Answer:
pH = 9.03
Explanation:
The equilibrium of the NH₄Cl / NH₃ buffer in water is:
NH₃ + H₂O ⇄ NH₄⁺ + OH⁻
Initial moles of both NH₃ and NH₄⁺ are:
0.100L ₓ (0.20 mol / L) = <em>0.0200 moles </em>
The NH₃ reacts with HCl producing NH₄⁺, thus:
NH₃ + HCl → NH₄⁺ + Cl⁻
<em>That means, moles of HCl added to the solution are the same moles are consumed of NH₃ and produced of NH₄⁺</em>
Moles added of HCl were:
0.025L ₓ (0.20mol / L) = 0.0050 moles of HCl. Thus, final moles of NH₃ and NH₄⁺ are:
NH₃: 0.0200 moles - 0.0050 moles = 0.0150 moles
NH₄⁺: 0.0200 moles + 0.0050 moles = 0.0250 moles.
Using H-H equation for bases:
pOH = pKb + log [NH₄⁺] / [NH₃]
<em>Where pKb is -log Kb =</em><em> 4.745</em><em>.</em>
Replacing:
pOH = 4.745 + log 0.0250mol / 0.0150mol
pOH = 4.967
As pH = 14- pOH
<em>pH = 9.03</em>
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