Question

At t°C, Kw for water is 6.4 x 10-13. The pH of water at
t°C will be​

Answer 1

Explanation:

Since, we know that

Kw= [H+] + [OH-]

Since, we know that in pure water, concentration of hydrogen ions is equal to hydroxide ions.

[H+] = [OH-]

Thus,

Kw = [H+]²

So,

[H+] = (6.4 x 10-13) ^ 1/2

[H+]  = 0.00000252982

Now as we know

pH= -log [H+]

pH = 5.6

Answer 2

TOPIC :- IONIC EQUILIBRIUM

CONCEPT :- Kw is called ionic product of water. Kw = [H+] [OH-]. And, pH = -log[H+]

SOLUTIONS

Given,

Kw = 6.4 × 10^-13

[H+]² = 6.4 × 10^-13

[H+] = 8 × 10-7

Thus,

pH = - log[8 × 10^-7]

=> pH = 7 - log8

=> pH = 7 - 2log2

=>pH = 7 - 0.6

=> pH = 6.4 Ans.