What is the Kb value for a 0.17M solution of unknown with a pH of 9.45?

Kb = 4.673 E-9

Explanation:

B (aq) + H2O (l) ↔ BH + (aq) + OH - (aq) Kb = [BH+][OH-] / [B] pH + pOH = 14

∴ C B = 0.17 M

∴ pH = 9.45

⇒ pOH = 14 - 9.45 = 4.55

⇒ pOH = - Log [OH-] = 4.55

⇒ [OH-] = 2.8184 E-5 M

mass balance:

⇒ C B = 0.17 M = [B] + [BH+]

charge balance:

⇒ [OH-] = [BH+] ... [H3O+] is neglected, it come frome the water

⇒ [B] = 0.17 - [BH+]

⇒ [B] = 0.17 - [OH-] = 0.17 - 2.8184 E-5

⇒ [B] = 0.16997 M

⇒ Kb = [OH-]² / [B]

⇒ Kb = (2.8184 E-5) ² / (0.16997)

⇒ Kb = 4.673 E-9