Question

calculate PH and Poh of.
0.0001M of Koh​

Answer 1

Answer:

pH =10

pOH = 4

Explanation:

KOH ----------> K+ + OH-

0.0001 - - at time =0

0 10^(-4) 10^(-4) at time = t

pOH = - log[OH-]

pOH = - log[10^(-4)] = - (-4)× log 10

pOH = 4

pH + pOH = 14

=> pH = 14 - pOH = 14-4

=> pH = 10

Answer 2

Explanation:

Given

OH = 0.0001

Kw = [ H+] [OH]

10^-14 = [H+] [ 10^-4]

10^-14/10^-4 = [ H+]

10^-10 = [H+]

now,

PH = - log [10^-10]

= 10 log 10

= 10