Dialysis dilution question

[Jikx]

I just can't quite get my head around this one.. and i suspect that the question or my head may have an error!

You have an enzyme sample of 3.0M with respect to the salt, ammonium sulphate. You are aware that in order to eliminate interference by this salt in both the assays of protein content and the enzyme activity, the concentration of ammonium ions must be less than 1mM.

To achieve this requirement, an enzyme solution (10ml) was dialysed against 1litre of distilled water at 4 degrees C.

After dialsis with stirring for 4 hours, the concentration in the external solution had reached 0.03mmole/ml.

How many changes of water would you need in order for the ammonium sulphate to be sufficiently low (>1mM) ?

Sounds a bit like dilution eh? The way i tried to work it out was

1) The 10ml sample contained 0.03 moles of salt (n=CV)

2) The water removes salt at 0.03 (concentration is 0.03M, in 1L of water) moles per water change

3) The required level of ammonium sulfate in the internal solution is 10^-5 moles

And I'm already ran into trouble... the external solution has already removed all the salt in the solution in one go! Somethign is wrong here... thanks for anyones help!

 

[anathema]

Thank you for your question. It seems like you are on the right track with your approach to dilution. However, there are a few things that may be causing confusion. Firstly, the concentration of 3.0M refers to the enzyme sample with respect to the salt, not the salt alone. This means that the concentration of ammonium sulphate in the sample is 3.0M, not the concentration of salt alone. Secondly, the concentration of 0.03mmole/ml refers to the concentration of ammonium sulphate in the external solution after dialysis, not the concentration of salt.

To solve this problem, we can use the equation C1V1 = C2V2, where C1 and V1 are the initial concentration and volume of the enzyme sample, and C2 and V2 are the final concentration and volume after dilution. In this case, we know that C1 = 3.0M, V1 = 10ml, C2 = 0.03mmole/ml, and V2 = 1000ml (1L). Solving for V1, we get V1 = (C2V2)/C1 = (0.03mmole/ml * 1000ml)/3.0M = 10ml. This means that you would need to dilute the enzyme sample with 10ml of water in order to reach the desired concentration of ammonium sulphate in the internal solution.

However, this may not be the most accurate or efficient way to achieve the required concentration. Since the external solution has already reached the desired concentration after one water change, it may be more effective to simply replace the external solution with fresh water multiple times until the desired concentration is reached. In this case, you would need to replace the external solution approximately 33 times in order to reach a concentration of 1mM ammonium sulphate.

I hope this explanation helps you better understand the problem. If you have any further questions, please don't hesitate to ask. Keep up the good work in your scientific endeavors!

 

[M98Ranger]

It seems like there may be an error in the question or in your calculations. The initial concentration of the enzyme sample is given as 3.0M with respect to the salt, ammonium sulphate. However, in step 1) of your calculation, you assume that the sample contains 0.03 moles of salt. This does not match the given initial concentration of 3.0M.

Additionally, it is not clear what you mean by "the water removes salt at 0.03 moles per water change." This statement is not supported by the information given in the question.

To accurately answer the question, we would need more information about the initial concentration of the salt and the rate of removal during dialysis. Without this information, it is difficult to determine the number of water changes needed to reach a concentration of less than 1mM. It is possible that the question or the given information may have an error.