Question

27. At 100°C, k, = 10-12. pH of pure water at
100° C will be
(1) 70
(2) 6.0
(3) 8.0
(4) 12.0​

Answer 1

Explanation:

At 100°c k = 10-12 pH of pure water is near about 7 .

Answer 2

Answer:2. 6.0

Explanation:kw of water(H+)(OH-)

10^-12= (H+)(OH-) at 100°C

Let H+ &OH- = X& X

10^-12= X.X

10^-12= X^2

X= 10^-6

(H+)= (OH-)= 10^-6

pH = - log (H+)

-log10^-6

6log10

pH = 6