Chapter 3, Problem 5P

Interpretation Introduction

Interpretation:

The concentration of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}$ at the end of dialysis should be determined along with the process for further lower the concentration of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}$.

Concept introduction:

Ammonium Sulfate or ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}$ is an inorganic salt. This compound is non-hazardous to human. It is also known as Sulfuric acid diammonium salt. It is extensively used as a fertilizer in soil that contains 21% nitrogen and 24% sulfur.

Answer to Problem 5P

At the end of dialysis, the concentration is 0.001 M ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}$ .The further lowering the concentration of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}$ can be done by dialyzing the sample to 1mM, in a buffer-free solution of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}$.

Explanation of Solution

Given:

Precipitation of a protein with 1 M ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}$.

1 ml of the sample is dialyzing in 1000 ml of buffer.

To find the concentration or volume of the concentrated or dilute solution the following equation is used:

${\text{M}}_{\text{1}}{\text{V}}_{\text{1}}{\text{ = M}}_{\text{2}}{\text{V}}_{\text{2}}$

${\text{M}}_{\text{1}}$ and ${\text{M}}_2$ = Molarity of the solution.

${\text{V}}_{\text{1}}$ and ${\text{V}}_2$ = Volume of the solution.

The concentration of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}$ after the dialysis can be calculated as follows:

$\begin{array}{l}{\text{M}}_{\text{1}}\text{=}\text{1M}\text{of}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}\\ {\text{V}}_{\text{2}}\text{=}\text{1}\text{mL}\\ {\text{M}}_{\text{2}}\text{=}x\text{M}\text{of}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}\\ {\text{V}}_{\text{2}}\text{=}\text{1000}\text{ mL}\end{array}$

Put the above values in equation:

$\begin{array}{l}{\text{M}}_{\text{1}}{\text{V}}_{\text{1}}{\text{=M}}_{\text{2}}{\text{V}}_{\text{2}}\\ \text{1 M}\text{of}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}\text{×}\left(\text{1 mL}\right)\text{ = }x\text{ M}\text{of}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}\text{×}\left(\text{1000 mL}\right)\\ x\text{ =}\text{}\frac{\text{1}}{\text{1000}}\\ \text{=}0.\text{001}\text{M}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}\end{array}$

Therefore, at the end of dialysis the concentration is 0.001 M ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}$.

The process for further lowering the concentration of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}$ is as follows:

The concentration could be further dropped by dialyzing the sample again. It can be done by placing the dialysis tube into a fresh 1000 mL solution (a buffer having no

${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}$ ). In this whole process the dialysis tube must have 0.001 M ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}$. This process can be recurring multiple times for significantly decrease the concentration of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}$ or any other small solute or salt.

So, it could be done by dialyzing the sample to 1mM, in a buffer-free solution of (NH₄)₂SO₄.

Conclusion

At the end of dialysis, the concentration is 0.001 M ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}$ .The further lowering the concentration of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}$ can be done by dialyzing the sample to 1mM, in a buffer-free solution of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}$.