How much volume I have to add of acetic acid and grams of sodium acetate?
Sodium acetate is the conjugate base of acetic acid . In order to have a buffering activity, the conjugate base should be in equilibrium with the corresponding acid (acetic acid) in this case.
If you want to prepare 0.1 M acetate buffer of pH 5.4, use the Hendersen-Hasselbalch equation:
pH=pKa + log { [sodium acetate]/[acetic acid] }
5.4 = 4.76 + log { [sodium acetate]/[acetic acid] }
0.64 = log { [sodium acetate]/[acetic acid] }
Take antilog on both sides
4.365 = [sodium acetate]/[acetic acid]
Thus, [sodium acetate] = 4.365*[acetic acid] (Equation 1)
I am assuming you want to prepare a 0.1 M buffer, so the total [sodium acetate]+[acetic acid] = 0.1 M (Equation 2)
Add [acetic acid] on both sides of Equation 1
[sodium acetate]+[acetic acid] = 4.365*[acetic acid] + [acetic acid]
Thus, 0.1 = 5.365*[acetic acid] (From Equations 1 and 2)
[acetic acid] = 0.01863 M
[sodium acetate] = 4.365*[acetic acid] = 0.08131 M
Now for the weighing part,
Molecular weight of acetic acid = 60 g/mol
Thus 1 Molar solution will contain 60 g of acetic acid.
0.01863 Molar solution will contain 60*0.01863 = 1.1178 g acetic acid
Since acetic acid is liquid, convert the weight into volume using Density = Mass/Volume. Density of acetic acid is 1050 g/1000 mL.
Thus Volume = (1.1178*1000)/1050 = 1.064 mL
Molecular weight of sodium acetate = 82 g/mol
Thus 1 Molar solution will contain 82 g of sodium acetate.
0.08131 Molar solution will contain 82*0.08131 = 6.6674 g acetic acid
If the sodium acetate is in its trihydrate form, consider the molecular weight of 3 water molecules in the calculation (82+54 = 136 g/mol)
All these calculations are for preparing 1 Litre of the buffer. If you want different volumes, make sure to adjust for it in the molarity equations above.
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