The given question is incomplete. The complete question is:
The pH of a 0.98 M solution of propanoic acid is measured to be 2.43. Calculate the acid dissociation constant of propanoic acid. Round your answer to two significant digits.
Answer: The acid dissociation constant of propanoic acid is [tex]1.4\times 10^{-5}[/tex]
Explanation:
[tex]CH_3CH_2COOH\rightarrow CH_3CH_2COO^-+H^+[/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= 0.98 M and pH = 2.43
[tex]pH=-log[H^+][/tex]
[tex]2.43=-log[c\times \alpha][/tex]
[tex]2.43=-log[0.98\times \alpha][/tex]
[tex]3.72\times 10^{-3}=0.98\times \alpha[/tex]
[tex]\alpha=3.79\times 10^{-3}[/tex]
[tex]K_a=\frac{(0.98\times 3.79\times 10^{-3})^2}{(0.98-0.98\times 3.79\times 10^{-3})}=1.4\times 10^{-5}[/tex]
The acid dissociation constant of propanoic acid is [tex]1.4\times 10^{-5}[/tex]
