I\'m working on a lab, and I seem to be stuck. My numbers don\'t seem to be maki
ID: 1017783 • Letter: I
Question
I'm working on a lab, and I seem to be stuck. My numbers don't seem to be making sense.
I discovered the half-equivelance point of an unknown acid is 5.5.
The lab states the half-equivelance point is equal to the pKa of the unknown weak acid.
-log(5.5)= -.74
I am given this table af possible acids, but can't see how that (5.5 pKa and -.74 Ka) makes it possible to select one. It seems like a pKa of 5.5 should result in a much smaller Ka value.
Not sure what I did wrong.
Possible weak acids:
Acid Name
Acid Formula
Ka Value
pKa Value
Chlorous acid
HClO2
1.2 x 10 -2
2.92
Monochloracetic acid
HC2H2ClO2
1.35 x10 -3
3.87
Hydrofluoric acid
HF
7.2 x 10 -4
4.14
Acetic acid
HC2H3O2
1.8 x 10 -5
4.75
Carbonic acid
H2CO3
4.3 x 10 -7
6.37
Hypochlorous acid
HOCl
3.5 x 10 -8
7.46
Hydrocyanic acid
HCN
6.2 x 10 -10
9.21
Acid Name
Acid Formula
Ka Value
pKa Value
Chlorous acid
HClO2
1.2 x 10 -2
2.92
Monochloracetic acid
HC2H2ClO2
1.35 x10 -3
3.87
Hydrofluoric acid
HF
7.2 x 10 -4
4.14
Acetic acid
HC2H3O2
1.8 x 10 -5
4.75
Carbonic acid
H2CO3
4.3 x 10 -7
6.37
Hypochlorous acid
HOCl
3.5 x 10 -8
7.46
Hydrocyanic acid
HCN
6.2 x 10 -10
9.21
Explanation / Answer
A:- It is Ka value of weak acid, Ka = 5.5, then the pKa value, pKa = -log (Ka)
= -log (5.5)
pKa = -0.74