Question

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

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 4.60 ?

Answer 1

Let volume of HNO3 be V mL

mol of HNO3 added = 10*V mmol

Before adding HNO3

Before Reaction:

mol of CH3COO- = 0.1 M *1000.0 mL

mol of CH3COO- = 100 mmol

mol of CH3COOH = 0.01 M *1000.0 mL

mol of CH3COOH = 10 mmol

10*V HNO3 will react with 10*V of CH3COO- to form extra 10*V of CH3COOH

After adding HNO3

mol of CH3COOH = 10+10*V mmol

mol of CH3COO- = 100-10*V mmol

we have below equation to be used:

This is Henderson–Hasselbalch equation

pH = pKa + log {[conjugate base]/[acid]}

4.6 = 4.75+log {[CH3COO-]/[CH3COOH]}

log {[CH3COO-]/[CH3COOH]} = -0.15

[CH3COO-]/[CH3COOH] = 0.7079

So,

(100-10*V)/(10+10*V) = 0.7079

100-10*V = 7.0795 + 7.0795*V

(10+7.0795)*V = 100 - 7.0795

V = 5.44 mL

Answer: 5.44 mL