Question

# What volume of Acetic acid (17.6 N) and what amount (g) of Sodium Acetate would you...

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 (pKa of Acetic acid = 4.76 and MW of Sodium Acetate = 136)?

Given, Concentration of Acetic acid = 17.6 N

pKa of Acetic acid = 4.76

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

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

Let [CH3COOH] = (x) mol/L

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

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

4.66 = logKa + 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

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