🤓 Based on our data, we think this question is relevant for Professor Dunning's class at TEXAS.
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?
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.