Question

How do you calculate the ${{K}_{a}}$ of an acid?

Answer 1

Hint The ${{K}_{a}}$ of an acid is calculated when the acid is dissolved in water. $K$ is an equilibrium constant and ${{K}_{a}}$is the equilibrium constant for an acid. When the acid is dissolved in water then there is the formation of a proton and an anion.

Complete step by step answer:
An acid is a compound whose pH is less than 7 and it can give protons in the solution.
K is the equilibrium constant and it is equal to the ratio of the concentration of the product to the concentration of the reactant.
Suppose, a reaction is given below:
$aA+bB\to cC+dD$
Where A and B are the reactants and a and b are the moles of A and B. C and D are the products, and c and d are the moles of C and D. The equilibrium constant for this reaction will be:
$K=\dfrac{{{[C]}^{c}}{{[D]}^{d}}}{{{[A]}^{a}}{{[B]}^{b}}}$
So, when this equilibrium constant is written for an acid then this is called the equilibrium constant of acid and it is represented as ${{K}_{a}}$. Suppose HA is acid and in water, its reaction will be:
$HA\rightleftharpoons {{H}^{+}}+{{A}^{-}}$
The equilibrium constant of acid for this reaction will be:
${{K}_{a}}=\dfrac{[{{H}^{+}}][{{A}^{-}}]}{[HA]}$
All the components must be in concentration.
For example, a 0.10 M of acid ionizes 15% in the solution, so the $[HA]$ will be $0.985\text{ M}$ and both the products will be $0.015\text{ M}$

The ${{K}_{a}}$ will be:
${{K}_{a}}=\dfrac{[{{H}^{+}}][{{A}^{-}}]}{[HA]}=\dfrac{(0.015)(0.015)}{0.985}=2.28\text{ x 1}{{\text{0}}^{-2}}$

Note: In the same process we can calculate the equilibrium constant for a base which is represented as ${{K}_{b}}$. Suppose BOH is a base and when it is dissolved in water, the reaction will be:
$BOH\rightleftharpoons {{B}^{+}}+O{{H}^{-}}$
The value of the equilibrium constant of the base will be:
${{K}_{b}}=\dfrac{[{{B}^{+}}][O{{H}^{-}}]}{[BOH]}$