Answer to Question #71387 in Inorganic Chemistry for Mayuresh Dhamanskar

Question

To determine the PH of acid solution like 0.1M HCL, 0.05 M HCL,0.025M HCL and 0.0125M HCL by using quinhydrone and calomel electrodes and to calculate their degree of dissociation and dissociation constant

Answer 1

HCl – strong acid, its degree of dissociation = 1
pH = -lg H+ = -lg 0.1 = 1
pH = -lg H+ = -lg 0.05 = 1.3
pH = -lg H+ = -lg 0.025 = 1.6
pH = -lg H+ = -lg 0.0125 = 1.9