Answer: The value of acid ionization constant [tex]K_a[/tex] for acetic acid is [tex]1.87\times 10^{-5}[/tex]
Explanation:
[tex]CH_3COOH\rightarrow H^+CH_3COO^-[/tex]
cM 0 0
[tex]c-c\alpha[/tex] [tex]c\alpha[/tex] [tex]c\alpha[/tex]
So dissociation constant will be:
[tex]K_a=\frac{(c\alpha)^{2}}{c-c\alpha}[/tex]
Give c= 0.85 M and [tex]pH[/tex] = 2.4
[tex]pH=-log[H^+][/tex]
[tex][H^=}=c\times \alpha=10^{-2.4}=3.98\times 10^{-3}[/tex]
[tex]K_a=?[/tex]
Putting in the values we get:
[tex]K_a=\frac{(3.98\times 10^{-3})^2}{(0.85-3.98\times 10^{-3})}=1.87\times 10^{-5}[/tex]
Thus the value of acid ionization constant [tex]K_a[/tex] for acetic acid is [tex]1.87\times 10^{-5}[/tex]
