Problem: Hydroxylamine, NH2OH, is a weak base. The following is the equilibrium equation for its reaction with water:NH2OH(aq) + H2O(l) ⇌ NH3OH+(aq) + OH -       Kb = 9.1 x 10 -9What is the pOH of a 2.37 M NH 2OH solution?a) 1.15b) 3.83c) 4.99d) 6.72e) 8.13

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FREE Expert Solution

We’re being asked to calculate the pOH of a 2.37 M NH2OH solution.


Since NH2OH has a low Kb value, it’s a weak base. Remember that weak bases partially dissociate in water and that bases accept H+ from the acid (water in this case). The dissociation of NH2OH is as follows:


NH2OH(aq) + H2O(l)  OH(aq) + NH3OH+(aq); Kb = 9.1 × 10–9


From this, we can construct an ICE table. Remember that liquids are ignored in the ICE table.




The Kb expression for NH2OH is:



Note that each concentration is raised by the stoichiometric coefficient: [NH2OH], [OH] and [NH3OH+] are raised to 1.


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Problem Details

Hydroxylamine, NH2OH, is a weak base. The following is the equilibrium equation for its reaction with water:

NH2OH(aq) + H2O(l) ⇌ NH3OH+(aq) + OH -       Kb = 9.1 x 10 -9

What is the pOH of a 2.37 M NH 2OH solution?

a) 1.15

b) 3.83

c) 4.99

d) 6.72

e) 8.13

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