Answer:
Will be more acidic
Step-by-step explanation:
The equilibrium of NH3 in water is:
NH3(aq) + H2O(l) ⇄ NH4⁺(aq) + OH⁻(aq).
Where equilibrium constant, Kb, is:
Kb = 1.85x10⁻⁵ = [NH4⁺] [OH⁻] / [NH3]
From 0.10M NH3, the reaction will produce X of NH4⁺ and X of OH⁻ and Kb will be:
1.85x10⁻⁵ = [X] [X] / [0.10M]
1.8x10⁻⁶ = X²
X = 1.34x10⁻³ = [OH⁻]
As pOH = -log[OH⁻] = 2.87
And as pH = 14 - pOH
pH of the 0.10M NH3 is 11.13
Now, to find the pH of the NH4Cl and NH3 we need to use H-H equation for bases:
pOH = pKb + log [NH4⁺] / [NH3]
Where pKb is -log Kb = 4.74 and [] are moles of both compounds.
Moles of [NH4⁺] = [NH3] = 60mL, 0.060L*0.22M = 0.0132moles:
pOH = 4.74 + log [0.0132] / [0.0132]
pOH = 4.74
pH = 14 - 4.74 = 9.26
That means the pH of the resulting solution will be more acidic