Question

The ionic product of water at 25C is 10 exponent 14.Its ionic product at 90C will be

Answer 1

At 25°C, ionic product of water (K_w) is 1* 10^-14 mol²L^-2.

As the temperature increases, dissociation of water also increases. Hence K_w increases.

But at all temperatures, [H⁺] remains equal to [OH^-].

It is given that at 90°C, [H₃0⁺]=10^-6mole/lite

So [OH^-] will be = 10^-6mole/lite

And product of concentration of [H⁺] and [OH^-] ions is equal to ionic product of water.

So K_w = [H⁺] [OH^-] = 10^-6 * 10^-6

K_w = 1*10^-12 mol² L^-2

So value of K_w at 90°C is 1*10^-12 mol² L^-2.