Question

Please type where possible and be very neat.

You have 1 M solutions of formic acid and sodium formate. Calculate the volume of sodium formate required to make 1 liter of a solution with a pH of 4.1 in which the concentration of formic acid is 0.15 M. The pKa of formic acid is 3.75.

Using the Henderson-Hasselbalch equation, explain why the pH of a solution of a weak acid and its conjugate base does not change upon dilution.

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.

now, pKa = 3.75

pH = 4.1

substitute

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

4.1 = 3.75 + log([Formate ion] / [Formic acid])

10^(4.1-3.75) = [Formate ion] / [Formic acid]

2.2387 = [Formate ion] / [Formic acid]

we know that

[Formic acid] = 0.15; then

2.2387 = [Formate ion] / 0.15

[Formate ion] = 0.15*2.2387 = 0.335805 M of Sodium formate solution required

This is 1 liter so

mol of Formate ion = 0.335805 mol

mol of Fromic acid =0.15

Now...

M = mol/V --> V= mol/M

Vacid= mol/M = 0.15/1 = 0.15 L = 150 mL of formica cid required

Vformat e= mol/M = 0.335805 /1 = 0.335805 L = 336 mL of sodium formate required

Then, add water up to V = 1000 mL, which is 1 Liter solution

B)

as stated before, pH won't change due to equilibirums of acid/conjugate:

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