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# 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 isA. 4.8 B. 5C. 5.6990D. 5.3010

Last updated date: 20th Jun 2024
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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.