The dissociation of hydroxylamine H2NOH in water Kb = 1.5x10-8 forms what two products. (Show ICE...

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The dissociation of hydroxylamine H2NOH in water Kb = 1.5x10-8forms what two products. (Show ICE Table).
You have a 0.340 M solution of the base above. You create an ICEtable and solve for x and get x = 1.45x10^-6
1. What is the hydroxide concentration of your base is[x]
2. the pH of your base is [y]

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The dissociation of Hydroxylamine is as follows:

NH₂OH(aq)Â Â +Â Â H₂O(l)Â Â <----->Â Â NH₃OH⁺(aq)Â Â +Â Â OH^-(aq)

initialÂ Â Â Â Â Â Â Â 0.34Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â

changeÂ Â Â Â Â Â -xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +x

eqmÂ Â Â Â Â Â Â Â Â Â Â 0.34-xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â x

Â Â Â Â Â Â Â Â Â Â Â Â Â The dissociation constant can be expressed as:

Â Â Â Â Â Â Â Â K_b = [Nh3OH+] [OH-] / [NH2OH]

Â Â Â Â Â Â Â Â Â Â Â Â Â Â 1.5 x 10-8 = x² / (0.34-x)

Â Â Â Â Â Â Â Â Â Â Â Given the value of x=1.45x10-6

Â Â Â Â Â Â Â Â Â Â Â a) Conc of OH- = x = 1.45 x 10-6

Â Â Â Â Â Â Â Â Â Â Â b) pH= 14-pOH = 14-[- log (1.45 x 10-6)] = 14-5.838 = 8.162

⁾

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The dissociation of hydroxylamine H2NOH in water Kb = 1.5x10-8 forms what two products. (Show ICE...

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