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# 50mL of 2N acetic acid mixed with 10mL of 1N sodium acetate solution will have an approximate pH: A) 7 B) 6 C) 5 D) 4

Last updated date: 16th Jun 2024
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Hint: Buffer solution is a mixture of a weak acid and its conjugate base or weak base and its conjugate acid. The pH of the solution changes by little value when a small amount of strong acid or base is added to the buffer solution. The Henderson-Hasselbalch equation relates the pH with dissociation constant and with the concentration of acid and its conjugate base or vice versa.

The Henderson-Hasselbalch equation establishes the relation between pH or pOH with a dissociation constant of weak acid or base with the concentration of acid and its conjugate base or a weak base and its conjugate acid. The Henderson-Hasselbalch equation for a weak acid is as follows:
$\text{pH=pKa+lo}{{\text{g}}_{\text{10}}}\dfrac{\left[ \text{conjugate Base} \right]}{\left[ \text{Acid} \right]}$
Where,
pH=measure of acidity of buffer solution
$\text{pKa=-logKa}$
$\text{Ka}$ is equal to the dissociation constant of acid
$\left[ \text{Acid} \right]$ is the concentration of acid
$\left[ \text{conjugate Base} \right]$ is the concentration of conjugate base of the acid
According to the given data,
$\text{Concentration of acetic acid }\!\!~\!\!\text{ =2N}$
$\text{Volume of acetic acid=50mL}$
$\text{Concentration of sodium acetate }\!\!~\!\!\text{ =1N}$
$\text{Volume of sodium acetate=10mL}$
Let’s first calculate the concentration of acetic acid and sodium acetate,
Here, the Total volume is ${{\text{V}}_{\text{1}}}=\text{50mL+10mL=60mL}$ as acetic acid and sodium acetate-mix together to form a buffer.

Use the formula of,${{\text{N}}_{\text{1}}}{{\text{V}}_{\text{1}}}\text{=}{{\text{N}}_{\text{2}}}{{\text{V}}_{\text{2}}}$ to find out the concentration of acid and its conjugate base
1) The concentration of acetic acid, ${{\text{N}}_{\text{1}}}\text{=}\dfrac{{{\text{N}}_{\text{2}}}{{\text{V}}_{\text{2}}}}{{{\text{V}}_{\text{1}}}}\text{=}\dfrac{\text{50 }\!\!\times\!\!\text{ 2}}{\text{60}}=\dfrac{5}{3}$
2) The concentration of sodium acetate, ${{\text{N}}_{\text{1}}}\text{=}\dfrac{{{\text{N}}_{\text{2}}}{{\text{V}}_{\text{2}}}}{{{\text{V}}_{\text{1}}}}\text{=}\dfrac{\text{10 }\!\!\times\!\!\text{ 1}}{\text{60}}=\dfrac{1}{6}$
3) The dissociation constant for acetic acid $\text{Ka}$ is $\text{1}{{\text{0}}^{\text{-5}}}$
Therefore,
\begin{align} & pKa=-{{\log }_{10}}Ka \\ & \text{ =}-{{\log }_{10}}({{10}^{-5}}) \\ & \text{ = 5} \\ \end{align}
Let’s put value in the Henderson-Hasselbalch equation. We get,
$\text{pH=5}\text{.0+lo}{{\text{g}}_{\text{10}}}\dfrac{\left[ \dfrac{1}{6} \right]}{\left[ \dfrac{5}{3} \right]}$
$\text{pH=5}\text{.0+lo}{{\text{g}}_{\text{10}}}\left[ \dfrac{1}{6} \right]\left[ \dfrac{3}{5} \right]$
$\text{pH=5}\text{.0+lo}{{\text{g}}_{\text{10}}}\left[ \dfrac{1}{10} \right]$
Since, $\text{lo}{{\text{g}}_{\text{10}}}\left[ \dfrac{1}{10} \right]=-1.0$
$\text{pH=5}\text{.0-1}\text{.0}$
$\text{pH=4}\text{.0}$
Therefore the pH of the acetic acid buffer is 4.0.
So, the correct answer is “Option D”.