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.
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Ответ:
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
Ответ:
Good for you :)
Explanation: