Consider the diprotic acid H2A with K1 = 1.00 x 10-4 and K2 = 1.00 x 10-8 . Find the pH and concentrations of H2A, HA-, and A2- in (a) 0.100 M H2A; (b) 0.100 M NaHA; (c) 0.100 M Na2A

Explanation:

H₂A    ⇄   HA⁺  +  A⁻

K₁ = [ HA⁺] [A⁻ ] / [ H₂A ]

HA⁺ ⇄  H⁺ + A⁻

K₂ = [ H⁺] [ A ⁻ ] / [ HA⁺ ]

H₂A           ⇄   HA⁺      +        A⁻

.100                     0                 0

.100 - C                C                C

K₁ = [ HA⁺] [A⁻ ] / [ H₂A ]

Putting the values

1.00 x 10⁻⁴ =   C X C / .1  -  C

10⁻⁵ - 10⁻⁴ C = C²

C²  + 10⁻⁴ C - 10⁵ = 0

C = 8 X 10⁻² M

So concentration of H₂A that is [ H₂A ]  = 1 - C =   0 .1 - 8 X 10⁻² M

= .02 = 2 x 10⁻² M

NaHA    ⇄   Na⁺ + HA⁻

NaHA is a strong acidic salt  so it will ionise 100 %

NaHA    ⇄   Na⁺ + HA⁻

.1                   .1         .1  

concentration of HA⁻  =  .1 M

Na₂A ⇄  2Na⁺  + A⁻²

Na₂A is also a strongly ionic  salt so it will dissociate 100 % .

Na₂A ⇄  2Na⁺  + A⁻²

.1               .1          .1  

concentration of A⁻²  = .1 M  

HA⁺    ⇄       H⁺   +     A⁻

8 X 10⁻²         C₁        C₁  

K₂ = [ H⁺] [ A ⁻ ] / [ HA⁺ ]

Putting the values

1.00 x 10⁻⁸ = C₁ X C₁ /  C

1.00 x 10⁻⁸ = C₁²  /  8 X 10⁻²

C₁² = 8 x 10⁻¹⁰

C₁ = 2.828 x 10⁻¹⁰

[ H⁺]  = 2.828 x 10⁻¹⁰

pH = - log  [ H⁺]  = - log 2.828 x 10⁻¹⁰

= 10 - log 2.828

= 10 - .45

= 9.55 .

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