When 60 mL of 0.22 M NH4Cl is added to 60 mL of 0.22 M NH3, relative to the pH of the 0.10 M NH3 solution the pH of the resulting solution will:.

Will be more acidic

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