PO}}_3^{-}$

d. ${\mathrm{HSO}}_4^{-}$

Step-by-Step Solution

Answer

The conjugate bases for the following acids are:

(Part a) ${\mathrm{F}}^{-}$

(Part b)  ${\mathrm{OH}}^{-}$

(Part c)  ${\mathrm{HPO}}_3^{-}$

(Part d)  ${\mathrm{SO}}_4^{-2}$

Step 1- Introduction (Part a)

According to the Bronsted-Lowery theory, an acid-base bond pair consists of molecules with differing  $\mathrm{H}+$ions. When an acid loses $1$${\mathrm{H}}^{+}$, it forms a conjugate base, and when an acid gains $1$${\mathrm{H}}^{*}$, it forms a conjugate acid. 

Step 2- Explanation (part -a)

a.

A conjugate base is formed in an acid when the $\mathrm{H}+$ ion is lost in the acid.

$HF$ loses one proton to form $F ¯$. This is the conjugate base of $HF$. 

$HF\rightleftarrows {\mathrm{H}}^{+}+{\mathrm{F}}^{-}$

$\left(acid\right)$            (conjugate base)

Step 3- Explanation (part b)

b.

Conjugate bases are formed in water when $\mathrm{H}+$ ions are lost in water.  loses one proton to form $\mathrm{OH} ¯$.

This is the conjugate base of ${\mathrm{H}}_{2}O$.

${\mathrm{H}}_2\mathrm{O}\rightleftarrows {\mathrm{H}}^{+}+{\mathrm{OH}}^{-}$

   (acid)            (conjugate base)

Step 4- Explanation (part c)

c.

A conjugate base is formed ${\mathrm{H}}_2{\mathrm{PO}}_3^{-}$at  when the $\mathrm{H}+$ ion is lost in a given acid. ${\mathrm{H}}_2{\mathrm{PO}}_3^{-}$ loses one proton to form ${\mathrm{HPO}}_3^{-}$ .

This is the conjugate base of  ${\mathrm{H}}_2{\mathrm{PO}}_3^{-}$.

${\mathrm{H}}_2{\mathrm{PO}}_3^{-}\rightleftarrows {\mathrm{H}}^{+}+{\mathrm{HPO}}_3^{-}$

$( acid ) (conjugate base)$

Step 5- Explanation (part d)

d.

A conjugate base is formed at ${\mathrm{HSO}}_4^{-}$ when the  $\mathrm{H}+$ ion is lost in a given acid. ${\mathrm{HSO}}_4^{-}$ loses one proton to form ${\mathrm{SO}}_4^{-2}$.

This is the conjugate base of ${\mathrm{HSO}}_{4.}^{-}$.

$\begin{array}{ll}{\mathrm{HSO}}_4^{-}\rightleftarrows {\mathrm{H}}^{+}+{\mathrm{SO}}_4^{-2}& \\ \left(\text{ acid }\right)& \text{ (conjugate base) }\end{array}$