Question

How does the conjugate base differ from the acid HBr?

Answer 1

Hint To answer this we should understand the Bronsted Lowry theory. Normally an acid means the substance that furnishes ${{H}^{+}}$ ion when dissolved in water. We know HBr is a strong acid .

Complete step by step solution:
Bronsted Lowry theory was introduced in 1923 by Danish chemist Johannes Nicolaus Bronsted and the English chemist Thomas Martin Lowry. This is known as proton theory of acids and bases. According to this theory, the compound that can transfer a proton to another compound is an acid and the compound that accepts the proton is a base. Basically an acid is a proton donor and a base is a proton acceptor.
Let's account for HBr and understand this concept better.
Firstly, HBr is allowed to react with water:
\[HBr+{{H}_{2}}O\to B{{r}^{-}}+{{H}_{3}}{{O}^{+}}\]
From the above mentioned reaction we can say that HBr has donated a proton (${{H}^{+}}$) to the water, and the water has accepted the proton. According to Bronsted Lowry theory HBr is a Bronsted acid, and the water is a Bronsted base.
When the acidic substance loses an ${{H}^{+}}$ ion that is a proton as per Bronsted Lowry theory, it forms a base known as conjugate base of an acid. Similarly, when a basic substance gains a proton it forms an acid known as conjugate acid of base.

Thus, HBr lost a proton to form $B{{r}^{-}}$. So, $B{{r}^{-}}$ is the conjugate base and ${{H}_{3}}{{O}^{+}}$ is the conjugate acid of HBr.

Note: An element loses or gains electrons in the outermost shell to complete its octet. As the elements that have fully filled octet will be stable chemically. In the Bronsted Lowry theory, a conjugate base is whatever is left over after the proton has left. So, in our case $B{{r}^{-}}$ is the conjugate base.