10-24. the diprotic acid h2a has pk1 5 4.00 and pk2 5 8.00. (a) at what ph is [h2a] 5 [ha2]? (b) at what ph is [ha2] 5 [a22]? (c) which is the principal species at ph 2.00: h2a, ha2, or a22? (d) which is the principal species at ph 6.00? (e) which is the principal species at ph 10.00?

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.

here is table 2-2:

country

percent of eligible voters that voted

rank

united states

44.9%

138

canada

60.1%

107

costa rica

84.6%

14

hungary

66.9%

82

japan

57.0%

118

egypt

27.7%

159

south africa

85.5%

10

ans is no, the data set cannot be plotted as a single circle graph