Question

If you add 0.7M Na2A to 0.5M H2A and pKa1= 3 and pKa2= 6 what is the pH?

Answer 1

Na2A --> 2Na + A-2

H2A --> H+ + HA

Note that, initially

[A-2] = 0.7 M

[H2A] = 0.5 M

[HA-] = 0

after reaction

[A-2] = 0.7-0.5 = 0.2 M

[H2A] = 0.5 -0.5 = 0 M

[HA-] = 0 + 0.5 = 0.5 M

now, we have

[A-2] = 0.2 M

[HA-] = 0.5 M

this is a buffer since A-2 and HA- are present:

A buffer is any type of substance that will resist pH change when H+ or OH- is added.

This is typically achieved with equilibrium equations. Both type of buffer will resist both type of additions.

When a weak acid and its conjugate base are added, they will form a buffer

The equations:

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,

apply Henderson Haselbach equations

pH = pKa2 + log(A-2/HA)

pH = 6 + log(0.2/0.5)

pH = 5.6020