Question

$1{\text{ mole}}$ of $C{H_3}COOH(p{K_a} = 4.7447)$ and $1{\text{ mole }}$of $C{H_3}COONa$ are dissolved in water to form $1{\text{ litre}}$ aqueous solution. The $pH$ of the resulting solution will be :
A.) $9.2553$
B.) $4.7447$
C.) $14$
D.) $7$

Answer 1

Hint: To solve this question, we need to know the concept of buffer solution which is a mixture of weak acid and its conjugate base or which contain weak base and its conjugate acid. Then we have to apply a $pH$ formula for them.

Complete step by step answer:
In this answer first we will discuss the concept of buffer solution. Buffer solution is an aqueous solution which consists of a mixture of weak base and conjugate acid of that weak base or it may be a mixture of weak acid and conjugate base of that weak acid also. The conjugate acid is formed when a weak base loses or donate its proton (${H^ + }$) and the conjugate base is formed when a weak acid gains protons.
In the given question, $C{H_3}COOH$ is the weak acid and $C{H_3}CO{O^ - }$ is its conjugate base. Hence, we can say that $C{H_3}COOH$ and $C{H_3}COONa$ acts as a buffer solution because it consists of a weak acid and conjugate base of that acid.
The formula for the $pH$of the given buffer solution can be given as :
$pH = p{K_a} + \log \dfrac{{{\text{[Conjugate Base]}}}}{{{\text{[Acid]}}}}$ $ - (1)$
As given in the question both conjugate base and weak acid have the same concentration that is $1mole$and also $p{K_a} = 4.7447$. Now, by putting these values in equation $ - (1)$ we get:
$
  pH = 4.7447 + \log \left( {\dfrac{1}{1}} \right) \\
  pH = 4.7447 + 0 \\
  pH = 4.7447 \\
 $
Thus we can say that the required $pH$ value is $4.7447$.
Hence, option B is the correct answer.


Note:
In such questions, we can remember that if we want to find the conjugate base of an acid then we just need to remove the $({H^ + })$ that is proton from that acid and if we want to find the conjugate acid of a base then we need to add a proton to that base.