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At 100°C the Kw of water is 55 times its value at 25°C. What will be the pH of neutral solution? (log 55 = 1.74)

1. 6.13
2. 7.00
3. 7.87
4. 5.13

Correct answer: 6.13

Solution

At 100°C, the Kw of water increases, so [H+] = √Kw. Since Kw is 55 times its value at 25°C (10^-14), Kw = 55 × 10^-14 = 5.5 × 10^-13. Thus, [H+] = √(5.5 × 10^-13) = 2.35 × 10^-7. pH = -log[H+] = -log(2.35 × 10^-7) = 6.13.

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