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We have to prepare a buffer solution whose concentration is already known as to be$0.063$molar. We also have been given the pH at which this buffer will be prepared. And we can calculate the $p{{K}_{a}}$of acetic acid from the acid dissociation constant given.

Let’s write the Henderson-Hasselbalch equation, so that you will get a better perspective of what we are trying to achieve here. It is as follows:

$pH=p{{K}_{a}}+\log \left( \dfrac{\text{Conjugate B}ase}{Acid} \right)$

The buffer consists of acetic acid and its sodium salt. Their respective equations will be as follows:

- For acetic acid

$C{{H}_{3}}COOH\rightleftharpoons C{{H}_{3}}CO{{O}^{-}}+{{H}^{+}}$

- For sodium acetate

$C{{H}_{3}}COONa\rightleftharpoons C{{H}_{3}}CO{{O}^{-}}+N{{a}^{+}}$

The acid here is obviously acetic acid ($C{{H}_{3}}COOH$) and the conjugate base here is the acetate ion ($C{{H}_{3}}CO{{O}^{-}}$). The concentration of the conjugate base is decided by the reaction at (1.3). This is so because unlike acetic acid which does not undergo a complete dissociation, the sodium acetate dissociates completely into its ions when dissolved in polar solvent. So the concentration of the conjugate base is equal to the concentration of sodium acetate.

Let the concentration of acetic acid be “x”, then the concentration of the conjugate base is $0.063-x$, as the total concentration of the buffer to be prepared is $0.063$molar.

$p{{K}_{a}}$of acetic acid is:

$\begin{align}

& =-\log \left( {{K}_{a}} \right) \\

& =-\log \left( 1.8\times {{10}^{-5}} \right) \\

& =4.744 \\

\end{align}$

Putting the respective values in equation (1.1), we get:

\[4.5=4.744+\log \left( \dfrac{0.063-x}{x} \right)\]

Solving the above equation we get $x=0.04M$, which is the concentration of acetic acid. Therefore the concentration of conjugate base is $0.063-x=0.023M$, which is the concentration of sodium acetate.

In an equilibrium reaction, the compound which dissociates the most is usually taken into consideration when required in calculations. As in the above case the concentration of conjugate base, which is the acetate ion, is equal to that of sodium acetate.

In the question the volume of solution is already mentioned to be one litre and so not extra calculations were needed. But if anything other than this value would have been mentioned, then we had to first calculate the actual number of moles of the buffer present and then do the other procedures.

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