How to prepare 0.1M 200mL HCl Solution using 6M HCl

In summary, the conversation discusses a problem with preparing a 200mL 0.1M solution using a 6M HCl solution and determining the molarity of HCl used in an experiment. The individual used 12 mL of HCl and added water to make a total volume of 200mL, resulting in a 0.36M solution instead of the desired 0.1M. The conversation also touches on potential sources of error and confusion in the calculations.
  • #1
AMan24
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2

Homework Statement


problem 1) so i have a 6M HCl solution, and i want to make it a 200mL 0.1M solution

problem 2) so for this experiment i was supposed to mix HCl and NaOH until they neutralized.
I have to confirm the molarity of HCl that was used using math. So i used 10mL of an unknown molarity of HCl, to neutralize 0.1M of either 15.2mL of NaOH or 5.1 mL of NaOH. I got 15.2mL of NaOH because its my data. I got 5.1 mL of NaOH because if i do 6M HCl x 12 mL HCl = 200 mL x (M2 HCl) = i get 0.36 M which is the molarity i used. So if i divide it by 3, i get 0.12 which is about what i should have used.

Homework Equations



M1V1 = M2V2

The Attempt at a Solution



1)
What i think it should be:

(0.1M)(0.2L) = (6M)(x)...

That gives 0.0033L or 3.3mL. Which i think means i should have used 3.3mL of the HCl and added water until it reached 200 mL

Now what i actually did and used in the lab was:

(0.2L)(6M) = (0.1M)(x)

Which gives 12L... or 12,000mL...2) The problem is; using the data from this calculation (0.1M)(0.2L) = (6M)(x)", the math doesn't add up. When i do 10mL HCl x (M1) = 0.1M NaOH x 5.1mL NaOH i get 0.051M HCl. Which doesn't make sense because it should be 0.1M

If i use the one that i think is wrong; "(0.2L)(6M) = (0.1M)(x)". 10mL HCl(M1)HCl = 0.1M NaOH(15.2)mL NaOH, i get 0.15M HCl, which is pretty close, and could be off due to experimental error. The 10mL, and 0.1M are constant. The M1 is always unknown, the v2 has two options, either 15.2mL (which should be wrong but its not), or the 5.1 mL (which should be right but is wrong).

This is bizarre... The only thing i can think of is that, to find the molarity of the HCl for the 2nd part, its not supposed to be M1V1 = M2V2 when its two different substances, but i have no idea what to do other than that.

- i just found out the proper way to determine part 2, with balancing equations and using stoichiometry. I get the same answer as doing m1v1 = m2v2, because the HCl and NaOH have a 1:1 ratio so it doesn't matter. But the answers are still not adding up. The wrong version, aka, (0.2L)(6M) = (0.1M)(x), is giving me the correct answer... This seriously makes 0 sense.If there is any data I am missing or something is wrong with what i said please let me know
 
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  • #2
Half of the problem is what you do and how you write about it is so chaotic it is impossible to follow. It is not clear what you did, why and what for.

AMan24 said:
so i have a 6M HCl solution, and i want to make it a 200mL 0.1M solution

No, you don't want to "make it a 200 mL 0.1 M solution. You want to use it to prepare 200 mL of 0.1 M solution.

And you are right you should use 3.3 mL of the concentrated acid and dilute it to 200 mL.

AMan24 said:
Now what i actually did and used in the lab was:

(0.2L)(6M) = (0.1M)(x)

Which gives 12L... or 12,000mL

Are you saying you have prepared 12 L of the diluted acid? Calculations are right, yes you can prepare 12 L of 0.1 M solution from 200 mL of 6M solution, but that's not what you were asked to do.
 
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  • #3
Borek said:
Half of the problem is what you do and how you write about it is so chaotic it is impossible to follow. It is not clear what you did, why and what for.
No, you don't want to "make it a 200 mL 0.1 M solution. You want to use it to prepare 200 mL of 0.1 M solution.

And you are right you should use 3.3 mL of the concentrated acid and dilute it to 200 mL.
Are you saying you have prepared 12 L of the diluted acid? Calculations are right, yes you can prepare 12 L of 0.1 M solution from 200 mL of 6M solution, but that's not what you were asked to do.

I used 12 mL of HCl and added 188 mL of water for a total of 200 mL of water. So if i plug it into the formula it should be (12 mL HCl) x (6M HCl) = (200mL HCl) x M2, and that gives me 0.36M. So the solution i made is 0.36M when it should have been 0.1M. And thanks for confirming the 3.3mL, that's going to narrow things down a lot.
 

1. How do I calculate the amount of 6M HCl needed to make a 0.1M solution?

To prepare a 0.1M HCl solution, you will need to dilute the concentrated 6M HCl solution. The formula for dilution is M1V1 = M2V2, where M1 and V1 are the initial molarity and volume, and M2 and V2 are the final molarity and volume. In this case, you will need 200mL of the 0.1M solution, so you can rearrange the formula to find the volume of 6M HCl needed: V1 = (M2V2)/M1 = (0.1M)(200mL)/(6M) = 3.33mL. Therefore, you will need to measure 3.33mL of the 6M HCl solution and add it to 196.67mL of water to make a 0.1M HCl solution with a total volume of 200mL.

2. How do I measure 3.33mL of the concentrated 6M HCl solution?

Measuring small volumes accurately can be challenging, especially with concentrated solutions. The best way to measure 3.33mL of the 6M HCl solution is to use a graduated pipette or a volumetric pipette. These types of pipettes have a specific volume marked on them and are designed to deliver that exact volume. You can also use a burette, which allows you to measure and deliver small volumes accurately through a stopcock.

3. Can I use any type of water to dilute the 6M HCl solution?

No, it is important to use distilled or deionized water when preparing solutions in the lab. Tap water or other types of water may contain impurities that can affect the accuracy of your solution. Distilled or deionized water has been purified and is free from any impurities, making it the best choice for diluting solutions.

4. How do I mix the 6M HCl solution and water to prepare the 0.1M solution?

To mix the 6M HCl solution and water, you can use a volumetric flask. First, measure out 196.67mL of distilled or deionized water and pour it into the volumetric flask. Then, add the 3.33mL of the 6M HCl solution to the flask. Swirl the flask gently to mix the solutions together. It is important to be gentle when mixing to avoid introducing any air bubbles into the solution.

5. How should I store the prepared 0.1M HCl solution?

The 0.1M HCl solution should be stored in a tightly sealed container, away from direct sunlight and heat sources. It is also important to label the container with the date, concentration, and any other relevant information. This solution should be used within a few days of preparation, as it may lose potency over time due to evaporation or reaction with air. If any changes in appearance or smell are noticed, the solution should not be used and should be properly disposed of.

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