We are being asked to calculate the pH for a 0.250 mol NH3 in 1 L solution
NH3 is a neutral amine, therefore we know that it is a weak base.
NH3(aq) + H2O(aq) ⇌ OH–(aq) + NH4+(aq); Kb = 1.8 × 10-5
Calculate the molarity:
Molarity = 0.250 M
From this, we can construct an ICE table. Remember that liquids are ignored in the ICE table.
The Kb expression for NH3 is:
Note that each concentration is raised by the stoichiometric coefficient: [NH3], [OH–] and [NH4+] are raised to 1.
What is the pH of a solution in which 0.250 mol of NH3 (Kb = 1.8 × 10-5) is dissolved in sufficient water to make 1.00 L of solution?
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