
12 g of $C{{H}_{3}}COOH$ and 4 g of NaOH are mixed and diluted to 1 liter solution. If ${{P}^{{{K}_{a}}}}$ of $C{{H}_{3}}COOH$ is 4.8 the pH of the solution is
A. 4.8
B. 5
C. 5.6990
D. 5.3010
Answer
504k+ views
Hint: There is a relationship between pH and Pka of the solution and it is as follows.
\[pH={{P}^{{{K}_{a}}}}+\log \left[ \dfrac{salt}{base} \right]\]
Here pH= pH of the solution
${{P}^{{{K}_{a}}}}$ = Dissociation constant
Salt = concentration of the salt
Base = concentration of the base
Complete step by step solution:
- In the question it is given that 12 g of $C{{H}_{3}}COOH$ and 4 g of NaOH are mixed and diluted to 1 liter solution.
- ${{P}^{{{K}_{a}}}}$ of the solution is 4.8 .
- We have to calculate the pH of the solution.
- The chemical reaction of acetic acid with sodium hydroxide is as follows.
\[C{{H}_{3}}COOH+NaOH\to C{{H}_{3}}COONa+{{H}_{2}}O\]
- For 12 g of acetic acid the number of moles
\[\begin{align}
&\implies \dfrac{12}{60} \\
& \implies 0.2mole \\
\end{align}\]
- For 4 g of sodium hydroxide the number of moles
- Therefore 0.2 moles of acetic acid reacts with 0.1 moles of sodium hydroxide and forms 0.1 mole of sodium acetate as the product.
\[\begin{align}
&\implies \dfrac{4}{40} \\
&\implies0.1mol \\
\end{align}\]
- Now substitute all the known values in the below formula to get the pH of the solution.
\[pH={{P}^{{{K}_{a}}}}+\log \left[ \dfrac{salt}{base} \right]\]
Here pH= pH of the solution
${{P}^{{{K}_{a}}}}$ = Dissociation constant = 4.8
Salt = concentration of the salt = 0.1
Base = concentration of the base = 0.1
\[\begin{align}
& pH={{P}^{{{K}_{a}}}}+\log \left[ \dfrac{salt}{base} \right] \\
& pH=4.8+\log \left[ \dfrac{0.1}{0.1} \right] \\
& pH=4.8+0 \\
& pH=4.8 \\
\end{align}\]
- Therefore the pH of the solution is 4.8.
- So, the correct option is A.
Note: If the concentration of both salt and base are the same in a chemical reaction then the pH of the solution is equal to ${{P}^{{{K}_{a}}}}$ of the solution. Sodium acetate is a salt going to form by the reaction of sodium hydroxide with acetic acid.
\[pH={{P}^{{{K}_{a}}}}+\log \left[ \dfrac{salt}{base} \right]\]
Here pH= pH of the solution
${{P}^{{{K}_{a}}}}$ = Dissociation constant
Salt = concentration of the salt
Base = concentration of the base
Complete step by step solution:
- In the question it is given that 12 g of $C{{H}_{3}}COOH$ and 4 g of NaOH are mixed and diluted to 1 liter solution.
- ${{P}^{{{K}_{a}}}}$ of the solution is 4.8 .
- We have to calculate the pH of the solution.
- The chemical reaction of acetic acid with sodium hydroxide is as follows.
\[C{{H}_{3}}COOH+NaOH\to C{{H}_{3}}COONa+{{H}_{2}}O\]
- For 12 g of acetic acid the number of moles
\[\begin{align}
&\implies \dfrac{12}{60} \\
& \implies 0.2mole \\
\end{align}\]
- For 4 g of sodium hydroxide the number of moles
- Therefore 0.2 moles of acetic acid reacts with 0.1 moles of sodium hydroxide and forms 0.1 mole of sodium acetate as the product.
\[\begin{align}
&\implies \dfrac{4}{40} \\
&\implies0.1mol \\
\end{align}\]
- Now substitute all the known values in the below formula to get the pH of the solution.
\[pH={{P}^{{{K}_{a}}}}+\log \left[ \dfrac{salt}{base} \right]\]
Here pH= pH of the solution
${{P}^{{{K}_{a}}}}$ = Dissociation constant = 4.8
Salt = concentration of the salt = 0.1
Base = concentration of the base = 0.1
\[\begin{align}
& pH={{P}^{{{K}_{a}}}}+\log \left[ \dfrac{salt}{base} \right] \\
& pH=4.8+\log \left[ \dfrac{0.1}{0.1} \right] \\
& pH=4.8+0 \\
& pH=4.8 \\
\end{align}\]
- Therefore the pH of the solution is 4.8.
- So, the correct option is A.
Note: If the concentration of both salt and base are the same in a chemical reaction then the pH of the solution is equal to ${{P}^{{{K}_{a}}}}$ of the solution. Sodium acetate is a salt going to form by the reaction of sodium hydroxide with acetic acid.
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