1.

Kw for water is 6.4 × 10–13. The pH of water at t°C will be

Answer»

- IONIC EQUILIBRIUMCONCEPT :- 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 × 10-7Thus, pH = - log[8 × 10^-7]=> pH = 7 - log8=> pH = 7 - 2log2=>pH = 7 - 0.6=> pH = 6.4 ANS.



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