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Grade:12

2 Answers

ABHISHEK JAIN AskiitiansExpert-IITD
20 Points
10 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

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Aman Bansal
592 Points
8 years ago

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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