Question #7a067

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Question #7a067

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What is the stoichiometric equation corresponding to the equilibrium mass action expression given below?
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Answer 1 · 2015-03-18T08:22:25

[H2NNH2]=2M [H2NNH3+]= $2.45 \cdot 10^{- 3}$ $2.45 *10^-3$ M [OH-]= $2.45 \cdot 10^{- 3}$ $2.45 *10^-3$ M

At equilibrium the concentration of N2H4 will be (2-x) M and the concentration of N2H5+ and OH- will be x M.
From the Kb we can write $x^{2}$ $x^2$ / $(2 - x)$ $(2-x)$ = $3 \cdot 10^{- 6}$ $3 *10^-6$
Neglecting $x$ $x$ in $(2 - x)$ $(2-x)$ since it is $⟨ ⟨ 2$ $<<<<2$ , the equation simplifies in $x^{2}$ $x^2$ / $2$ $2$ = $3 \cdot 10^{- 6}$ $3 *10^-6$ and resolving $x$ $x$ = $2.45 \cdot 10^{- 3}$ $2.45 *10^-3$

[OH-]= $2.45 \cdot 10^{- 3}$ $2.45 *10^-3$ pOH= $2.61$ $2.61$ pH= $11.39$ $11.39$

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Question #7a067

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