Answer: The concentration of acetic acid in the vinegar is [tex]7.10\times 10^{-2}M[/tex]
Explanation:
To calculate the pH of the solution, we use the equation:
[tex]pH=-\log[H^+][/tex]
We are given:
pH = 2.95
Putting values in above equation, we get:
[tex]2.95=-\log[H^+][/tex]
[tex][H^+]=1.122\times 10^{-3}M[/tex]
The chemical equation for the ionization of acetic acid follows:
[tex]CH_3COOH\rightleftharpoons H^++CH_3COO^-[/tex]
The expression of [tex]K_a[/tex] for above equation follows:
[tex]K_a=\frac{[H^+][CH_3COO^-]}{[CH_3COOH]}[/tex]
We are given:
[tex]K_a=1.8\times 10^{-5}[/tex]
[tex][H^+]_{eq}=[CH_3COO^-]_{eq}=1.122\times 10^{-3}M[/tex]
Putting values in above expression, we get:
[tex]1.8\times 10^{-5}=\frac{(1.122\times 10^{-3})\times (1.122\times 10^{-3})}{[CH_3COOH]}[/tex]
[tex][CH_3COOH]_{eq}=0.0699M[/tex]
So, the concentration of acetic acid = [tex][CH_3COOH]_{eq}+[H^+]_{eq}=(0.0699+0.001122)=7.10\times 10^{-2}M[/tex]
Hence, the concentration of acetic acid in the vinegar is [tex]7.10\times 10^{-2}M[/tex]
The concentration of acetic acid in vinegar is 6.9 * 10^-2 M.
Let us represent acetic acid with the general formula AH;
The dissociation of the acid is given by;
AH + H2O ⇄ H3O^+ + A^-
pH = 2.95 hence;
[H3O^+] = Antilog (-2.95) = 1.12 * 10^-3 M
Note that [H3O^+] = [ A^-]
Recall that;
Ka = [H3O^+] [ A^-]/[AH]
So;
1.8×10^−5 = (1.12 * 10^-3)^2/[AH]
[AH] = (1.12 * 10^-3)^2/1.8×10^−5
[AH] = 1.25 * 10^-6/1.8×10^−5
[AH] = 6.9 * 10^-2 M
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