Question

What is the final pH of 1.0 L of a 10 mM buffered acetic acid solution (at its pKa ) after the addition of 10 µL of 1 M HCl?

Answer 1

if it is at its pKa, 4.75

then

pH = pKa = 4.75

recall that:

The Weak acid equilibrium:

HA(aq) <-> H+(aq) + A-(aq)

Weak acid = HA(aq)

Conjugate base = A-(aq)

Neutralization of H+ ions:

A-(aq) + H+(aq) <-> HA(aq); in this case, HA is formed, H+ is neutralized as well as A-, the conjugate

Neutralization of OH- ions:

HA(aq) + OH-(aq) <-> H2O(l) + A-(aq) ; in this case; A- is formed, OH- is neutralized as well as HA.

Now,

pH = pKa + log(A-/HA)

initially

4.75 = 4.75 + log(A-/HA)

then

10^0 = A-/HA

A-/HA = 1

total mmol:

mmol of buffer = M*V = 1*(10*10^-3) = 0.01 mol of buffer

A- + HA = 0.01 mol

A-/HA = 1

0.01 /2 = 0.005 mol

then:

A- = 0.005 mol

HA = 0.005 mol

After adding:

10 microL of 1M of HCl --> (10*10^-6)(1) = 10^-5 mol of H+

A- + H+ --> HA

A- decreases, HA increases

A- = 0.005 - 10^-5 = 0.00499

HA = 0.005 +10^-5 = 0.00501

Now...

substitute

pH = pKa + log(A-/HA)

pH = 4.75 + log(0.00499/0.00501)

pH = 4.74826