R the titration of 70.0 mL of 0.300 M NH 3 with 0.500 M HCl at 25 ∘ C, determine the relative pH at each of these points. Determine the relative pH before the addition of any HCl .

pH = 11.36

Explanation:

Before the addition of any HCl, you have 0.300M NH₃ in equilibrium with water:

NH₃(aq) + H₂O(l) ⇄ NH₄⁺(aq) + OH⁻(aq)

When system is in equilibrium:

Kb = 1.8x10⁻⁵ = [NH₄⁺] [OH⁻] / [NH₃]

Concentrations in equilibrium are:

[NH₃] = 0.300M - X

[NH₄⁺] = X

[OH⁻] = X

1.8x10⁻⁵ = [X] [X] / [0.300M - X]

5.4x10⁻⁶ - 1.8x10⁻⁵X = X²

X² + 1.8x10⁻⁵X - 5.4x10⁻⁶ = 0

Solving for X:

X = -0.00233 → False answer, there is no negative concentrations

X = 0.00231 → Right answer.

As [OH⁻] = X; [OH⁻] = 0.00231

As pOH = -log [OH⁻] = 2.636

14 = pOH + pH

14 - pOH = pH

pH = 11.36

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