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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
Ka =Kw/Kb
=10^-14/2.5×10^-5
=4×10^-10
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Abhishek Jain
AskiitiansExpert-IITD
Dear Subhadeep,
According to Henderson – Hasselbalchequation:pOH = pKb + log ([salt]/ [base])pKb = – log Kb = – log .85 x 10-5 = 4.733Therefore, pOH = 4.733 + log (0.2 / 0.1)= 4.733 + 0.301 = 5.034pH = 14 – pOH = 14 – 5.034 = 8.966
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