Question

At 100C the Kw of water 1s 55 times its value at 25C.What will be the pH of neutral solution ?(log55=1.74)

Answer 1

Kw  = [H+] [OH−] = (1 × 10^14) at 25'C

Kw  = [H+] [OH−] = (55 × 10^14) at 100'C 

[H+]^2 = 55 × 10^14 at 100'C 

2 pH  = − log (55 × 10−14)       
         
          = − log 55 +14 log 10 
           
           = −1.74 + 14 
 
             =12.262

pH = (12.262)/2 = 6.13

Answer 2

Answer:

6.13

Explanation:

at 100 degree celcius the kW of water