I'd like to do a titration test, I've got a sulphuric acid solution @ 1M (2N) but I need one @ 0.12N. How many ml of the 1M (2N) solution do I need to add to distilled water to make up 100ml please?
I need 0.12N not M, I'm not sure of the scale change between M and N?
I made a couple of typos and I don't want this wrong info to stay around; therefore, I'm deleting my other post.
You want 0.12N.
Same set up.
2N*xmL = 0.12N x 100
xmL = 0.12*100/2 = 6 mL
I'm sure you understand that you add 6.00 mL of the 2N stuff to a 100 mL volumetric flask, make to the mark with DI water, mix thoroughly. You do NOT add 6.00 mL of the 2N stuff to 94 mL H2O
check:
2N x (6 mL/100 mL) = 0.12N
The proper conversion between M and N is
since 1M = 2N then 0.12N = 0.06M
I'm not a chemistry student and it was 40 years ago that I did any equations but...
A little more research has reveled that Sulphuric Acid has 2 protons (H+) in a molecule of the acid. So in this case the scales of Normality and Molarity 1M=2N or N=M/2
I was somewhat familiar with the molarity term but not normality.
Why can't you just add 6ml of 2N(1M) to 94ml DI water to get 100ml of 0.12N solution? Just curious :)
You can add 6 mL to 94 mL, but the total volume may or may not be exact. The method that Dr. Bob222 recommended is the correct method. If you were to add 6mL of the acid to volumetric flask and then add 94 mL of DI water to it, then the bottom of the meniscus may be above or below the 100mL line, making the concentration inexact. You want the proper amount of acid, or H+ ions, and you want the volume to be enough so that the finial volume will reflect the change to 0.12N from 2N.
Oh I see what you're saying, but my old chemistry teacher would have killed me if I'd added water to acid.
He would have told me to put at least 75% of the expected DI water content into the flask first then the measured amount of acid, then top up to 100ml.
Though I suppose it wouldn't really matter with the strength of these dilutions, doing it this way is best practice though.
Your chemistry teacher is correct. The reaction is exothermic, so adding water to acid will cause it to splash around and it may get on you, or someone else in the lab, which isn't good lab safety protocol. The proper way is to add the acid to water, and yes, you should add about 70mL of water before adding the acid, and then fill the volumetric flask to the 100mL line. I totally forgot to explain the method that way and was only concerned with the concentration.
One other thing: when you do this, make sure to shake it vigorously for at least 45 seconds if doing it by hand, but I would prefer that you use a vortexer for at least 45 seconds to make sure that it is mixed properly. I know when I needed to make dilutions or needed a dilute concentration for a lab protocol or an experiment that I was running in lab, that if my numbers weren't correct, that I probably messed up on one of the dilutions because I didn't mix it properly, which would cause me to stay in the lab all night repeating the experiment or protocol.
I've been absent for the last several hours; just to catch up I didn't bother with the 75% H2O first before adding the acid. 1M H2SO4 is so dilute it shouldn't cause any problems. However, your teacher and Devron are correct; that is the PROPER procedure.
Yes, 1M = 2N for H2SO4 BUT only if you are neutralizing BOTH H ions. For example,
H2SO4 + 2NaOH ==> Na2SO4 + 2H2O and the normality of H2SO4 in this case twice the molarity. However, if the reaction is
H2SO4 + NaOH ==> NaHSO4 + H2O then the M is the same as N because only one H is neutralized. And for whatever it's worth, my chemistry was more than 60 years ago. cheers.
Thanks for all your help, it's much appreciated and my titration of nicotine solution for my ejuice worked a treat :)
To determine the amount of 1M (2N) sulphuric acid solution needed to prepare a 0.12N solution, we can use the concept of normality (N) and the equation:
N1 * V1 = N2 * V2
where N1 is the normality of the starting solution, V1 is the volume of the starting solution, N2 is the desired normality, and V2 is the desired volume.
In this case, we have:
N1 = 2N
V1 = unknown (we need to find this)
N2 = 0.12N
V2 = 100 mL (as stated in the question)
Rearranging the equation, we have:
V1 = (N2 * V2) / N1
Substituting the given values:
V1 = (0.12N * 100 mL) / 2N
V1 = (12 mL * 100 mL) / 2N
V1 = 1200 mL/N
Therefore, you will need to add 1200 mL of the 1M (2N) sulphuric acid solution to distilled water in order to make up 100 mL of a 0.12N solution.