Question

How much 5.60M NaOH must be added to 410.0 mL of a buffer that is 0.0205 Macetic acid and 0.0235 M sodium acetate to raise the pH to 5.75?

Answer 1

According to Henderson Hasselbach equation for acetic acid dissociation,

pKa = pH - log [CH3COO-] / [CH3COOH]

4.75 = pH - log [0.0235] / [ 0.0205]

pH = 4.81 This is the starting pH of the buffer.

Now if we did change the pH to 5.75, what will happen to the ratio;

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

[CH3COO-] / [CH3COOH] = 10

now, we know when a base is introduced,

CH3COOH + OH- -----------------> CH3COO- + H20

By calculating moles of the two species,

initial CH3COOH = 8.405 x 10 ^ -3 mol, initial CH3COO- = 9.635 X 10 ^ -3 mol

now if we introduce a x amount of base, these amounts will modify as the equillibrium shifts to the right. as;

final CH3COOH = 8.405 x 10 ^ -3 - X and final CH3COO- = 9.635 X 10 ^ -3 + X

From the relationship we built earlier we know;

9.635 X 10 ^ -3 + X / 8.405 x 10 ^ -3 - X is 10.

By solving this simple relationship; x = 6.765 x 10 ^ -3 mol

Therefore the amount of 5.60 M NaOH to be added;

1.208 ml