Question

What volume of Acetic acid (17.6 N) and what amount (g) of Sodium Acetate would you need to prepare 100 ml of 0.2 M buffer, pH 4.66 (pK_a of Acetic acid = 4.76 and MW of Sodium Acetate = 136)?

Answer 1

Given, Concentration of Acetic acid = 17.6 N

pK_a of Acetic acid = 4.76

Molecular weight of Sodium Acetate = 82g/mol [Not 136]

For acetic acid, pH = pK_a + log[CH3COO-]/[CH3COOH]

Let [CH₃COOH] = (x) mol/L

Then [CH3COO^-] = (0.2 - x) mol/L

Therefore, pH = pK_a + log[(0.2 - x)/x]

4.66 = logK_a + log[(0.2 - x)/x]

4.66 = 5.146 + log[(0.2 - x)/x]

-0.486 = log[(0.2 - x)/x]

Taking antilog both sides; 0.326 = (0.2 - x)/x

0.326 x = 0.2 - x

1.326x = 0.2

x = 0.15 mol/L = [CH3COOH]

0.2 - x = 0.05 mol/L = [CH3COO^-]

Now, Molarity = Moles/Volume

And Moles = Weight/Molecular Weight

For 100 ml of solution:

0.1L x 0.15 mol/L x l/17.6 mol = 0.00085 L CH₃COOH

0.1 x 0.05 mol/L x 82 g/mol = 0.41 g CH₃COONa