Question

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

Answer 1

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.

Complete answer:
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, \[\text{lo}{{\text{g}}_{\text{10}}}\left[ \dfrac{1}{10} \right]=-1.0\]
Let’s substitute the value of log in the equation we get,
\[\text{pH=5}\text{.0-1}\text{.0}\]
\[\text{pH=4}\text{.0}\]
Therefore the pH of the acetic acid buffer is 4.0.
So, the correct answer is “Option D”.

Additional Information:
Limitations of Henderson-Hasselbalch equation:
It is assumed that at equilibrium the concentration of acid and its conjugate base remains constant.
The hydrolysis of water and its dependency on the pH for the solution was neglected.
For the sake of simplicity, Hydrolysis of base and acid dissociation was neglected.
The equation is applicable for weak acid and base and it may fail when dealing with strong acid and base.

Note: pKa values are quantitative measurements for the strength of the acid. A weak acid has pKa values in the range of -2 to 12 for strong acid it is less than -2. In solving such types of questions always remember to add the volumes of acid and its conjugate base which is provided into the question to get the final volume of the buffer solution.