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how would you prepare 250ml of 0.2M sodium phosphate buffer of PH 6.4? if the following chemicals are available : dibasic sodium phosphate, monobasic sodium phosphate and 0.129M solution of NaOH. Assume that the pKa1, pKa2, and pKa3, of phosphoric acid are 2.12, 6.8, and 12.3 respectively. molecular weights of dibasic and monobasic sodium phosphate salts are 142 and 138 respectively

User Mitsuko
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Step-by-step explanation:

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how would you prepare 250ml of 0.2M sodium phosphate buffer of PH 6.4? if the following chemicals are available : dibasic sodium phosphate, monobasic sodium phosphate and 0.129M solution of NaOH. Assume that the pKa1, pKa2, and pKa3, of phosphoric acid are 2.12, 6.8, and 12.3 respectively. molecular weights of dibasic and monobasic sodium phosphate salts are 142 and 138 respectively

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To prepare a 250ml of 0.2M sodium phosphate buffer solution at pH 6.4, we need to use the Henderson-Hasselbalch equation:

pH

=

pKa

+

log

(

[

A

]

[

HA

]

)

pH=pKa+log(

[HA]

[A

]

)

where

A

A

is the conjugate base (dibasic sodium phosphate), and

HA

HA is the weak acid (monobasic sodium phosphate).

First, we need to determine the ratio of

A

A

to

HA

HA using the Henderson-Hasselbalch equation and the given pKa values.

6.4

=

6.8

+

log

(

[

dibasic sodium phosphate

]

[

monobasic sodium phosphate

]

)

6.4=6.8+log(

[monobasic sodium phosphate]

[dibasic sodium phosphate]

)

Solving for

[

dibasic sodium phosphate

]

[

monobasic sodium phosphate

]

[monobasic sodium phosphate]

[dibasic sodium phosphate]

:

0.4

=

log

(

[

dibasic sodium phosphate

]

[

monobasic sodium phosphate

]

)

−0.4=log(

[monobasic sodium phosphate]

[dibasic sodium phosphate]

)

[

dibasic sodium phosphate

]

[

monobasic sodium phosphate

]

=

1

0

0.4

[monobasic sodium phosphate]

[dibasic sodium phosphate]

=10

−0.4

[

dibasic sodium phosphate

]

[

monobasic sodium phosphate

]

0.3981

[monobasic sodium phosphate]

[dibasic sodium phosphate]

≈0.3981

Next, we can set up two equations based on the molarity and molecular weights of the salts to solve for the concentrations of

dibasic sodium phosphate

dibasic sodium phosphate and

monobasic sodium phosphate

monobasic sodium phosphate in the final solution.

For dibasic sodium phosphate:

Molarity

=

Number of moles

Volume in liters

Molarity=

Volume in liters

Number of moles

0.2

M

=

0.3981

���

×

142

g/mol

0.25

L

0.2M=

0.25L

0.3981x×142g/mol

���

=

0.2

×

0.25

0.3981

×

142

x=

0.3981×142

0.2×0.25

For monobasic sodium phosphate:

0.129

M

=

���

×

138

g/mol

0.25

L

0.129M=

0.25L

y×138g/mol

���

=

0.129

×

0.25

138

y=

138

0.129×0.25

After calculating the values of

���

x and

���

y, you can prepare the buffer solution by mixing the appropriate masses of dibasic and monobasic sodium phosphate salts in 250 ml of water.

User Colymore
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