How much acetic acid (of a 17.4 M stock) and sodium acetate (in g) are needed to prepare 1 L of a 0.1 M acetate buffer at pH 5.3?
pKa of acetic acid: 4.76
Molecular mass of acetic acid: 60.05 g/mol
Molecular mass of sodium acetate: 82.03 g/mol
According to the question -
Given values are -
the mass of sodium acetate neeeded to prepare 1 L of 0.1 M sodium acetate.
The molarity of the given solution is 0.1 M that means 0.1 mole of sodium acetate is dissolved in 1 liter. The molar mass of sodium acetate is 82 g/mol therefore 0.1 mole will be
mass of 1 mole of sodium acetate =82g
The Henderson-Hasselbalch for pH of the buffer solution is
by plugging in the data in the abve expression for pH
0.54 = log 0.1 M - log [acid]
0.54 = -1.000 M - log [acid]
Add log[acid] on both sides
log [acid] = -1.540 M
[acid] = 0.0288 M
Concentration of acid needed = 0.0288 M
volume of acetic acid needed.
This can be found by using dilution law
C1V1 = C2 V2
Mass of sodium acetate needed = 8.2 g and
Volume of stock solution of acetic acid needed = 1.66 mL