Sunday, December 24, 2017

Dilution Lab Problems and Solutions (Answers)

Dilution Lab Problems and Solutions (Answers)

In the laboratory, you are often asked to prepare dilutions of solutions. These dilution lab problems can sometimes leave you grasping for solutions. Well, fret no more, in this post, simple dilution problems are given with the corresponding answers below. I hope you enjoy this learning process.

Instructions:


For the following problems, identify the given and the unknown. State what formula could be used, and show your computations.

1.    How do you prepare a 1:4 dilution of HCl?
2.    What’s the volume of diluent (NSS) needed to prepare a 1:5 serum dilution with a total of 5 mL?
3.    How do you prepare a Normal Saline Solution (NSS)?
4.    What is the resulting dilution for each tube in this serial dilution?

Tube No.    Volume of stock standard solution in mL    Volume of diluent in mL    Resulting Dilution
1           
2           
3           

5.    How much volume do you need to prepare a 5 ml of a 1:10 dilution?


ANSWERS

1.    How do you prepare a 1:4 dilution of HCl?

A 1:4 dilution indicates that there for every 1 part of the solute, there are 3 parts of the solvent. 4 indicate the total of the parts of the solute and the solvent. The solute is HCl (hydrochloric acid) and the solvent would be distilled water.

The easiest method is to assume that I part = 1 mL (milliliter), hence, 1 part is = 1mL HCl

If 1 part = 1 mL
Hence:
3 parts = 3 mL

When you add the parts, the total is 4, hence, the dilution 1:4

So, to prepare 1:4 dilution, you add 1 mL of HCL to 3 mL of distilled water.


2.    What’s the volume of diluent (NSS) needed to prepare a 1:5 serum dilution with a total of 5 mL?

NSS or Normal Saline Solution is also 85% saline. This is also very simple to solve, since 5 is the total of the solute and diluent parts, you can assume that 1 part = 1 mL.

So, 4 parts is required to complete the 5 parts.

Since 1 part =1 mL
So, 4 parts = 4 mL

Since the total volume is also 5 mL, then by adding 1 mL of serum + 4 mL of NSS; you, therefore, need 4 mL of NSS to prepare the dilution.

3.    How do you prepare a serum dilution of 1:3, if your available serum is only 0.25 mL?

1:3 – indicates 1 part of serum + 2 parts of diluent

Since the available solute or serum volume is only 0.25 mL, you have to equate this to 1 part of the dilution.

Hence, if 1 part = 0.25 mL

Therefore, 2 parts diluent = 0.25 mL x 2 parts = 0.50 mL

Hence, to prepare a serum dilution of 1:3, you can add:
0.25 mL of serum + 0.50 mL of diluent.

1 part + 2 parts = 3 parts

So, the dilution is 1:3


4.    What is the resulting dilution for each tube in this serial dilution?

Tube No.    Volume of standard solution in mL    Volume of diluent in mL    Resulting Dilution
1    0.5 (pure stock soln.)    1     1:3
2    0.5 from tube #1    1    1:9
3    0.5 from tube #2 (mix and discard 0.5 mL)    1    1:27

5.    How much volume do you need to prepare a 5 ml of a 1:10 dilution of standard solution?
1:10 dilution indicates what?

Yes, it indicates that 1 part of standard stock solution is added to 9 parts of diluent.

You can do the easiest method by adding 1 mL of the standard stock solution plus 9 mL of the diluent.

However, since the total volume is stated, which is 5 mL, you have to determine how many parts would the solution consists of.

You can divide 5 mL by 10, to determine the volume of each part.

Hence, 5/10 = 0.5 mL

So, you can now equate 1 part with 0.5 mL
The 9 parts diluent would therefore be = 9 x 0.5 = 4.5 mL

So, you add 0.5 mL standard stock solution to 4.5 mL diluent to come up with a 5 mL total volume of 1:10 standard dilution.

There you go! It’s relatively easy, if you always remember that the dilution factor (DF) is the total of 1 part of the solute and the designated parts of the solvent.

If I say the DF is 5, I’m also stating that the dilution is 1:5. So, there are 1 part of solute + 4 parts of solvent.  

For 1:9

There is one part of solute + 8 parts of the solvent


Dilution is different from ratio because in ratio the numbers remain the same. You don’t add them. Unlike in dilution, you add the parts of the solute plus parts of the solvent.

If you’re performing serial dilution, remember to multiply the previous dilution of the tube from where you got the solute.

Good luck with your laboratory math during your exams or when you’re working.

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