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

the answer is b because, investigating whether another species of moth also drinks tears might be interesting, but it would not directly explain why this species of moth does.

to explain why this species of moth drinks tears, the scientist could ask: "what nutrients that moths need are in elephant tears? "

the scientist could experiment with moth repellent or oter ways to keep the moths away, but that would not answer his original question.