Question

Given stocks of 1 M Lactic Acid (pKa 3.86) at pH 3.5, 2 M naOH, 2 M HCl and water: how would you prepare 4 liters of 0.05 M Lactic Acid at pH 3.25? Give 2 M Tris, pH 7.1 (pKa 8.0), 5 M NaOH, 5 M HCl, and water: how would you prepare 2 L of 0.2 M Tris, pH 7.4?

Answer 1

Question 1.

For the lactic acid buffer:

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:

get pKa

pKa = 3.86

Now,

V total = 4 Liter

[Lactic acid] = [HA] = 0.05 M

pH = 3.25

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

3.25 = 3.86 + log([A-]/0.05)

[A-] = 0.05*10^(3.25-3.86)

[A-] = 0.012273 M

we need 0.012273 M of lactte

total mol:

mol of A- required = 0.012273M*4L = 0.049092 mol of lactate

mol of HA required = 0.05*4 = 0.02 mol of lactic acid

Initially:

total HA required = 0.02 + 0.049092 = 0.069092 mol of HA

then, add 0.049092 mol of NaOH so it neutralizes

0.049092 mol of OH- required

M = mol/V

V = mol/M = 0.049092 /2 = 0.024546 Liters of NaOh required = 24.55 mL of NaOH will be requried

V of lactic acid:

V = mol/M = (0.069092 )/(1) = 0.069092 L = 69.10 mL of lactic acid

Then

form it via:

Vacid = 69.10 mL of lactic acid

Vbase = 24.55 mL of NaOH

then, add total volume of water (ionized) until V = 4 L