Question

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

** I saw this question answered in another post where the answer given was 0.28 mL, but my homework program says that is wrong...

Answer 1

The pKa of acetic acid is 4.75.

Initial moles of CH3COOH = 1.00 x 0.0100 = 0.0100 mol

Initial moles of CH3COONa = 1.00 x 0.100 = 0.100 mol


Let a be the volume of HNO3 that must be added

Moles of HNO3 added = a/1000 x 10.0 = 0.01a mol

CH3COONa + HNO3 => CH3COOH + NaNO3


Moles of CH3COOH = 0.0100 + 0.01a

Moles of CH3COONa = 0.100 - 0.01a


pH = pKa + log([CH3COONa]/[CH3COOH])

= pKa + log(moles of CH3COONa/moles of CH3COOH)


5.00 = 4.75 + log((0.100 - 0.01a)/(0.0100 + 0.01a))

log((0.100 - 0.01a)/(0.0100 + 0.01a)) = 0.25

(0.100 - 0.01a)/(0.0100 + 0.01a) = 10^0.25 = 1.78

(0.100 - 0.01a) = 1.78 (0.0100 + 0.01a)

0.100 - 0.01a = 0.0178 +0.0178 a

0.0278 a = 0.0822

a = 2.97

Volume of HNO3 = a = 2.97 mL