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Given 0.5 M solution of acetic acid (pKa = 4.76) and solid sodium acetate (Mr =...

Given 0.5 M solution of acetic acid (pKa = 4.76) and solid sodium acetate (Mr = 82 g/mol), describe how you would go about preparing 500 mL of 0.2M acetate buffer pH 4.0.

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Answer #1

Given that ([A-] + [HA])= 0.2 M in 500 ml

Use the Hendersen-Hasselbalch equation to calculate the amount of Na+CH3COO- and CH3COOH as follows:

pH = pka + log ([A-] / [HA])

4.0 = 4.76 + log ([A-] / [HA])

- 0.76 = log ([A-] / [HA])

([A-] / [HA])= 10^-0.76= 0.173

Therefore [A-] = 0.173 [HA]

[A-] + [HA] = 0.2

Then;

[A-] + [HA]= 0.2

0.173 [HA] + [HA]= 0.2

1.173 [HA]= 0.2

[HA] = 0.171 M

And [A-] = 0.2 –[HA]= 0.2- 0.171= 0.029 M

Number of mole = molarity * volume in L

Moles of sodium acetate =0.029* 0.500= 0.0145 moles

Amount in g = molar mass * number of moles

=82 g/mole* 0.0145 moles

=1.189 g sodium acetate

sodium acetate ,Mr = 82 g/mol)

mole of acetic acid = molarity * volume in L

=0.171 *0.500= 0.0855 moles

Volume in L = Number of moles/ MOLARITY

= 0.0855 /0.5

= 0.171 ml acetic acid

Cross check

pH = pka + log ([A-] / [HA])

= 4.76 + log (1.189 g sodium acetate /82 )/ 0.0855

= 4.76 + log 0.0145/0.0855

= 4.76- 0.771

= 3.989

=4.00

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