Question

You wish to make a buffer solution with a pH of 3.5. Which of the following acids (and a salt of its conjugate base) would be the most ideal choice?

Acid A (Ka = 3.2 x 10^-3)

Acid B (Ka = 7.1 x 10^-4)

Acid C (Ka= 1.4 x 10^-5)

Acid D (Ka = 8.5 x 10^-6)

Answer 1

This is an acidic buffer; since there is a weak acid + conjugate base:

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 that explain this phenomena are given below:

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.

Given this:

assume [A-] = [HA] = 1 approx

so

log(1) = 0

then

pH = pKa

if pH = 3.5, then

pKa = 3.5 approx

which is about

Ka = 10^-pKA = 10^-3.5 = 0.000316 or 3.16*10^-4

best answer will be

Acid B (Ka = 7.1 x 10^-4); pKa = -log(7.1*10^-4) = 3.15