1) What is the concentration of acetic acid in the 0.100M pH 5.000 buffered solu
ID: 891841 • Letter: 1
Question
1) What is the concentration of acetic acid in the 0.100M pH 5.000 buffered solution?
Note 1: Based on the notation you use, you may write that the acetic acid concentration as either [HOAc] or Ca.
Note 2: Don't forget that the formal buffer concentration is 0.1M.
Note 3: Enter your answer below in three signficant figures. Don't worry about writing the units.
2) What mass of sodium acetate (NaOAc, 82.03 g/mol) would be needed to prepare 1L of the pH 5.000 0.100 M acetic acid/acetate buffer?
Note: Again, your answer should have three significant figures.
3) Glacial acetic acid is 99.5% acetic acid (f.w. 60.052) by weight and has a density of 1.05g/mL. Calculate the volume of glacial acetic acid needed to make 1L of the 0.100M pH 5.000 buffer.
***pKa of acetic acid is 4.76
Explanation / Answer
1) According to Henderson-Hasselbalch equation:
pH = pKa + log[CH3COO-] / [CH3COOH] , let us plugin the given values in the equation
5.00 = 4.76 + log[CH3COO-] / [CH3COOH]
0.24 = log [CH3COO-] / [CH3COOH]
[CH3COO-] / [CH3COOH] = 1.74
From given data buffer pH = 0.100M
Hence, [CH3COOH] + [CH3COO-] = 0.100
Let us consider x = [CH3COO-]
Then 0.100 –x = [CH3COOH]
x/ 0.100 –x = 1.74
x = 0.174 – 1.74x
2.74x – 0.174 = 0
x = 0.0635 = [CH3COO-]
The concentration of acetic acid = 0.1-x = 0.10 – 0.0635 = 0.037 M
2)
From the above solution, sodium acetate concentration = 0.0635 M
Let us consider required mass of sodium acetate = x
Molarity = (w/MW) * (1000/V)
0.0635 = (x/82.03)*(1000/1000)
x = 5.209 g
3) For acetic acid density = 1.05 g/ml, and the molecular weight is 60.052 g/mol.
So in 1000 ml there are 17.5 moles so the molarity of glacial acetic acid is 17.5M
Required concentration of acetic acid = 0.037 M in 1000 ml
The volume of glacial acetic acid required = 0.037 x 1000 / 17.5 = 2.114 ml
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