Asked by L.Bianchessi
Find the pH of each mixture of acids.
0.115 M HBr and O.125 M HCHO2.
I TRIED DOING THIS PROBLEM THE SAME WAY AS THE ONE BEFORE BUT IT DIDN'T WORK.
THE PH SHOULD BE 0.939.
I did HBr as stronger acid and got concentration of H+ and put it into the ka expression for formic acid and my pH turned out to large.
Do I do a different process for this one?
0.115 M HBr and O.125 M HCHO2.
I TRIED DOING THIS PROBLEM THE SAME WAY AS THE ONE BEFORE BUT IT DIDN'T WORK.
THE PH SHOULD BE 0.939.
I did HBr as stronger acid and got concentration of H+ and put it into the ka expression for formic acid and my pH turned out to large.
Do I do a different process for this one?
Answers
Answered by
DrBob222
No, the only difference is that HBr is a strong acid that ionizes 100% versus the other problem where both were weak acids with a Ka value for each.
HBr = H^+ + Br^- and (H^+) = 0.115
You probably need to see if the HCOOH contributes anything to add.
..........HCOOH ==> H^+ + COO^-
initial...0.125.....0......0
change.....-x........x......x
equil....0.125-x.....x.......x
1.77E-4 = (0.115+x)(x)/(0.125-x)
I called 0.115-x = 0.115 and 0.125-x = 0.125 and x = about 0.00019 which adds nothing to the 0.115 so
pH from 0.115 = 0.939. Voila! again.
These problems are a little easier than they could be. Most often I see a volume of each mixed and the problem want a pH. In those mixtures, the acids dilute each other so the molarity in the problem isn't the molarity you use in calculating (H^+).
HBr = H^+ + Br^- and (H^+) = 0.115
You probably need to see if the HCOOH contributes anything to add.
..........HCOOH ==> H^+ + COO^-
initial...0.125.....0......0
change.....-x........x......x
equil....0.125-x.....x.......x
1.77E-4 = (0.115+x)(x)/(0.125-x)
I called 0.115-x = 0.115 and 0.125-x = 0.125 and x = about 0.00019 which adds nothing to the 0.115 so
pH from 0.115 = 0.939. Voila! again.
These problems are a little easier than they could be. Most often I see a volume of each mixed and the problem want a pH. In those mixtures, the acids dilute each other so the molarity in the problem isn't the molarity you use in calculating (H^+).
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