Question

The equilibrium constants (Ka) for HCN and HF in H2O at 25 degrees are 6.2E-10 and 7.2E-4,respectively. The relative order of base strengths including water is:

Answer 1

1) HCN

2) HF

Step-by-step explanation:

• HCN ↔ H+ + CN-

∴ Ka1 = [H+]*[CN-] / [HCN] = 6.2 E-10

⇒ pKa1 = - log Ka1 = 9.2076

• HF ↔ H+ + F-

∴ Ka2 = [H+]*[F-] / [HF] = 7.2 E-4

⇒ pKa2 = - Log Ka2 = 3.1426

• 2H2O ↔ H3O+ + OH-

∴ Kw = [H3O+]*[OH-] = 1 E-14

∴ pKw = 14

∴ pKw = pKa + pKb

⇒ pKb1 = 14 - pKa1 = 14 - 9.2076 = 4.792

⇒ pKb2 = 14 - pKa2 = 14 - 3.1426 = 10.857

∴ pKb2 > pKb1

⇒ the relative order of base strengths:

1) HCN

2) HF

Because a reducide pKb is associated with a stronge base, while a large pKb revels a weak base.