CHRIS25 - 24-5-2012 at 07:34
Hi, after I have put a 1mL drop of Copper Chloride into a glass with 25mL of warm tap water. I place 1mL of a 1 mole NaOH via syringe - squirting
slowly into the solution. Ok - Question: It turns milky BUT, do I swirl the glass to see how long it takes to become clear again OR do I keep the
syringe slowly moving without agitating the solution too much. I have done this so many times but am unsure how to best interpret the amount of HCL
left in solution, agitate and it dissolves very quickly or leave it be and it dissolves slowly? Which interpretation is correct please?
bbartlog - 24-5-2012 at 11:26
I'm surprised that the liquid CuCl didn't cause spattering when you added it to the water... mp is north of 400C...
More seriously: what exactly did you add to the tap water? 1 milliliter of some solution of CuCl2 (or was it actually CuCl)? How concentrated? And how
would adding NaOH solution titrate it? You would end up with some sort of precipitate almost immediately. Are you trying to determine the amount of
excess HCl in a CuCl2 solution?
CHRIS25 - 24-5-2012 at 12:29
I added 1 mil Copper chloride solution. this solution at the moment would contain a small amount of Copper 1 ions and a fair bit of copper 2 ions,
also plenty of dissolved oxygen and a certain amount of concentrated HCL, the amount is unknown. To test whether the HCL amount is exhausted or low I
add 1mil of this solutin to ordinary faitrly warm tap water. I then add, via a syringe, the same amount of a 1 mole NAOH. It turns milky where the
syringe output meets the CUCL solution, this is copper hydroxide forming at the outlet. The speed at which it dissolves indicates whether there is
still HCL in the solution. BUT my question is: It dissolves rapidly if i stir. It dissolves slowly if I do not agitate the solution, which
situation is the correct reading? Hope this explains everything.
Vargouille - 24-5-2012 at 14:26
Titration procedure recommends that you continue to add NaOH with agitation until the precipitate will no longer dissolve, no matter how hard you
agitate it. Otherwise it would take too long. When it starts taking longer and longer to dissolve, go dropwise.
CHRIS25 - 24-5-2012 at 14:33
ok then, but then how exactly do I interpret the results? For example, let's say I add 10mils to a 30mil solution and it stops dissolving at this
point, what does that tell me? I suppose there must be calculations or maths equations related to this kind of procedure?
Vargouille - 24-5-2012 at 15:49
With the molarity (mol/L) of the NaOH solution, you can multiply it by the volume used to find moles NaOH used. This is a 1:1 ratio with HCl, so using
the volume of the original solution, you can find the original concentration of HCl.
CHRIS25 - 25-5-2012 at 09:11
Sorry But this is not making any sense at all here. I used 1 mol of NAOH, I multiply 1 by the volume (30mL of water)? to find the moles of NAOH used?
When I know the moles of the NAOH that I used already. This is 1:1 ratio with HCL? So using the volume of the original solution - which? the cucl
or naoh? I find original concentration of HCL - which I already know to be 36%....
Magpie - 25-5-2012 at 10:47
V1N1 = V2N2
where V1 equals the volume of solution 1 and V2 equals the volume of solution 2, using any consistent unit of volume, eg, mLs.
N1 = the normality of solution 1, and N2 = the normality of solution 2.
Vargouille - 25-5-2012 at 12:23
M1V1=M2V2 works here because 1 mol NaOH reacts with 1 mol HCl. Magpie's equation works in normals, but I
prefer to work in molars. Same math, though.
EDIT: If you know that you used exactly 1 mole of NaOH, then that's your moles of HCl reacted. However, generally you only know the molarity and
volume of the NaOH solution you use to titrate, which is then used to find the moles of HCl. Once you have that, you divide that by the volume of the
original CuCl2 solution, even if you diluted it for the titration. As long as the stock solution of CuCl2 is not diluted, you can use this method.
You will find that [HCl] (meaning concentration of HCl; [ ] means concentration of whatever is within the brackets) is not whichever molarity
corresponds to 36% because some of the HCl will have been consumed in order to create the CuCl2 from CuO.
[Edited on 25-5-2012 by Vargouille]
CHRIS25 - 25-5-2012 at 14:19
I have to be honest - I obviously need to study this whole thing in more detail because your explanations, while excellent and very simple for the
established chemist, still leave me with the feeling that I'll just judge this part of my operation by feeling. I know that's un-scientific but I
suppose if I could see a demonstration it would make absolute sense and I would understand it. But as it goes, I am failing to grasp what you guys
have kindly explained.
Vargouille - 25-5-2012 at 15:36
Ask and you shall receive. It's quite easy to find titration procedures through Google, however, but I'll give you these links:
The actual procedure, if you need it. Your end-point is reached not by an indicator like phenolphthalein, but by the point where Cu(OH)2 remains as a
precipitate.
Titration Calculations, not quite what you're trying to do, but you can probably figure it out.
watson.fawkes - 25-5-2012 at 17:07
Two issues have been raised here, but they aren't the same.
The first is how to tell when you've added the right about of titrant (the NaOH solution in this case) to the solution. The term "end point" is used
to describe the point when you stop adding titrant and take a measurement of how much titrant you added. End points are often visual, but could also
be taken with a pH meter. Since you're trying to tell how much base it requires to neutralize an acid, you want good mixing. So to answer your
original question, go ahead and stir the mixture. Now how quickly an individual titrant addition goes to its final state tells you something about how
close to your endpoint you are. There's some finesse required to do a good job and not overshoot. If you do overshoot, you can back-titrate with acid.
The second issue is how to interpret the measurement of the amount of titrant to get to the end point. That depends on what the end point is. If the
end point is neutralization, then you can just balance things out. In this case that means the amount of H+ matches the amount of
OH-. Sometimes the end point is not a pH-neutral state, in which case the calculation is more complicated. In the present case you have a
fairly simple calculation. The H+ came only from the HCl used to make up the solution. The OH- came only from the titrant. Given
the molecular weights of HCl vs. H+ and NaOH vs. OH-, you can calculate how much of each ion species you have. (Hint: disregard
the mass of the electron; it's only about 1/2000 of an atomic unit of mass)
CHRIS25 - 26-5-2012 at 03:34
Thankyou Vargouille and Watson.fawkes. I'm going to have to really study your answers because I am trying to relate everything to discovering whether
the HCL end point in my solution is dissipated (which I know it's not) but to also discover more accurately how much HCL is in solution. I think at
this point I am getting far to far ahead of myself and do not have enough background knowledge to be able to comprehend and absorb what you are
saying.
Vargouille - 26-5-2012 at 05:38
If you need the background, I can suggest taking a look through the SM Chem textbook (it's somewhere in here), and "acquiring" a copy of "Chemistry:
Matter and Change" and for more advanced learning, "Chemistry: The Central Science".
If you don't want to be on that grey line, there are multiple chemistry virtual textbooks, such as this one.
CHRIS25 - 26-5-2012 at 05:50
That looks like a great site thanks.