NezukoLawliet
Answered

Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Ask your questions and receive accurate answers from professionals with extensive experience in various fields on our platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.

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

Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.