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abhishek arora Grade: 12
        

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7 years ago

Answers : (2)

ABHISHEK JAIN AskiitiansExpert-IITD
20 Points
										

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


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Abhishek Jain


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7 years ago
Aman Bansal
592 Points
										

Dear Subhadeep,


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


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Aman Bansal


Askiitian Expert



5 years ago
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