Acetic acid has a Ka of 1.8*10^-5. Three acetic acid/ acetate buffer solutions, A,B, and C, wer made using varying conce...

Given, ${K_a} = 1.8 \times {10^{ - 5}}$

The $p{K_a}$ of the acid is calculated as follows:

$\begin{array}{l}\\p{K_a} = - \log \left( {1.8 \times {{10}^{ - 5}}} \right)\\\\{\rm{ = 4}}{\rm{.74}}\\\end{array}$

Given that, for buffer A,

$\left[ {{\rm{Acetic acid}}} \right]$ is 10 times greater the concentration of $\left[ {{\rm{Acetate}}} \right]$ .

So,

$\begin{array}{l}\\\left[ {{\rm{Acetic acid}}} \right] = 10x\\\\\left[ {{\rm{Acetate}}} \right] = x\\\end{array}$

Substitute the values in the Henderson-Hasselbalch equation and calculate the ${\rm{pH}}$ of the buffer as follows:

$\begin{array}{l}\\{\rm{pH}} = 4.74 + \log \frac{x}{{{\rm{10}}x}}\\\\{\rm{ = 3}}{\rm{.74}}\\\end{array}$

Given that, for buffer B,

$\left[ {{\rm{Acetate}}} \right]$ is 10 times greater the concentration of $\left[ {{\rm{Acetic acid}}} \right]$ .

So,

$\begin{array}{l}\\\left[ {{\rm{Acetic acid}}} \right] = x\\\\\left[ {{\rm{Acetate}}} \right] = 10x\\\end{array}$

Substitute the values in the Henderson-Hasselbalch equation and calculate the ${\rm{pH}}$ of the buffer as follows:

$\begin{array}{l}\\{\rm{pH}} = 4.74 + \log \frac{{10x}}{x}\\\\{\rm{ = 5}}{\rm{.74}}\\\end{array}$

Given that, for buffer C,

$\left[ {{\rm{Acetic acid}}} \right] = \left[ {{\rm{Acetate}}} \right]$ .

So,

$\begin{array}{l}\\\left[ {{\rm{Acetic acid}}} \right] = x\\\\\left[ {{\rm{Acetate}}} \right] = x\\\end{array}$

Substitute the values in the Henderson-Hasselbalch equation and calculate the ${\rm{pH}}$ of the buffer as follows:

$\begin{array}{l}\\{\rm{pH}} = 4.74 + \log \frac{x}{x}\\\\{\rm{ = 4}}{\rm{.74}}\\\end{array}$

The ${\rm{pH}}$ of the three buffer solutions are as follows:

$\begin{array}{l}\\{\rm{Buffer A}}:3.74\\\\{\rm{Buffer B}}:5.74\\\\{\rm{Buffer C}}:4.74\\\end{array}$