Chemistry, asked by dianaxmet01, 9 months ago

how to obtain CCL3COOH

Answers

Answered by yashk1770
0

Explanation:

Answer

anor277

Apr 7, 2016

With difficulty, inasmuch you neglected to include the acid dissociation constant,

K

a

, of

C

l

3

C

(

=

O

)

O

H

. I got

p

H

=

0.804

Explanation:

From this site I learn that

K

a

,

C

l

3

C

(

=

O

)

O

H

=

2.2

×

10

1

.

This is fairly large as an organic acid, and likely the method of successive approximations will be inadequate:

C

l

3

C

(

=

O

)

O

H

+

H

2

O

C

l

3

C

(

=

O

)

O

+

H

3

O

+

So

K

a

=

2.2

×

10

1

=

[

H

3

O

+

]

[

C

l

3

C

(

=

O

)

O

]

[

C

l

3

C

(

=

O

)

O

H

]

If I put

[

H

3

O

+

]

=

x

, then

K

a

=

2.2

×

10

1

=

x

2

0.268

x

Solving this quadratic equation, I get two solutions:

x

1

=

0.157

;

x

2

=

0.377

. I had to use the quadratic equation to get this answer.

Clearly,

x

1

=

0.157

is the required answer.

Thus

[

H

3

O

+

]

=

0.157

m

o

l

L

1

=

[

C

l

3

C

(

=

O

)

O

]

AND

[

C

l

3

C

(

=

O

)

O

H

]

=

(

0.268

0.157

)

m

o

l

L

1

=

0.111

m

o

l

L

1

.

Thus

p

H

=

log

10

{

0.157

}

=

0.804

Please check the validity of this calculation. I am still using a web based calculator because my calculator (which has this programmed) has gone walkabout. Note that the

p

H

of this solution is low because trichloroacetic acid is strongish, certainly stronger than acetic acid. As far as I know this is an entropy rather than a enthalpy phenomenon. Trichloroacetate is a much less polarizing ion than acetate, and is entropically favoured.

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