posted by Jonah on .
If acetic acid is the only acid that vinegar contains [Ka = 1.8*10-5] , calculate the concentration of acetic acid in the vinegar.
pH of vinegar is 3.20
First I got the antilog of the pH given, which turned out to be 6.309E-4.
Then I did [(6.309E-4)^(2)]/(1.8E-5)to get .022113, but when I typed in that answer I was told to check my calculations because I might be missing a 'term.' I have no idea what to do after this point.
The concentration should be in molarity M. and it looks like you put in too many significant figures in the answer part. 0.022 M is what I would have put in.
Yeah that is what I put in as my answer since it asked for two sigfis, but it was still wrong.
HAc+ H2O---> Ac + H3O+
10^(-3.10)=6.310 x 10^-4 M
Ka=[6.310 x 10^-4 M][6.310 x 10^-4 M]/x
Ka=1.8 x10^-5=[6.310 x 10^-4 M][6.310 x 10^-4 M]/x
3.982 x 10^-7/1.8 x10^-5=[x]
x= 0.022 M
Not sure what to tell you.
Unless they wanted 3 sig figs, which would be 0.0221M.
I got the same answer and had the same problem but it worked when I put in .0221... it says the right answer is 2.3×10−2