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.
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
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.
Unless they wanted 3 sig figs, which would be 0.0221M.
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
To calculate the concentration of acetic acid in vinegar, you can use the formula for the dissociation constant (Ka) of acetic acid:
Ka = [H3O+][CH3COO-] / [CH3COOH]
Given that the Ka value for acetic acid is 1.8 × 10^-5, and the pH of vinegar is 3.20, you correctly obtained the antilog (inverse logarithm) of the pH as 6.309 × 10^-4.
Now, let's break down the calculation step by step:
Step 1: Calculate the concentration of H3O+ ions using the pH value:
Since pH is defined as the negative logarithm (base 10) of the concentration of H3O+ ions in a solution, we can determine H3O+ concentration using the antilog of the pH:
[H3O+] = 10^(-pH)
[H3O+] = 10^(-3.20)
[H3O+] ≈ 6.309 × 10^-4
Step 2: Calculate the concentration of CH3COO- ions:
The concentration of CH3COO- ions is equal to the concentration of H3O+ ions since both are produced in a 1:1 ratio during the dissociation of acetic acid.
Therefore, [CH3COO-] ≈ 6.309 × 10^-4
Step 3: Calculate the concentration of acetic acid (CH3COOH):
Using the formula for Ka, we can rearrange it and solve for [CH3COOH]:
Ka = [H3O+][CH3COO-] / [CH3COOH]
1.8 × 10^-5 = (6.309 × 10^-4)(6.309 × 10^-4) / [CH3COOH]
Now, solve for [CH3COOH]:
[CH3COOH] = (6.309 × 10^-4)^2 / (1.8 × 10^-5)
[CH3COOH] ≈ 0.022113
So, the approximate concentration of acetic acid in vinegar is 0.022113 M.
If you received an error message about possibly missing a "term," it could be due to rounding errors during the calculation. Try using more decimal places or follow the significant figure rules while rounding off the intermediate values.