Answer to Question #143599 in General Chemistry for Johannes Steven

Question

Hydrazine N2H4 can interact with water in two steps

N2H4(aq) + H2O (I) = N2H5^+(aq) + OH^-(aq) kb1 = 8.5 x 10^-7

N2H5^+(aq) + H2O (I) = N2H6^2+(aq) + OH^-(aq) kb2 = 8.9 x 10^-16

(a) what is the concentration of OH^-, N2H6^2+ in a 0.010 M solution of hydrazine?
(b) what is the pH 0.010 M solution of hydrazine?

Answer 1

a) Equllibrium concentration$=$ $N_2H_4(x)\times N_2H_5^+OH^-(x)$

Equation for equilibrium constant

$=$ $[N_2H_6^+][OH^-]\over N_2H_4$

Substituting values in the equation yields;

$x\times x\over0.01-x$ $=8.9\times 10^{-16}$

$=x^2+(8.9\times10^{-16}\times0.01M)^{1\over 2}$

Using the quadratic formula;

$a=1 b=8.6\times 10^{-7} c=-8.9\times 10^{-16}$

$-b \stackrel{+}{-}$ $\sqrt {{b^2-4ac}} \over2a$

$=$$-8.6\times 10^{-7} \overset{+}{-}$ $\sqrt{(8.6\times10^{-7})^2-4(1)(8.9\times 10^{-16}} \over 2(1)$

$=-1.699\times 10^{-6}M$

b) $pH=-log\sqrt{K_b\times C_o}$

$=-log (K_b\times C_o)^{1\over2}$

$=-log (8.9\times10^{-16}\times 0.01M)^{1\over2}$

$=-log 2.983\times 10^{-9}$

$pH=8.5$