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The pH of 0.004 M hydrazine solution is 9.7. Calculate its ionization constant Kb and pKb. |
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Answer» Given, pH = 9.7 \(\therefore\) pH = -log10[H+] or [H+] = antilog (-pH) = antilog (-9.7) = 1.67 x 10-10 \(\therefore\) [H+][OH-] = Kw or [OH-] = \(\frac{K_w}{(H^+)}\) = \(\frac{1\times10^{-14}}{1.67\times10^{-10}}\) = 5.98 x 10-5 The concentration of the corresponding hydrazinium ion is also the same as that of OH− ion. The concentration of both these ions is very small so the concentration of the undissociated base can be taken equal to 0.004 M. \(\therefore\) For the reaction, NH2NH2 + H2O ⇌ NH2NH4⊕+ OH⊕ Kb = \(\frac{[NH_4^+][OH^-]}{[NH_3]}\) = \(\frac{Ca.Ca}{C(1-a)}\) = 8.96 × 10−7 pKb = -logKb = -log(8.96 x 10-7) = log 8.96 + 7 log 10 = - 0.9523 + 7 = 6.047 |
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