Answer:
a) pH = 9.44
b) 25mL of HBr are required.
c) pH = 5.37
d) pH = 9.26
Explanation:
The equilibrium of NH₃ in water occurs as follows:
NH₃(aq) + H₂O(l) ⇄ NH₄⁺(aq) + OH⁻(aq)
Kb = 1.8x10⁻⁵ = [NH₄⁺] [OH⁻] / [NH₃]
The reaction of NH₃ with HBr is:
NH₃ + HBr ⇄ NH₄⁺ + Br⁻
That means when you add X moles of HBr before to reach the equivalence point, the moles are:
Moles NH₃ = Initial moles - moles HBr
Moles NH₄⁺ = Moles HBr
Initial moles NH₃ = 0.050L * (0.05mol / L) = 2.5x10⁻³ moles NH₃
Moles HBr = 0.010L * (0.10mol / L) = 1x10⁻³ moles HBr
Thus, after the reaction:
Moles NH₃ = 1.5x10⁻³moles
Moles NH₄⁺ = 1x10⁻³ moles
Using Henderson-Hasselbalch equation, pH of this solution is:
pOH = pKb + log [NH₄⁺] / [NH₃]
Where pKb is -log Kb = 4.74
And [] could be taken as moles of each compound:
pOH = 4.74 + log [1x10⁻³ moles] / [1.5x10⁻³ moles]
pOH = 4.56
And pH = 14 - 4.56
b) To reach equivalence point, we need to add 2.5x10⁻³ moles HBr:
2.5x10⁻³ moles HBr * (1L / 0.10mol) = 0.025L = 25mL of HBr are required
c)At equivalence point, you will have just 2.5x10⁻³ moles of NH₄⁺ in 50.0mL + 25mL = 75mL. The molar mass of NH₄⁺ is:
[NH₄⁺] = 2.5x10⁻³mol / 0.075L = 0.0333M
The equilibrium of NH₄⁺ is:
NH₄⁺ ⇄ NH₃ + H⁺
Ka = [NH₃] [H⁺] / [NH₄⁺] = 5.556x10⁻¹⁰
[NH₃] = [H⁺] because both comes from the same equilibrium and [NH₄⁺] = 0.0333M:
[X] [X] / [0.0333M] = 5.556x10⁻¹⁰
[X]² = 1.85x10⁻¹¹M
[X] = [H⁺] = 4.3x10⁻¹¹M
pH = -log [H⁺] = 5.37
d) At midpoint, [NH₃] = [NH₄⁺]. Using H-H equation:
pOH = 4.74 + log [NH₃] / [NH₄⁺]
pOH = 4.74
And pH = 14 - 4.74
