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
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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.
6 answers
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+
3.20=-log[H30+]
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.
3.20=-log[H30+]
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
That just means there was a rounding error somewhere, might've been on the calculators side if you approximated some numbers instead of using the real values, since 2.3*10^-2 = 0.023, which is similar to your answer of 0.022
2 answers