Question

How to calculate the pKa of HCl?

Answer 1

Hint The substance that furnishes $H^{+}$ ion when dissolved in water is known as acid. We know HCl is a strong acid. pKa measures how tightly a proton is held by the Bronsted acid.

Complete step by step answer:
According to Bronsted Lowry 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. We know that $p{K}_a$ is the measure of how tightly a proton is held by the Bronsted acid. The lower the$p{K}_a$ , the more easily it gives up its proton.
Let's dissolve HCl in water:
$$HCl+H_{2}O\to C{l}^{-}+H_3{O}^{+}$$
From the above mentioned reaction we can say that HCl has donated a proton ($H^{+}$) to the water, and the water has accepted the proton. According to Bronsted Lowry theory HCl is a Bronsted acid, and the water is a Bronsted base.
We can calculate $K_a$ of the above reaction by taking the ratio of concentration of product by reactant.
$$K_a={\displaystyle \frac{[H_3{O}^{+}][C{l}^{-}]}{[HCl]}}$$
We know that HCl is a strong acid, so the value of $K_a$will be very large.
$$K_a={\displaystyle \frac{{10}^7}{1}}$$
The relationship between $K_a$ and $p{K}_a$is given by:
$p{K}_a=-\log K_a$
$p{K}_a=-\log ({10}^7)$
$$p{K}_a=-7$$

Thus, $p{K}_a$ is -7.

Note: $p{K}_a$ is the measure of the strength of the bronsted acid. In this case the $p{K}_a$ is very low which means HCl will easily give out its proton. 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.