What is the molarity of an aqueous ammonia solution that has an OH- concentration of 0.0011 M? Kb = 1.8x10-5
To determine the molarity of the aqueous ammonia solution, we need to use the equilibrium expression for the reaction of ammonia with water:
NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
We are given the concentration of OH⁻, which is 0.0011 M. We can assume that the concentration of NH₄⁺ is negligible compared to the concentration of NH₃, so we can represent the concentration of NH₃ as (x - 0.0011) M, where x is the initial molarity of NH₃.
The equilibrium constant expression for this reaction is:
Kb = [NH₄⁺][OH⁻] / [NH₃]
The equilibrium constant, Kb, is given as 1.8x10⁻⁵.
To calculate the molarity of the aqueous ammonia solution, we need to rearrange the equilibrium constant expression as follows:
Kb = [OH⁻] / [NH₃]
Rearranging further, we get:
[NH₃] = [OH⁻] / Kb
Plugging in the values, we have:
[NH₃] = 0.0011 M / 1.8x10⁻⁵
Simplifying the expression, we find:
[NH₃] = 61.1 M
Therefore, the molarity of the aqueous ammonia solution is 61.1 M.