Answer:
Q1. At what pH is [H₂A] = [HA²]?
This is a common ion effect on pH question
and this is the 1st ionic equation for the diprotic acid H2A
H₂A -------- H⁺ + HA²⁻
pK1=4, then, Log K1 = 4 , K1 = 10⁴
The equilibrium equation becomes K1 = (H⁺)( HA²⁻)/H₂A ---------- Eqn 1
similarly, for the 2nd equation, we have
HA²⁻ -------- H⁺ + A²⁻
pK2=8, then, Log K2 = 8 , K2 = 10⁸
The equilibrium equation becomes K2 = (H⁺)( A²⁻)/HA²⁻ ---------- Eqn 2
Given, if HA²⁻ = H₂A, pK1=4
Using Hasselbach Equation, pH =pK1 + Log ((H⁺)( HA²⁻)/H₂A)
since both concentrations HA²⁻ and H₂A are cancelled out and we have
pH = 4 + pH, 2pH=4
pH=2
Q2. At what pH is [HA²⁻] = [A²⁻]?
Similarly, using the hasselbach equation for the 2nd reaction, we have
pH = pK2 + Log ((H⁺)( A²⁻)/HA²⁻)
Given, if A²⁻ = HA²⁻, pK1=8
we have, pH = 8 + Log ( H⁺)
pH=8+pH
2pH=8
pH=4, Now pH = 4 when [HA²⁻] = [A²⁻]
Q3. Which is the principal species at pH 2.00: H₂A, HA²⁻, or A²⁻?
At pH = 2, and pK1 = 4
from eqn 1, K1 = (H⁺)( HA²⁻)/H₂A
K1/(H⁺) = HA²⁻/H₂A
AntiLog of both sides gives, K1/H⁺ = 10⁻⁴/0.01 = 0.01 = HA²⁻/H₂A (Concentration ratio)
Here it means the acid H₂A, is more dominant or a principal specie in the 1st reaction.
Similarly from eqn2, K2 = (H⁺)( A²⁻)/HA²⁻
taking ratios, we have K2/(H⁺) = A²⁻/HA²⁻
Antilog PK2 and pH, gives 10-8/0.01 = 0.000001 = A²⁻/HA²⁻
Here it means the base HA²⁻, is more dominant or a principal specie in the 2nd reaction.
Q4. Which is the principal species at pH = 6.00: H₂A, HA²⁻, or A²⁻?
At pH = 6, and pK1 = 4
from eqn 1, K1 = (H⁺)( HA²⁻)/H₂A
K1/(H⁺) = HA²⁻/H₂A
AntiLog of pK1 & pH value, K1/H⁺ = 10⁻⁴/10⁻⁶ = 100 = HA²⁻/H₂A (Concentration ratio)
Here it means the base HA²⁻, is more dominant or a principal specie in the 1st reaction.
Similarly from eqn2, K2 = (H⁺)( A²⁻)/HA²⁻
taking ratios, we have K2/(H⁺) = A²⁻/HA²⁻
Antilog PK2 and pH, gives 10⁻⁸/10⁻⁶ = 0.01 = A²⁻/HA²⁻
Here it means the base HA²⁻, is more dominant or a principal specie in the 2nd reaction.
Q5. Which is the principal species at pH = 10.00: H₂A, HA²⁻, or A²⁻?
At pH = 10, and pK1 = 4
from eqn 1, K1 = (H⁺)( HA²⁻)/H₂A
K1/(H⁺) = HA²⁻/H₂A
AntiLog of pK1 & pH value, K1/H⁺ = 10⁻⁴/10⁻¹⁰ = 10⁶ = HA²⁻/H₂A (Concentration ratio)
Here it means the base HA²⁻, is more dominant or a principal specie in the 1st reaction.
Similarly from eqn2, K2 = (H⁺)( A²⁻)/HA²⁻
taking ratios, we have K2/(H⁺) = A²⁻/HA²⁻
Antilog PK2 and pH, gives 10⁻⁸/10⁻¹⁰ = 0.01 = A²⁻/HA²⁻
Here it means the base HA²⁻, is more dominant or a principal specie in the 2nd reaction.
