Question

1. What do you think the graphs might have looked like if you added strong acid (HCl) to the water and acetate buffer? Sketch a graph to illustrate your answer.

2. If you needed to prepare a buffer solution of pH 7.4, what conjugate acid/base system might you choose and why?

3. What would be the most optimal pH range for a buffer system made of TRIS-acetate buffer? Its pKa = 8.3.

Answer 1

Q1.

Water alone:

If we added HCl; which is a strong acid, this is, HCl dissociates as H+ and Cl- at 100% we will expect a drastic increase in H+ concentration.

Recall that if H+ increases, then pH decreases, meaning that acidity increased

For acetic acid buffer

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.

Therefore; the pH will not increase drastically; since the [H+] addition is buffered by Acetate ion (A-)

Graphs:

Water

Buffer

Q2.

If pH = 7.4

the best option is to get an acid with pKa values nearest to that

recall that from buffer equation:

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.

Note that the equilibirum equation can be mathematically manipulated in order to favour the "buffer" construction

Recall that, in equilibrium

Ka = [H+][A-]/[HA]

Multiply both sides by [HA]

[HA]*Ka = [H+][A-]

take the log(X)

log([HA]*Ka) = log([H+][A-])

log can be separated

log([HA]) + log(Ka) = log([H+]) + log([A-])

note that if we use "pKx" we can get:

pKa = -log(Ka) and pH = -log([H+])

substitute

log([HA]) + log(Ka) = log([H+]) + log([A-])

log([HA]) + -pKa = -pH + log([A-])

manipulate:

pH = pKa + log([A-]) - log([HA])

join logs:

pH = pKa + log([A-]/[HA])

which is Henderson hasselbach equations.

then, the best value is when A- = HA approx,

for this, best value will be pKa

pH = pKa approx for

Hypochlorous acid;

pKa for HClO = 7.46

use any salt with ClO- anion, such as NaClO or KClO for instance

Q3

Most optimal pH range is typically +/- 1 units of the pKa, since ratio between Tris base/acid is 1:10

then

pH range = 7.3 to 9.3