1 Answer

Martinfleis · Answer 1 · 2021-07-06T12:00:51+0000

Answer: The pH of an aqueous solution of .25M acetic acid is 2.7

Step-by-step explanation:

$HC_2H_3O_2\rightarrow H^+C_2H_3O_2^-$

cM 0 0

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

So dissociation constant will be:

$K_a=((c\alpha)^(2))/(c-c\alpha)$

Give c= 0.25 M and
$\alpha$ = ?

K_a=1.8* 10^(-5)

Putting in the values we get:

$1.8* 10^(-5)=((0.25* \alpha)^2)/((0.25-0.25* \alpha))$

$(\alpha)=0.0084$

$[H^+]=c* \alpha$

[H^+]=0.25* 0.0084=0.0021

Also
pH=-log[H^+]

pH=-log[0.0021]=2.7

Thus pH is 2.7

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