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Ionic Equilibrum question

2013 · Q112
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Ionic Equilibrum question

2013 · Q112

NEETChemistryIonic EquilibrumMCQ+4 / −1
At 100oC the Kw of water is 55 times its value at 25oC. What will be the pH of neutral solution? (log 55 = 1.74)
  1. A
    7.00
  2. B
    7.87
  3. C
    5.13
  4. D
    6.13
View written solutionFree

Correct answer: D

Kw at 25oC = 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

⇒\Rightarrow⇒ pH = 12{1 \over 2}21​ - log55 - 14log10

⇒\Rightarrow⇒ pH = 6.13

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