moles C2H5NH2=0.150mol C2H5NH3L 20.0 mL x 10-3 L1 mL
moles C2H5NH2 = 0.003 mol
moles HCl=0.0981mol HClL 5.0 mL x 10-3 L1 mL
moles HCl = 4.905×10‒4 mol
Step 2: build an ICF chart to identify the last amount of every species.
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A 20.0 mL sample of 0.150 M ethylamine is titrated through 0.0981 M HCl. What is the pH after the addition of 5.0 mL that HCl? for ethylamine, pKb= 3.25.
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