Question

$K_b$ of $N_2{H}_4$ is $4\times {10}^{-4}$.Then what is the acid dissociation constant of $N_2{H}_5^{+}$ and $N_2{H}_6^{2+}$ respectively?

• A. $Datainsufficient,4\times {10}^{-6}$
• B. $Datainsufficient,2.5\times {10}^{-8}$
• C. $2.5\times {10}^{-9},Datainsufficient$
• D. $2.5\times {10}^{-9},4\times {10}^{-6}$

Answer 1

Answer: (D)

$2.5\times {10}^{-9},4\times {10}^{-6}$

$N_2{H}_4+H_{2}O\rightleftharpoons N_2{H}_5^{+}+O{H}^{-}$

$K_a{K}_b={10}^{-14}=K_w$

$\Rightarrow K_a=\frac{{10}^{-14}}{K_a}=\frac{{10}^{-14}}{4\times {10}^{-4}}=2.5\times {10}^{-9}⟶$ for $N_2{H}_5^{+}$

$N_2{H}_5^{+}+H_{2}O\rightleftharpoons N_2{H}_6^{2+}+O{H}^{-}$

$K_a=\frac{{10}^{-14}}{2.5\times {10}^{-9}}=4\times {10}^{-6}⟶$ for $N_2{H}_6^{2+}$