1.

Derive ionic product of water. Also find the value at 25°C

Answer»

When high current is passed through water it under goes partial dissociation.

H2O ⇌ H+ + OH-

Applying law of mass action,

K = \(\frac{[H^+][OH^-]}{[H_2O]}\); 

K[H2O] = [H+][OH-]

But k[H2O] = Kw

∴ Kw = [H+][OH-]

Where, Kw is ionic product of water.

Value of Kw at 25°C:

It is found that at 25°C [H+] = [OH–] = 10-7 mol/dm3

∴ Kw = [H+][OH–] = 10-7 × 10-7 = 10-14 (mol/dm3)2

At 25°C the value of Kw is 10-14(tmol/dm3)2


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