Answer
Verified
394.5k+ views
Hint: We can calculate the$p{{K}_{a}}$of acetic acid from the given data. Then use the Henderson-Hasselbalch equation to solve the problem.
Complete step by step answer:
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.
Note:
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.
Complete step by step answer:
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.
Note:
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.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
Trending doubts
State the differences between manure and fertilize class 8 biology CBSE
Why are xylem and phloem called complex tissues aBoth class 11 biology CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
What would happen if plasma membrane ruptures or breaks class 11 biology CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What precautions do you take while observing the nucleus class 11 biology CBSE
What would happen to the life of a cell if there was class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE