Ethylamine → C2H5NH2
C2H5NH2(aq) + HCl(aq) → C2H5NH3+(aq) + Cl-(aq)
Step 1: Calculate the initial amount (in moles) of each species:
• 20.0 mL of 0.150 M C2H5NH2
• 5.0 mL of 0.0981 M HCl
moles C2H5NH2 = 0.003 mol
moles HCl = 4.905×10‒4 mol
Step 2: Construct an ICF chart to determine the final amount of each species.
A 20.0 mL sample of 0.150 M ethylamine is titrated with 0.0981 M HCl. What is the pH after the addition of 5.0 mL of HCl? For ethylamine, pKb= 3.25.
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