Question

The pH of 0.004 M hydrazine solution is 9.7. Calculate its ionization constant K_b and pK_b.

Answer 1

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^⊕

K_b = \(\frac{[NH_4^+][OH^-]}{[NH_3]}\)

= \(\frac{Ca.Ca}{C(1-a)}\)

= 8.96 × 10⁻⁷

pK_b = -logK_b

= -log(8.96 x 10^-7)

= log 8.96 + 7 log 10

= - 0.9523 + 7

= 6.047