Given,
[H+] = 7.62 x 10-2 M
Autoionization constant of water, Kw = 10-14 at 25 oC
Now,
Kw = [H+][OH-] = 10-14
or, [OH-] = 10-14/(7.62 x 10-2)
= 1.31 x 10-13 M
Hence, the concentration of OH- = 1.31 x 10-13 M
---
Now, pOH = - log[OH-]
= - log(1.31 x 10-13)
= 12.9
Hence, the pOH of the solution = 12.9
---
pH = -log[H+]
or, pH = -log(7.62 x 10-2)
or, pH = 1.12
Hence, the pH of the solution = 1.12
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