Then we would have BH2 2+ <> BH+ H+
F-x x x+ [H+]
F= 0.1 X (0.05/0.27)= 0.0185 M
[H+] = 0.05 x (0.02/0.27)=3.703e-3 M
using ka= 3.467e-3, I got a ph of 2.28, which is wrong again
what am i doing wrong?
I solved the pH at the second equivalence point and that's the answer I got thank you. Now, past the equivalence point, there's only HCl left, right? so we would have HCl <> H+ + Cl-. The concentration of H+ would no longer be x but would be x + the concentration of H+ stemming from the excess...
The problem is my book and my notes don't go into too much detail for polyprotic systems, which is why I am having trouble with this problem. Ill try to work on it myself but can you please give me a quick guideline on how to do it whenever you have the time? It is truly appreciated! you have...
So what you're saying is that after the first equivalence point, I only take into account the excess 45 mL of HCl? That means that 45mL x 0.05 M=2.2E-3 mol of HCl and there's a total volume of 195 mL
[BH+]=(5E-3 - 2.25E-3)/0.195 L= 0.0141 M
[BH2 2+]= 2.25E-3/0.195= 0.0115 M
pH=2.46 +log...
F is the formal concentration, so in this case it would be 5E-3 M. the formula I used is from the book and we use it in class. I agree that ph=pka1+pka/2 but I can't figure out how to get that answer using the formula
The pka values are for tryptophan so they can't be wrong
Homework Statement
For the titration of 50.0 mL of a 0.100 M solution of a dibasic compound with 0.0500 M HCl, calculate the pH at 100mL (equivalence point) and at 145 mL.
pka1=2.46
pka2=9.41
Homework Equations
(1)pH=pka + log[BH+]/[BH2+]
(2)[h+]=sqrt((K1K2F)+(K1Kw)/(K1+F))
The...