Question

How would you write the formula of the conjugate base of the Bronsted-Lowry acids $H_{2}SO_{3}$ and HI ?

Answer 1

Hint: According to the Bronsted Lowry acid base theory, acid is a proton donor and base is a proton acceptor. When an acid loses a proton, a conjugate base is obtained. When a base accepts a proton, a conjugate acid is obtained.

Complete answer:
- The Bronsted-Lowry theory is based on an acid and base reaction.
- As we know that an acid is a species which has the capacity of donation a proton that is hydrogen ion is called Bronsted-Lowry acid.
- On the other hand, a base is a species which has the capacity of accepting protons and it needs to have a lone pair of electrons on the base which bonds to the hydrogen ion.
- According to this theory, the conjugate acid is a chemical species which is formed after the base accepts the hydrogen atom. A conjugate base is formed when an acid gives a proton.
- Since the conjugate base of an acid is determined by that compound which is left behind after the acid donates one of its acidic hydrogen atoms.
For $H_{2}SO_{3}$, the conjugate base is $HSO^{-}_{3}$
For HI, the conjugate base is $I^{-}$
The Bronsted-Lowry acid-base equation is shown below
B+HA\Leftrightarrow $HB^{+}+A^{-}$
Here, B = base proton receiver
     HA = acid proton donor
$HB^{+}$ = the conjugate acid
$HA^{-}$ = the conjugate base
For $H_{2}SO_{3}$,
${H_{2}}{SO_{3}}_{aq}+ {H_{2}}{O}_{l} \leftrightarrows {H_{3}O^{+}}_{aq}+{HSO^{-}_{3}}_{aq}$
Here, 2 protons are available to donate. $HSO^{-}_{3}$ is the compound that is left behind after the first hydrogen ion is donated. Thus, the conjugate base is $HSO^{-}_{3}$.
For HI ,
$HI_(aq)+H_{2}O_{l}\Leftrightarrow H_{3}O^{+}_(aq) +I^{-}_{aq}$
Similarly, the compound left behind is $I^{-}$.
Thus, the conjugate base is $I^{-}$.

Note: Bronsted-Lowry acid base theory is similar to Arrhenius concept of acid and base. The Arrhenius concept is limited to aqueous solution. However, Bronsted-Lowry theory can also be applied to nonaqueous solutions.