Question

Given a diprotic acid, h2a, with two ionization constants of ka1 = 2.3× 10–4 and ka2 = 3.0× 10–12, calculate the ph for a 0.107 m solution of naha.

Answer 1

Answer:

pH = 7.581

Explanation:

• NaHA → NA+  +  HA-
• HA- + H2O ↔ H2A + OH-  ∨  HA- + H2O ↔ H3O+  +  A2-
• H2A + H2O ↔ HA-  +  H3O+
• 2H2O ↔ H3O+  +  OH-

∴ Ka1 = [HA-][H3O+] / [H2A] = 2.3 E-4

∴ Ka2 = [H3O+][A2-] / [HA-] = 3.0 E-12

∴ C NaHA = 0.107 m

mass balance:

⇒ C NaHA = [NaHA] + [HA-] + [A2-] = 0.107 m............(1)

charge balance:

⇒ [Na+] + [H3O+] = [HA-] + 2[A2-] + [OH-]

∴ {OH-] is neglected, come from water

∴ [Na+] = C NaHA = 0.107 m

⇒ C NaHA + [H3O+] = [HA-] + 2[A2-]...........(2)

(2) - (1):

⇒ [H3O+] = [A2-] - [NaHA]

∴ [NaHA] = [H2A]

⇒ [H3O+] = [A2-] + [H2A]

∴ [H2A] = [A2-][H3O+] / Ka1

∴ [A2-] = Ka2*[HA-] / [H3O+]

replacing:

⇒ [H3O+] = (Ka2*[HA-] / [H3O+]) - ([HA-][H3O+] / Ka1)

⇒ [H3O+] = (Ka2*[HA-]*Ka1 - [HA-]*[H3O+]²) / ([H3O+]*Ka1)

asumming: C NaHA ≅ [HA-]

⇒ [H3O+] = √((Ka2*Ka1*CNaHA)/(Ka1 + CNaHA))

⇒ [H3O+] = √(((3.0 E-12)*(2.3 E-4)*(0.107))/(2.3 E-4 + 0.107))

⇒ [H3O+] = √((7.383 E-17)/(0.10723))

⇒ [H3O+] = 2.624 E-8 m

∴ pH = - Log[H3O+]

⇒ pH = - Log(2.624 E-8)

⇒ pH = 7.581