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frez [133]
3 years ago
11

The pH of a solution of propanoic acid is measured to be . Calculate the acid dissociation constant of propanoic acid. Round you

r answer to significant digits.
Chemistry
1 answer:
Genrish500 [490]3 years ago
6 0

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}

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