How much 10.0 M HNO3 must be added to 1.00 L of a buffer that is 0.0100 M acetic acid and 0.100 M sodium acetate to reduce the pH to 5.30 ?

Question

Acetic acid and its conjugate base acetate can form an acid-base buffer.  The pK_a of acetic acid is 4.75.

Answer 1

HNO3 will react with the base, sodium acetate, as follows:

H+ + CH3COO- ==> CH3COOH

0..........0.100....................0.0100.........Initial

+x..........-x.......................+x.................Change

x........0.100 - x.............0.0100 + x.......Equilibrium

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

5.30 = 4.75 + log (0.1-x / 0.01+x)

log (0.1-x / 0.01+x) = 0.55

(0.1-x / 0.01+x) = 3.55

0.0355 + 3.55x = 0.1-x

x = 0.0142 M = [HNO3] = 0.0142 mol / L

0.0142 mol x 1 L / 10.0 mol = 0.00142 L = 1.42 mls

Check:

H⁺ + CH3COO- ==> CH3COOH

0.0142....0.100...............0.01..........Initial

-0.0142...-0.0142...........+0.0142......Change

0......0.0858.............0.0242...........Equilibrium

Solve for pH and see if it turns out to be 5.30...

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

pH = 4.75 + log (0.0858 / 0.0242)

pH = 4.75 + 0.55

pH = 5.29 (Check: close enough)