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At 100^oC the K_{w} of water is 55 times its value at 25^\circC. What will be the pH of neutral solution? (log 55=1.74)

Option: 1

6.13


Option: 2

7.00


Option: 3

7.87


Option: 4

5.13


Answers (1)

best_answer

Pure water is always neutral, at different temperatures the pH scale can change but water is always neutral. 
On an increase of temperature range of the pH scale decreases. 

H_2O \rightleftharpoons H^++OH^-

At 25^oC

K_w=[H^+][OH^-]=10^{-14}

At 100^oC

K_w=[H^+][OH^-]=55 \times10^{-14}

pK_w=-log(55 \times 10^{-14})=12.26\\ pH=\frac{12.26}{2}=6.13

the COrrect answer is option 1.

Posted by

Ajit Kumar Dubey

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