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.
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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