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

1. Mar 26, 2016

### AMan24

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

M1V1 = M2V2

3. 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 doesnt 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 im missing or something is wrong with what i said please let me know

Last edited by a moderator: Mar 26, 2016
2. Mar 26, 2016

### Staff: Mentor

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.

3. Mar 26, 2016

### AMan24

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, thats going to narrow things down a lot.