Question

Given a solution of acetic acid. How many times of the acid concentration, acetate salt should be added to obtain a solution with pH$=7$? (${{K}_{a}}$ for dissociation of $C{{H}_{3}}COOH$ $=1.8\times {{10}^{-5}}$)

Answer 1

Calculate the $p{{K}_{a}}$ of the given acid first. Then use the Henderson-Hasselbalch equation to obtain the answer. The equation is as follows:
$pH=p{{K}_{a}}+\log \frac{\left[ CB \right]}{\left[ Acid \right]}$

Complete answer:
Before diving right into the methods and equations required to solve this kind of question, first let us understand what it is actually asking. As you can see, it’s a buffer of a weak acid ($C{{H}_{3}}COOH$) and the sodium salt of that acid ($C{{H}_{3}}COONa$). The concentration of both of these compounds is unknown to us. But actually that is not required to find its answer. We just have to find the ratio of the concentration of salt to the acid.
At equilibrium, there would be two reactions taking place inside the mixture, they are as follows:
$C{{H}_{3}}COOH\rightleftharpoons C{{H}_{3}}CO{{O}^{-}}+{{H}^{+}}$
$C{{H}_{3}}COONa\to C{{H}_{3}}CO{{O}^{-}}+N{{a}^{+}}$
The equation (1.1) is at equilibrium because complete dissociation of the weak acids does not take place. The reaction (1.2) is a forward reaction where almost all of the reactants are changed into products. That means, the concentration of $C{{H}_{3}}CO{{O}^{-}}$, which is the conjugate base or CB will be same as the concentration of the sodium salt of acetic acid.
Let’s take the concentration of acid to be “x”. The concentration of the salt should be something multiplied to “x”. Let that something be the constant “A”, so the concentration of the salt is “Ax”, which is the same as that of the CB. We have to find the value of “A” in this question.
The given${{K}_{a}}$of the acid is$1.8\times {{10}^{-5}}$. It’s$p{{K}_{a}}$ can be calculated as follows:

& p{{K}_{a}}=-\log \left[ {{K}_{a}} \right]=-\log \left( 1.8\times {{10}^{-5}} \right) \\\ & \Rightarrow p{{K}_{a}}=4.745 \\\ \end{aligned}$$ The Henderson-Hasselbalch equation is as follows: $$pH=p{{K}_{a}}+\log \frac{\left[ CB \right]}{\left[ Acid \right]}$$ Putting the respective value in places, the above equation becomes: $$\begin{aligned} & 7=4.745+\log \frac{Ax}{x}=4.745+\log \left( A \right) \\\ & \Rightarrow \log \left( A \right)=7-4.745=2.255 \\\ & \Rightarrow A=\operatorname{antilog}\left( 2.255 \right) \\\ & \Rightarrow A=179.887\approx 180 \\\ \end{aligned}$$ So, the concentration of the acetate salt should be $180$ times that of the acid. **Note:** The conjugate base concentration that is included in the Henderson-Hasselbalch equation is specific to a particular acid. Therefore, if there is a mixture of many acids in a solution, each should have their own conjugate base forms. Approximation of a number should only be done when that number represents the final answer. This avoids unnecessary errors in the calculation.