## Guest

20 Points
13 years ago

Dear Abhishek,

(i)  H2O ↔ [H+] [OH-]

so K of water = [H+] [OH-]/H2O                      (concentration of [H+] and [OH-] is 10^-7 M in water)

= 10^-7 × 10^-7/55.5

= 1.8 × 10^-16

(ii) It is a conjugage acid-base pair

so Kw = Ka × Kb

ionic product of water (Kw) = [H+] [OH-]

= 10^-14

Kb =Kw/Ka

=10^-14/6×10^-10

=1.66×10^-5

(iii)

It is also a conjugage acid-base pair

so Kw = Ka × Kb

ionic product of water (Kw) = [H+] [OH-]

= 10^-14

Ka =Kw/Kb

=10^-14/2.5×10^-5

=4×10^-10

Abhishek Jain

Aman Bansal
592 Points
11 years ago

According to Henderson – Hasselbalchequation:
pOH = pKb + log ([salt]/ [base])
pKb = – log Kb = – log .85 x 10-5 = 4.733
Therefore, pOH = 4.733 + log (0.2 / 0.1)
= 4.733 + 0.301 = 5.034
pH = 14 – pOH = 14 – 5.034 = 8.966

Cracking IIT just got more exciting,It s not just all about getting assistance from IITians, alongside Target Achievement and Rewards play an important role. ASKIITIANS has it all for you, wherein you get assistance only from IITians for your preparation and win by answering queries in the discussion forums. Reward points 5 + 15 for all those who upload their pic and download the ASKIITIANS Toolbar, just a simple  to download the toolbar….

So start the brain storming…. become a leader with Elite Expert League ASKIITIANS

Thanks

Aman Bansal