Question

Calculate the pH of 100.0 mL of a buffer that is 0.0550 M NH4Cl and 0.180 M NH3 before and after the addition of 1.00 mL of 5.80 M HNO3.

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 = 9.25 for NH4+ ((from NH4Cl))

then

pH = pKa + log(NH3/NH4+)

initially

mmol of NH4+ = MV = (100*0.055) = 5.5 mmol

mmol of NH3 = MV = (0.18*100) = 18 mmol

the addition of HNO3 acid;

mmol of HNO3 = MV = 1*5.8 = 5.8 mmol

there is reaction of NH3 + H+ = NH4+

so

mmol of NH3 left = 18-5.8 = 12.2

mmol of NH4 formed = 5.5+5.8 = 11.3

substitute

pH = 9.25 + log(11.3/12.2)

pH = 9.21671