What volume (mL) of 0.2650 N H3PO4 is required to neutralize:

Question

a) 35.25 mL of 0.2300 M KOH
b) 19.14 mL of 0.2300 M NH4OH

Unfortunately, the teacher didn't get the opportunity to teach us titrations in class today, but this problem (and more) are still due by 11 pm tonight (we have to send the answers electronically). I would really appreciate any help I can get. In the problem above, each "M" means "molarity" and the long capital "N" in the question ...well, i don't know what that means, I was hoping to find out. Thank you

Answer 1

M stands for molarity and is = mols/L solution
N stands for normality and is = equivalents/L of solution.

Most schools have stopped teaching normality and I applaud your teacher for including it. I think it's a huge mistake to dismiss normality BUT the IUPAC has spoken, they don't recommend it, so it doesn't get done.
Unfortunately, when you specify the normality of H3PO4 you MUST specify the reaction. This problem doesn't do that and I don't know if you are neutralizing 1, 2, or all 3 hydrogen ion of the H3PO4. Do you have any answers or anything that might help me know how to go about this.

Answer 2

Unfortunately, that's all the information I have. The point of doing these problems on the site called "CalmWeb" is so that we can't look up the answers in the back of the book, because we only know we have the answer correct once we type it in, so sorry about that.
I'd like to know how you'd try to solve it if all the hydrogen ions had to be neutralized, because the problem doesn't specify whether or not only a portion of them were neutralized, so for now, I'd like to assume they all were.

Answer 3

I was waiting for your reply to suggest that the best thing to do is just that; i.e., assume all of three H^+ are neutralized.
If that is so we have this equation.
H3PO4 + 3KOH ==> K3PO4 + 3H2O
millimols KOH = mL x M = 35.25 x 0.2300 = ?
mmols H3PO4 used = 1/3 of that = x

Then M H3PO4 = mmols/mL = xfrom previous step/mL and rearrange to mL = x/M = x/0.08833. The 0.08833 comes from changing 0.2650 N H3PO4 to 0.2650/3 = 0.08833 M. I obtained 30.59 mL but you should confirm each step above. By the way, a volume of 30.59 mL is perfectly reasonable. I would try this first and see how that fits in your database. If it works then do the b part exactly the same way except for the values used for each step. Keep me posted.

Answer 4

Yes, thank you very much!!
and for part b I ended up getting 16.61 mL. Thank you for your help

Answer 5

15.61 mL should be right. By the way, I should point out that NH4OH doesn't exist. Years ago we thought NH3 + H2O ==> NH4OH which then ionized into NH4^+ and OH^- and that made a lot of sense.(and the ions DO exist.) But years of looking for the NH4OH MOLECULE has proved elusive and in the late 70s it was confirmed that no EXTRA spectra showed up when liquid NH3 was diluted with H2O except for the NH3 bands and the H2O bands. So we now write a solution of NH3 as NH3 + H2O and let it go at that.

Answer 6

For whatever it's worth I learned something tonight that I wasn't aware of. It makes no difference how many H ions are neutralized. When mL and M are boxed in with the KOH and NH4OH, the volume comes out to be the same number no matter which assumption one makes. When I worked it three different ways to accommodate whatever assumption I made and the volume came out to be the same I knew the answer would be right. You could have worked it much simpler had I realized that by a simple
mL x N = mL x N
35.25 x 0.2300 = mL x 0.2650
mL = 30.59 mL
Wow. Isn't chemistry fun?