In: Chemistry
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.
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