Answer:
2.9 is the initial pH of the analyte solution.
Explanation:
The dissociation constant of acetic acid as per theoretical value = [tex] K_a[/tex]
[tex]K_a=10^{-4.756}=1.8\times 10^{-5}[/tex]
The initial concentration of acetic acid = c = 0.0900 M
[tex]HAc\rightleftharpoons Ac^-+H^+[/tex]
initially
c 0 0
At equilibrium
(c-x) x x
The expression of dissociation constant :
[tex]K_a=\frac{[Ac^-][H^+]}{[HAc]}[/tex]
[tex]1.8\times 10^{-5}=\frac{x\times x}{(c-x)}[/tex]
[tex]1.8\times 10^{-5}=\frac{x\times x}{(0.0900-x)}[/tex]
Solving for x:
x = 0.001264 M
[tex][H^+]=0.001264 M[/tex]
The pH of the solution :
[tex]pH=-\log[0.001264]=2.898\approx 2.9[/tex]
2.9 is the initial pH of the analyte solution.
