A buffer solution contains 0.292 M ammonium chloride and 0.335 M ammonia.

Question

A buffer solution contains 0.292 M ammonium chloride and 0.335 M ammonia.  

If 0.0568 moles of hydroiodic acid are added to 250 mL of this buffer, what is the pH of the resulting solution?  

(Assume that the volume does not change upon adding hydroiodic acid.)

pH =

Answer 1

Initial moles NH₃ = 0.335 mol/L x 0.250 L = 0.08375 mol NH₃

Initial moles NH₄⁺ = 0.292 mol/L x 0.250 L = 0.073 mol NH₄⁺

After addition of 0.0568 moles HCl...

Final moles NH₃ = 0.08375 - 0.0568 = 0.02695 mol NH₃

Final moles NH₄⁺ = 0.073 + 0.0568 = 0.1298 mol NH₄⁺

Final [NH₃] = 002695 mol/0.25 L = 0.1078 M

Final [NH₄⁺] = 0.1298 mol/0.25 L = 0.5192 M

pOH = pKb + log [NH₄⁺]/[NH₃]

pOH = 4.74 + log (0.5192/0.1078)

pOH = 4.74 + 0.68

pOH = 5.42

pH = 14 - 5.42

pH = 8.58