[yh09]

Hi, how are you all doing?
I'm just doind this chemistry homework but it's getting me going nuts 

The question is:
How many mL of HCl 0.0195 are required to neutlalize 10 mL of Ca(OH)2 0.0116
I know that I have to use c1v1=c2v2, but I also know that there is a 2:1 ratio... so therefore, it should be:  Cca(oh)2 * Vca(oh)2 = 2 * Chcl * Vhcl because hcl's got the half of ions that Ca(OH)2's got.

Is it right? because I remember using that same method on similar questions and it worked perfectly...but now it does not.

Thaks for your precious help people 

[yh09]

Just a little indication:
to get to the answer, I had to do
2 * Cca(oh)2 * Vca(oh)2 =  Chcl * Vhcl
because if Ca(OH)2 = (for example) 1 mol/L, then [Ca2+]=1mol/L and [OH-]=2mol/L, right?
Therefore, it should be [Ca(OH)2)] * 2 because we want the concentratio n of the OH- ion, not the whole molecule...still right?
So concerning what I said always worked perfectly was :
5Fe2+ + MnO4- -> [...]
in this case it was Cfe * Vfe =  Cmno4 * Vmno4 * 5
because if [Fe2+]=1 mol/L than 5*1mol/L=5mol/L=[MnO4]

Do I have it right? If I do, can I conclude that (in c1v1=c2v2) when the number (coefficient) is in front of the molecule, we multiply the opposite molecule by this coefficient, whereas when the number is in the molecule (as in Ca(OH)2), we multiply the concentration of the molecule by the number.

Thanks again all, I appreciate your help.

[enahs]

Forget about C1V1 = C2V2 for a minute.

How many mols of Ca(OH)₂ do you have? How many mols of OH- does that give you?
And so in order to neutralize that, how many mols of HCl do you need?


Concentration X Volume gives you mols.
But you are interested in the mols of OH- from Ca(OH)₂; not just the mols of Ca(OH)₂.


Edit: tags corrected: sup -> sub

[yh09]

Hi,3
I have 10 mL (0.01L) of Ca(OH)2 0.0116mol by litre, which means 0,000116 mols of Ca(OH)2:
OH- = 0,000232 mols
Therefore, we need 0,000232 mols of HCl... from 0.0195 mol/L = 0,0119 Liters = 11,9 mL, and it`s the answer!

But, in general, is my conclusion right? (just as a trick to check my answers):

If I do, can I conclude that (in c1v1=c2v2) when the number (coefficient) is in front of the molecule, we multiply the opposite molecule by this coefficient, whereas when the number is in the molecule (as in Ca(OH)2), we multiply the concentration of the molecule by the number.

Thanks for your help.v

[yh09]

Can anybody please confirm (or tell me if it`s wrong) my last post?
This would be really appreciated and helpful.
Thanks all

[Borek]

11.9 is OK.

[yh09]

Thanks Borek, but I was reffering to the conclusion I wrote:

If I do, can I conclude that (in c1v1=c2v2) when the number (coefficient) is in front of the molecule, we multiply the opposite molecule by this coefficient, whereas when the number is in the molecule (as in Ca(OH)2), we multiply the concentration of the molecule by the number.

It`s just to help me check my answers next times....

Thanks again 

[Borek]

Honestly, it is so convoluted that I prefer to not touch it