You need to prepare an acetate buffer of pH 5.67 from a 0.841 M acetic acid...

You need to prepare an acetate buffer of pH 5.67 from a 0.841 M acetic acid solution and a 2.94 M KOH solution. If you have 675 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 5.67? The pKa of acetic acid is 4.76.

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Expert Solution

The pH of required buffer is

pH = pKa + log [salt] / [acid] = pKa + log [CH3COO-]/[CH3COOH]

pKa of acetic acid = 4.75

pH = 5.67 = 4.75 + log [CH3COO-] / [CH3COOH]

log [CH3COO-] / [CH3COOH] = 0.91

[CH3COO-] / [CH3COOH] = 8.128

Therefore [CH3COO-] = 8.317 X [CH3COOH] ...............(1)

The initial moles of acetic acid = Molarity X volume = 0.841 X 0.675 = 0.568 Moles

Let volume of KOH required = V mL

so moles of KOH added = V X 2.94

The moles of salt formed = 2.94 V

The moles of acid left = 0.568 - 2.94V

Putting these value in (1)

2.94V = 8.317 X [0.568-2.94V]

0.353 V = 0.568- 2.94V

3.293V = 0.568

V = 0.172 L = 172mL

queen_honey_blossom answered 1 year ago

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