Question

$0.16g{N}_2{H}_4(K_b=4\times {10}^{-6})$ are dissolved in water and the total volume made up to $500mL$. Calculate the percentage of $N_2{H}_4$ that has reacted to water in this solution.

Answer 1

The molar mass of hydrazine is $32$ g/mol. $0.16$ g of hydrazine in $500$ mL corresponds to molarity of $\frac{0.16}{32\times 0.5}=0.01M$

	$N_2{H}_4$	$N_2{H}_5^{+}$	$O{H}^{-}$
Initial molarity (M)	$0.01$	$0$	$0$
Equilibrium (M)	$001-x$	$x$	$x$

The expression for base dissociation constant $K_b$ is $K_b=\frac{\left[N_2{H}_5^{+}\right]\left[O{H}^{-}\right]}{\left[N_2{H}_4\right]}$
Substitute values in the above expression.
$4\times {10}^{-6}=\frac{x\times x}{0.01-x}$
Since $K_b$ is very small, $0.01-x\approx 0.01$
Hence, $x=2\times {10}^{-4}$
The percentage dissociation of hydrazine is $\frac{2\times {10}^{-4}}{0.01}\times 100=2$%