Adiprotic acid has k, values of 8.0 x 10-6 and 2.0 x 10-12 calculate the ph and the equilibrium concentrations of h2a, ha and a2 in 0.100 m solution of the diprotic acid.

Explanation:

(1) H2A + H2O ↔ H3O+  +  HA-

∴ K1 = 8.0 E-6 = ( [ HA- ] * [ H3O+ ] ) / [ H2A ]

(2) HA-  +  H2O   ↔  H3O+  +  A2-

∴ K2 = 2.0 E-12 = ( [ H3O+ ] * [ A2- ] ) / [ HA- ]

(3) 2 H2O ↔ H3O+  +  OH-

∴ Kw = [ H3O+ ] * [ OH- ] = 1 E-14

⇒ K1 >> 1 E3 K2the equilibrium of importance is the (1);  so from de above we can assume that:  [ H2A ] << [ HA- ] and [ A2- ] << [ HA- ]

mass balance:

⇒ M H2A = 0.100 M = [ H2A ] + [ HA- ] + [ A2- ]where we despise [ A2- ], from the assumptopns.

⇒ 0.100 M = [ H2A ] + [ HA- ]... (4)

charge balance:

⇒ [ H3O+ ] = [ HA- ] + 2[ A2- ] + [ OH- ]where we des´pise [ A2- ] and

[ OH-] it come from water

⇒ [ H3O+ ] = [ HA- ](5)

(5) in (4):

⇒ 0.100 = [ H2A ] + [ H3O+ ]

⇒ [ H2A ] = 0.100 - [ H3O+ ](6)

(6) in K1:

⇒ K1 = 8.0 E-6 = [ H3O+ ]² / ( 0.100 - [ H3O+ ]

⇒ 8.0 E-6 * ( 0.100 - [ H3O+ ] ) = [ H3O+ ]²

⇒ [ H3O+ ]² =  8.0 E-7 - 8.0 E-6 [ H3O+ ]

⇒ [ H3O+ ]² + 8.0 E-6 [ H3O+ ] - 8.0 E-7 = 0quadratic formula

⇒ [ H3O+ ] = 8.904 E-4 M

⇒ pH = - Log [ H3O+ ] = - Log ( 8.9 E-4 )

⇒ pH = 3.05

⇒ [ H2A ] = 0.100 - 8.9 E-4

⇒ [ H2A ] = 0.09911 M

⇒ [ HA ] ≅ [ H3O+ ] = 8.9 E-4 M

⇒ [ A2- ] = ( K2 * [ HA- ] ) / [ H3O+ ]

∴ [ H3O+ ] ≅ [ HA- ]

⇒ [ A2- ]≅ K2 = 2.0 E-12

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