Question #143599
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?
1
Expert's answer
2020-11-14T13:53:57-0500

a) Equllibrium concentration== N2H4(x)×N2H5+OH(x)N_2H_4(x)\times N_2H_5^+OH^-(x)

Equation for equilibrium constant

== [N2H6+][OH]N2H4[N_2H_6^+][OH^-]\over N_2H_4

Substituting values in the equation yields;

x×x0.01xx\times x\over0.01-x =8.9×1016=8.9\times 10^{-16}

=x2+(8.9×1016×0.01M)12=x^2+(8.9\times10^{-16}\times0.01M)^{1\over 2}

Using the quadratic formula;

a=1b=8.6×107c=8.9×1016a=1 b=8.6\times 10^{-7} c=-8.9\times 10^{-16}

b+-b \stackrel{+}{-} b24ac2a\sqrt {{b^2-4ac}} \over2a

==8.6×107+-8.6\times 10^{-7} \overset{+}{-} (8.6×107)24(1)(8.9×10162(1)\sqrt{(8.6\times10^{-7})^2-4(1)(8.9\times 10^{-16}} \over 2(1)

=1.699×106M=-1.699\times 10^{-6}M

b) pH=logKb×CopH=-log\sqrt{K_b\times C_o}

=log(Kb×Co)12=-log (K_b\times C_o)^{1\over2}

=log(8.9×1016×0.01M)12=-log (8.9\times10^{-16}\times 0.01M)^{1\over2}

=log2.983×109=-log 2.983\times 10^{-9}

pH=8.5pH=8.5


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