Answered

An aqueous solution contains 0.399 M hypochlorous acid. How many mL of 0.310 M sodium hydroxide would have to be added to 125 mL of this solution in order to prepare a buffer with a pH of 7.600?



Answer :

Answer:

Explanation:

To solve this problem, we will use the Henderson-Hasselbalch equation for buffer solutions and the concept of neutralization reactions.

The Henderson-Hasselbalch equation is:

pH

=

p

+

log

(

[

A

]

[

HA

]

)

pH=pK

a

+log(

[HA]

[A

]

)

Where:

pH

pH is the desired pH of the buffer.

p

pK

a

 is the negative logarithm of the acid dissociation constant

K

a

 of hypochlorous acid.

[

A

]

[A

] is the concentration of the conjugate base (hypochlorite ion,

ClO

ClO

).

[

HA

]

[HA] is the concentration of the weak acid (hypochlorous acid,

HOCl

HOCl).

Given:

Desired

pH

=

7.600

pH=7.600

Concentration of hypochlorous acid (

HOCl

HOCl) in solution = 0.399 M

Volume of hypochlorous acid solution = 125 mL

Concentration of sodium hydroxide (

NaOH

NaOH) solution = 0.310 M

The

K

a

 of hypochlorous acid (

HOCl

HOCl) is

3.0

×

1

0

8

3.0×10

−8

, thus

p

=

log

(

3.0

×

1

0

8

)

=

7.52

pK

a

=−log(3.0×10

−8

)=7.52

Calculate the required ratio of

[

A

]

[A

] to

[

HA

]

[HA]:

Using the Henderson-Hasselbalch equation:

7.600

=

7.52

+

log

(

[

ClO

]

[

HOCl

]

)

7.600=7.52+log(

[HOCl]

[ClO

]

)

7.600

7.52

=

log

(

[

ClO

]

[

HOCl

]

)

7.600−7.52=log(

[HOCl]

[ClO

]

)

0.080

=

log

(

[

ClO

]

[

HOCl

]

)

0.080=log(

[HOCl]

[ClO

]

)

1

0

0.080

=

[

ClO

]

[

HOCl

]

10

0.080

=

[HOCl]

[ClO

]

1.202

=

[

ClO

]

[

HOCl

]

1.202=

[HOCl]

[ClO

]

Calculate the moles of hypochlorous acid in 125 mL of solution:

Moles of HOCl

=

0.399

M

×

0.125

L

=

0.049875

moles

Moles of HOCl=0.399M×0.125L=0.049875moles

Let

x be the moles of

NaOH

NaOH added. Since

NaOH

NaOH neutralizes

HOCl

HOCl to form

ClO

ClO

, the moles of

ClO

ClO

 formed will also be

x and the moles of

HOCl

HOCl remaining will be

0.049875

0.049875−x:

Using the ratio from step 1:

0.049875

=

1.202

0.049875−x

x

=1.202

Solving for

x:

=

1.202

(

0.049875

)

x=1.202(0.049875−x)

=

0.0598935

1.202

x=0.0598935−1.202x

+

1.202

=

0.0598935

x+1.202x=0.0598935

2.202

=

0.0598935

2.202x=0.0598935

=

0.0598935

2.202

x=

2.202

0.0598935

0.0272

moles

x≈0.0272moles

Calculate the volume of 0.310 M

NaOH

NaOH solution required to get 0.0272 moles:

Volume of NaOH solution

=

0.0272

moles

0.310

M

0.0877

L

=

87.7

mL

Volume of NaOH solution=

0.310M

0.0272moles

≈0.0877L=87.7mL

So, you would need to add approximately 87.7 mL of 0.310 M sodium hydroxide to 125 mL of 0.399 M hypochlorous acid solution to prepare a buffer with a pH of 7.600.