Answer:

Explanation:
N2(g)+O2(g)⇌2NO(g), 
N2(g)+2H2(g)⇌N2H4(g), 
2H2O(g)⇌2H2(g)+O2(g), 
If we add above reaction we will get:
2N2(g)+2H2O(g)⇌2NO(g)+N2H4(g) Eq (1)
Equilibrium constant for Eq (1) is 
Divide Eq (1) by 2, it will become:
N2(g)+H2O(g)⇌NO(g)+1/2N2H4(g) Eq (2)
Equilibrium constant for Eq (2) is 

You answer this by using the pH formula and and the relation of pH and pOH, pH = -log[H+] and 14 = pH + pOH. The correct classification are as follows:
<span>A. [H2O+]=6.0x10^-12
basic
B. [H3O+]=1.4x10^-9
basic
C. [OH-]=5.0x10^-12
acidic
D. {OH-]=3.5x10^-10
acidic
Hope this answers the question.
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