Question

The pH of a solution of propanoic acid is measured to be . Calculate the acid dissociation constant of propanoic acid. Round your answer to significant digits.

Answer 1

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 $1.4\times 10^{-5}$

Explanation:

$CH_3CH_2COOH\rightarrow CH_3CH_2COO^-+H^+$

 cM                                0             0

$c-c\alpha$                           $c\alpha$          $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= 0.98 M and pH = 2.43

$pH=-log[H^+]$

$2.43=-log[c\times \alpha]$

$2.43=-log[0.98\times \alpha]$

$3.72\times 10^{-3}=0.98\times \alpha$

$\alpha=3.79\times 10^{-3}$

$K_a=\frac{(0.98\times 3.79\times 10^{-3})^2}{(0.98-0.98\times 3.79\times 10^{-3})}=1.4\times 10^{-5}$

The acid dissociation constant of propanoic acid is $1.4\times 10^{-5}$