Question

In decimolar solution, $\text{C}{{\text{H}}_{3}}\text{COOH}$ is ionised to the extent of 1.3 %. If log 1.3 = 0.11, pH value of the solution?
(A) 3.89
(B) 4.89
(C) 2.89
(D) Unpredictable

Answer 1

For this problem, firstly we have to write the reaction of ionisation of acetic acid after which we can write the equilibrium constant for the reaction. Due to which we can calculate the concentration of acid and then we will calculate the pH of the solution.

Complete step by step solution:
- In the given question, we have to calculate the pH of the solution of the acetic acid in which it ionises.
- As we know that in ionisation reactions, the reactant gives its respective ions. So, when acetic acid ionises it gives hydrogen ion and acetate ion as shown below:
$\text{C}{{\text{H}}_{3}}\text{COOH }\to \,\text{C}{{\text{H}}_{3}}\text{CO}{{\text{O}}^{-}}\text{ + }{{\text{H}}^{+}}$
- Now, it is given in the question that the concentration of acetic acid in solution is 1 Decimolar which means 0.01N.
- So, initially, the concentration of acetic acid is 0.1 and the concentration of the product is 0.
- But after some time, the concentration of the reactant will be $\text{0}\text{.01(1 - }\alpha \text{)}$ and the concentration of the products, $\text{C}{{\text{H}}_{3}}\text{CO}{{\text{O}}^{-}}\text{ and }{{\text{H}}^{+}}$ will be $\text{0}\text{.01}\alpha$ and $\text{0}\text{.01}\alpha$ respectively.
- Now, the expression of the equilibrium constant for the reaction will be:
${{\text{K}}_{\text{a }}}=\text{ }\dfrac{(\text{C}{{\text{H}}_{3}}\text{CO}{{\text{O}}^{-}})({{\text{H}}^{+}})}{(\text{C}{{\text{H}}_{3}}\text{COOH})}$
- Now, we can see that the concentration of the acid which is ${{\text{H}}^{+}}$ is $\text{C}\alpha$. So, by putting the value of C which is given 0.1N and alpha which is given 1.3% or $\text{1}\text{.3 }\times \text{ 1}{{\text{0}}^{-2}}$, we will get
$\text{C}\alpha \text{ = 0}\text{.01 }\times \text{ 1}\text{.3 }\times \text{ 1}{{\text{0}}^{-2}}\,\text{= 1}\text{.3 }\times \text{ 1}{{\text{0}}^{-3}}$.
- Now, we will calculate the pH of the solution that is:
$\text{pH = log(}{{\text{H}}^{+}}\text{)}$
$\text{pH = log(1}\text{.3 }\times \text{ 1}{{\text{0}}^{-3}}\text{)}$
$\text{pH = -log1}\text{.3 - log1}{{\text{0}}^{-3}}$
$\text{pH = 2}\text{.89}$

Therefore, option (C) is the correct answer.

Note: The dissociation constant tells us about the strength of the acid with which it can dissociate in the solution into its respective ions. Generally, the change in temperature and catalyst can affect the rate of dissociation of the acid.