the pH of 0.04M hydrazine solution is 9.7.Calculate ionization constant kb and PKb
Asked by | 4th Feb, 2012, 03:50: PM
NH2NH2 + H2O ------> NH2NH3+ + OH
From the given pH the Hion concentration can be measured.
So, we have:[H+] = antilog (pH) = antilog (9.7) = 1.67 × 1010
Now, [OH] = Kw / [H+] = 1 × 1014 / 1.67 × 1010
= 5.98 × 105
The concentration of the corresponding hydrazinium ion is also the same as that of hydroxyl ion. The concentration of both these ions is very small so the concentration of the undissociated base can be taken equal to 0.004M.
Thus, Kb = [NH2NH3+][OH] / [NH2NH2]= (5.98 × 105)2 / 0.004 = 8.96 × 107
So, pKb = logKb = log(8.96 × 107) = 6.04.
Answered by | 5th Feb, 2012, 10:21: AM
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