Question

The pH of a weak monobasic acid is 5. The degree of ionisation of acid in 0.1 M solution is:
[A] ${{10}^{-4}}$
[B] ${{10}^{-3}}$
[C] ${{10}^{-2}}$
[D] ${{10}^{-1}}$

Answer 1

Hint: To solve this, firstly write down the reaction of dissociation of a weak monobasic acid. Assume the degree of dissociation to be alpha and from the definition of pH you can find out the alpha by the help of the given pH.

Complete answer:
To answer this, firstly let’s understand what a monobasic acid is and what it means by its ionisation.
Ionisation is the process by which atom or a molecule gains or loses an electron and obtains a negative or a positive charge. The degree of ionization (or dissociation) is basically a way of representing the strength of an acid.
On ionisation with water the acids which produce a single hydronium ion i.e. readily ionises only one proton, are known as monobasic acids.
Now, to find the correct answer to this question, let us take an example of a monobasic acid like HA.
We can write its dissociation equation as-
     \[HA\to {{H}^{+}}+{{A}^{-}}\]
Let us assume its degree of dissociation is $\alpha $ , therefore, we can write that-

	HA	${{H}^{+}}$	${{A}^{-}}$
Moles before dissociation	0.1	0	0
Moles after dissociation	0.1 (1 - $\alpha $)	0.1$\alpha $	0.1$\alpha $

Now, we know that pH is the measure of acidity of the solution. It is equal to the negative log of the molar concentration of hydronium ions released in the solution. We can write is as-
     $pH=-\log [{{H}^{+}}]$
The pH of the solution is given to us as 5. It means,
     $[{{H}^{+}}]={{10}^{-5}}=0.1\alpha $
From here, we can calculate the degree of dissociation/ ionisation that will be ${{10}^{-4}}$.

Therefore, the correct answer is option [A] ${{10}^{-4}}$ .

Note:
While solving such questions remember to take into account the nature of the acid i.e. whether it is monobasic, dibasic or tribasic. The number of ${H^+}$ ions in the solution affect the acidity of the solution and hence the pH. So, if a dibasic acid was given in the question, then the degree of ionisation for ${H^+}$ would be double of what it is right now.