HBr + NaOH ==> NaBr + H2O
moles HBr = M x L
moles NaOH = M x L
There can be three scenarios:
a. If moles HBr exactly equals moles NaOH, then you have the salt, NaBr, in water and the pH = 7 because neither Na^+ nor Br^- hydrolyze.
b. If more moles HBr than NaOH are present, there will be excess HBr and the pH will be determine by the molarity of the HBr (it is a strong acid.)
c. If more moles NaOH than HBr are present, there will be an excess of NaOH and the pH will be determined by the molarity of the NaOH (it is a strong base).
Post your work if you get stuck.
If 30.0 mL of 4.8 10-2 M HBr is added to 21.5 mL of 3.6 10-2 M NaOH, what is the pH of the solution?
6 answers
I did:
(4.8e-2*30.0mL)/1000mL=.0014
(3.6e-2*21.5mL)/1000mL=7.74e-4
Then I added the two to get .005474 then divided that by the total L to get .106
I then took the -log(.106) to get .975, but that's the wrong answer
(4.8e-2*30.0mL)/1000mL=.0014
(3.6e-2*21.5mL)/1000mL=7.74e-4
Then I added the two to get .005474 then divided that by the total L to get .106
I then took the -log(.106) to get .975, but that's the wrong answer
Why in the world did you add them? And where did the 1000 mL come from? I don't see that in my instructions. Nor do I see anything about taking the log in my instructions.
so (4.8e-2*30.0mL)=1.44
(3.6e-2*21.5mL)=.774
now what do I do
(3.6e-2*21.5mL)=.774
now what do I do
Great. Now look at the equation.This is an acid/base reaction. Acids and bases react to give a salt + water.
HBr + NaOH ==> NaBr + H2O
I can't write these across the board because it ignores the spaces. Let me use periods (like this .......) to make it space correctly. Just ignore the periods.
HBr + NaOH = NaBr + H2O
1.44...774....0.....0 initially
-.774..-.774..+.774..+.774 change
.666....0.......774.......final
The HBr is in excess, the concn is now .666/51.5 mL = ??
Now you can do the pH = -log(H^+) thing)
I just figured out where your 1000 came from. You were changing mL to L. I didn't do that above; I left everything in mL and concn at the end is millimoles/mL. I think the pH is about 1.8 (a guess) or so but check me out on that.
HBr + NaOH ==> NaBr + H2O
I can't write these across the board because it ignores the spaces. Let me use periods (like this .......) to make it space correctly. Just ignore the periods.
HBr + NaOH = NaBr + H2O
1.44...774....0.....0 initially
-.774..-.774..+.774..+.774 change
.666....0.......774.......final
The HBr is in excess, the concn is now .666/51.5 mL = ??
Now you can do the pH = -log(H^+) thing)
I just figured out where your 1000 came from. You were changing mL to L. I didn't do that above; I left everything in mL and concn at the end is millimoles/mL. I think the pH is about 1.8 (a guess) or so but check me out on that.
ohh..I understand now, I forgot all about the ice table...I got 1.8 as well and that is the correct answer...Thanks!
1 answer