Question

100ml of 0.1M NaOH is added to 100ml of a 0.2M $\text{C}{{\text{H}}_{3}}\text{COOH}$solution. The pH of resulting solution will be (${{\text{K}}_{a}}$ for $\text{C}{{\text{H}}_{3}}\text{COOH=1}\text{.8 x 1}{{\text{0}}^{-5}}$)
(a). 4.47
(b). 5.74
(c). 3.74
(d). 7.00

Answer 1

Hint: This problem can be solved by using Henderson-Hasselbach equation in which we can determine the exact amount of acid and conjugate base needed to make a buffer of a certain pH. Here, addition of NaOH and $\text{C}{{\text{H}}_{3}}\text{COOH}$ makes a buffer solution.

Complete step by step solution:
> A buffer is a solution that can resist pH change upon the addition of acidic or basic components. It is able to neutralize small amounts of added acid or base, thus maintaining the pH of the solution relatively stable.
> There are a couple of ways to prepare a buffer solution of a specific pH. Here, the method i.e. using the Henderson-Hasselbach equation:
\[\text{pH=p}{{\text{K}}_{a}}+\log \dfrac{[{{\text{A}}^{-}}]}{[HA]}\]
> Where pH is the concentration of $[{{\text{H}}^{+}}]$, $\text{p}{{\text{K}}_{a}}$ is the acid dissociation constant, and ${{\text{A}}^{-}}$ and $\text{H}{{\text{A}}^{-}}$ are concentrations of the conjugate base and starting acid.
> The pH of the solution by mixing 100ml of 0.2M $\text{C}{{\text{H}}_{3}}\text{COOH}$ and 100ml of 0.1M NaOH will be by using the formula-
\[pH=p{{K}{a}}+\log\dfrac{[salt]}{[acid]}\]
Now, \[\text{C}{{\text{H}}_{3}}\text{COOH+NaOH}\to \text{C}{{\text{H}}_{3}}\text{COOH+}{{\text{H}}_{2}}\text{O}\]
Now, we will use the formula to calculate pH
\[pH=4.74+\log1 =4.74\]
Therefore, the correct option is (a) 4.47

Note: We should also know that higher the value of ${{\text{K}}_{a}}$, more will be the formation of ${{\text{H}}^{+}}$ ion and lower will be the pH of the solution. For $\text{p}{{\text{K}}_{a}}$, it is the opposite case. Higher the $\text{p}{{\text{K}}_{a}}$, lower will be the pH that i.e. the molecule will be acidic in nature.