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.${\mathrm{HSO}}_4^{-}$

Short 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-by-step solution

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

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

$(acid)(conjugatebase)$

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 role="math" localid="1652597213992" ${\mathrm{SO}}_4^{-2}$.

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

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