Question

50.0 ml of an acetic acid (CH₃COOH) of unknown concentration is titrated with 0.100 M NaOH. After 10.0 mL of the base solution has been added, the pH in the titration flask is 5.30. What was the concentration of the original acetic acid solution? (Ka(CH₃COOH) = 1.8 x 10^-5)

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:

get pKa

pKa = -log(Ka) = -log(1.8*10^-5) = 4.7447

now,

initial conditions:

mmol of HA = MV = M*50

mmol of A- = 0

after addition of NaOH:

mmol of NaOH = MV = 10*0.1 = 1 mmol

there is reaction

HA + NaOH = NaA + H2O, neutralization

mmol of HA = MV = M*50 - 1

mmol of A- = 0 + 1

now, substitute in buffer equation

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

since it is the SAME volume M = mol/V then

M1/M2 = mol 1/ mol 2, since V cancels out

pH = pKa + log(mmol of A- / mmol of HA)

substitute data

5.30 = 4.75 + log( 1 / (M*50 - 1))

10^(5.30-4.75) = 1 / (M*50 - 1)

M*50 -1 = 0.281838

M =( 0.281838+1 )/(50) = 0.02563676 mol of HA per liter

then

the original concentration was = 0.02563676 M of CH3COOH