Ionic EquilibriumHard
Question
Kw of H2O at 373 K is 1 × 10-12. Identify which of the following is/are correct.
Options
A.pKw of H2O is 12
B.pH of H2O is 6
C.H2O is neutral
D.H2O is acidic
Solution
pKw = - logKw = - log 1 × 10-12 = 12.
Kw = [H+][OH-] = 10-12.
[H+] = [OH-]
⇒ [H+]2 = 10-12 ; [H+] = 10-6 ; pH = - log [H+] = - log 10-6 = 6.
H2O is neutral because [H+] = [OH-] at 373 K even when pH = 6.
(D) is not correct at 373 K. Water cannot become acidic.
Kw = [H+][OH-] = 10-12.
[H+] = [OH-]
⇒ [H+]2 = 10-12 ; [H+] = 10-6 ; pH = - log [H+] = - log 10-6 = 6.
H2O is neutral because [H+] = [OH-] at 373 K even when pH = 6.
(D) is not correct at 373 K. Water cannot become acidic.
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