pH = 9.26 shows that [NH4OH] > HCl ∴ The mixture is a buffer Let HCl = x mL = x mmol ∴ NH4OH = (300 - x) = (300 - x) m mol and NH4Clformed = x m mol NH4OH unreacted pKb = 14 - 9.26 = 4.74 pOH = pKb + log
[NH4+]
[NH4OH]
4.74 = 4.74 + log
x
(300−2x)
x
300−2x
= 1 ∴ x = 100 mL = Volume of HCl 300 - x = 200 mL = Volume of NH4OH Hence,