mols HBr = 0.025 from your work
mols KOH = 0.015 from yur work.
.......HBr + KOH ==> KBr + H2O
I....0.025..0.015.....0.....0
C...-0.015.-0.015....0.015..0.015
E....0.010....0......0.015
HBr is in excess.
(HBr)= mols/L = 0.010/(0.06+0.05) = approx 0.09
Then pH = -log(H^) = ?
If you're still stuck post your work and I'll find the error.
By the way, when you first posted the problem you could have shown your work and I could have spotted the problem instantly. My best guess is that you didn't convert mols to M. Also if you need more help post the choices available. Sometimes they don't give exact numbers on choices.
A 50.0 mL sample of 0.50 M HBr is titrated with 0.25 M KOH. What is the pH after the addition of 60.0 mL of KOH?
A) 4.1
B) 13.4
C) 1.7
D) 2.0
E) 7.7
I'm confused on the process of solving this problem. My work so far:
.050L X .50 mol/1L= .025 mol HBR
.060L X .25 mol/1L= .015 mol KOH
I looked at a similar problem and it said then if HBR was in excess to take mol HBR-mol KOH/total L then the -log of that answer. The answer Im getting doesn't match the multiple choice. Where am I going wrong? Thanks.
1 answer
1 answer