Question

Determine the poh of a 0. 188 m nh3 solution at 25°c. the kb of nh3 is 1. 76 × 10-5.

Answer 1

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.

Learn more about NH₃ here:

brainly.com/question/17021611

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