# 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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83% (182 ratings)
###### 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.

83% (182 ratings) ###### 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

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Weak Bases concept. If you need more Weak Bases practice, you can also practice Weak Bases practice problems.

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