The pOH of the ammonia (NH₃) solution is 2.74.
Calculation:
We will start by composing a balanced ionization of NH₃ in aqueous.
NH₃(aq) + H₂O (l) → NH₄⁺(aq) + OH⁻(aq)
We'll also measure each compound's concentration at the end of the process.
NH₃ will partially dissociate. The quantity of NH3's dissociates can be disregarded. The concentration of NH₃ there at the final reaction is then nearly identical to that at the beginning reaction. NH₃ has a final concentration of 0.188 M.
NH₃≈ 0.188 M
As NH₄⁺ and OH ⁻ have the same ratio, then
[NH₄⁺]= [OH ⁻]
To determine the expression for the concentration of [OH] ions, we shall organize the accordance with the applicable (base constant) for the process into a written form.
Kb = ([NH₄][OH⁻]) / [NH₃]
Hence, the concentration of [NH₄⁺] can be written in the form of [OH⁻].
Kb = ([OH-] [OH⁻]) / [NH₃]
Kb = [OH⁻]² / [NH₃]
[OH⁻]² = Kb *[NH₃]
[OH⁻] = √(Kb *[NH₃])
Next, to find out the OH⁻ concentration we have to express the concentration as a form of the Kb equation :
[OH⁻] = √(1.76 * 10⁻⁵ * 0.188)
[OH⁻] = √3.31 * 10⁻⁶
[OH⁻] = √1.82 * 10⁻³
Hence,
pOH = -log[OH⁻]
pOH = - log1.82 * 10⁻³
pOH = 3 - log1.82
pOH = 2.74
So, the pOH of the NH₃ solution is 2.74.
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