NezukoLawliet
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CH3COOH  CH3COO– + H+
You start with 0.05 moles of acetic acid in 500 mL of water. At equilibrium, the pH of the solution is 2.873. What is the equilibrium constant of this reaction? Hint: You will need to calculate an antilog using a scientific calculator.

Sagot :

(a)

pH = 4.77

; (b)

[

H

3

O

+

]

=

1.00

×

10

-4

l

mol/dm

3

; (c)

[

A

-

]

=

0.16 mol⋅dm

-3

Explanation:

(a) pH of aspirin solution

Let's write the chemical equation as

m

m

m

m

m

m

m

m

l

HA

m

+

m

H

2

O

⇌

H

3

O

+

m

+

m

l

A

-

I/mol⋅dm

-3

:

m

m

0.05

m

m

m

m

m

m

m

m

l

0

m

m

m

m

m

l

l

0

C/mol⋅dm

-3

:

m

m

l

-

x

m

m

m

m

m

m

m

m

+

x

m

l

m

m

m

l

+

x

E/mol⋅dm

-3

:

m

0.05 -

l

x

m

m

m

m

m

m

m

l

x

m

m

x

m

m

m

x

K

a

=

[

H

3

O

+

]

[

A

-

]

[

HA

]

=

x

2

0.05 -

l

x

=

3.27

×

10

-4

Check for negligibility

0.05

3.27

×

10

-4

=

153

<

400

∴

x

is not less than 5 % of the initial concentration of

[

HA

]

.

We cannot ignore it in comparison with 0.05, so we must solve a quadratic.

Then

x

2

0.05

−

x

=

3.27

×

10

-4

x

2

=

3.27

×

10

-4

(

0.05

−

x

)

=

1.635

×

10

-5

−

3.27

×

10

-4

x

x

2

+

3.27

×

10

-4

x

−

1.635

×

10

-5

=

0

x

=

1.68

×

10

-5

[

H

3

O

+

]

=

x

l

mol/L

=

1.68

×

10

-5

l

mol/L

pH

=

-log

[

H

3

O

+

]

=

-log

(

1.68

×

10

-5

)

=

4.77

(b)

[

H

3

O

+

]

at pH 4

[

H

3

O

+

]

=

10

-pH

l

mol/L

=

1.00

×

10

-4

l

mol/L

(c) Concentration of

A

-

in the buffer

We can now use the Henderson-Hasselbalch equation to calculate the

[

A

-

]

.

pH

=

p

K

a

+

log

(

[

A

-

]

[

HA

]

)

4.00

=

−

log

(

3.27

×

10

-4

)

+

log

(

[

A

-

]

0.05

)

=

3.49

+

log

(

[

A

-

]

0.05

)

log

(

[

A

-

]

0.05

)

=

4.00 - 3.49

=

0.51

[

A

-

]

0.05

=

10

0.51

=

3.24

[

A

-

]

=

0.05

×

3.24

=

0.16

The concentration of

A

-

in the buffer is 0.16 mol/L.

hope this helps :)

The equilibrium constant of this reaction is 1.80×10-5

Given data,

pH of solution = 2.873

Number of moles of acetic acid (m) = 0.05 moles

Volume of water (V) =  500 mL = 0.5L

So, concentration (C) = m/V in lit = 0.05/0.5 = 0.1 M

Equilibrium constant ( K ) = [tex][CH_{3} COO-][/tex]×[tex][H+_{} ][/tex]/[tex][CH_{3} COOH][/tex]

Since, acetic acid is weak acid,

So, Equilibrium constant ( K ) = [tex][H+]^{2}[/tex]/[tex][CH_{3} COOH][/tex]   ....(i)

As the pH = 2.873, the [tex][H+_{} ][/tex] is antilog of -2.873 or 1.34×10-3 M.

Putting the value of concentration of [tex]H+_{}[/tex] and [tex]acetic_{} acid[/tex] in equation (i).

Equilibrium constant ( K ) = 1.80×10-5

What is weak acid ?

The acid which is partially dissociates into ions on dissolving in aqueous solution is called weak acid.

Example: [tex]acetic_{} acid[/tex].

To learn more about weak acid here.

https://brainly.com/question/12811944

#SPJ3

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