Answer:
[tex]pH=2.88[/tex]
Explanation:
Hello,
In this case, since dissociation of acetic acid is:
[tex]CH_3COOH\rightleftharpoons CH_3COO^-+H^+[/tex]
The equilibrium expression in terms of the acid dissociation constant is:
[tex]Ka=\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}[/tex]
In such a way, in terms of the reaction extent, we write:
[tex]Ka=\frac{x*x}{0.1-x}[/tex]
And Ka is computed from the pKa:
[tex]Ka=10^{-pKa}=10^{-4.75}=1.78x10^{-5}[/tex]
[tex]1.78x10^{-5}=\frac{x*x}{0.1-x}[/tex]
Thus, solving for [tex]x[/tex] we obtain:
[tex]x=0.001325M[/tex]
Which is also equal to the concentration of H⁺ so the pH is:
[tex]pH=-log(0.001325)\\\\pH=2.88[/tex]
Regards.
