At 100 the Kw of water is 55 times its value at 25 . What will be the pH of neutral solution? (log 55 = 1.74)
- 7.00
- 7.87
- 5.13
- 6.13
Answer
(4) 6.13
Solution
Kw(100 ) = 55 10-14
pKw = 14 - 55 = 14 - 1.74 = 12.26
For a neutral solution, [H+] = [OH-]
⇒ pH = pKw2 = 12.262 = 6.13
Answer: (4)