Question

What percentage of total ammonia (NH3 + NH4+) is present as ammonia (NH3) at pH of 7 and pH of 9 if pKa for NH4+ is 9.3. Based on your answer, explain whether low pH is more favorable or high pH for aquatic organisms.

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

pH = pKa + log(NH3/NH4+)

a)

pH = 7

7= 9.3+ log(NH3/NH4+)

NH3/NH4+ = 10^(7-9.3) = 0.0050

NH3 = 0.0050*NH4+

NH3 = 1

NH4+ = 0.0050

% ionization = NH4+ / NH3 *100= 0.0050/1 *100 = 0.5%

B)=

pH = 9

9= 9.3+ log(NH3/NH4+)

NH3/NH4+ = 10^(9-9.3) = 0.50

NH3 = 0.50*NH4+

NH3 = 1

NH4+ = 0.50

% ionization = NH4+ / NH3 *100= 0.50/1 *100 = 50%