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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) [NEET Kar. 2013]
  • a)
    6.13
  • b)
    7.00
  • c)
    7.87
  • d)
    5.13
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
At 100°C the Kw of water is 55 times its value at 25°C. What w...
Kw at  25°C = 1 × 10–14 At 25ºC
Kw = [H+] [OH] = 10–14 At 100°C (given) Kw = [H+] [OH] = 55 × 10–14
∵ for a neutral solution [H+] = [OH]
∴ [H+]2 = 55 × 10–14
or [H+] = (55 × 10–14)1/2
∵  pH = – log [H+]
On taking log on both side – log [H+] = –log (55 × 10–14)1/2
pH =  – 0.87 + 7      
=  6.13
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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) [NEET Kar. 2013]a)6.13b)7.00c)7.87d)5.13Correct answer is option 'A'. Can you explain this answer?
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