Problem: 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.

FREE Expert Solution

Ethylamine → C2H5NH2

Reaction:

C2H5NH2(aq) + HCl(aq) → C2H5NH3+(aq) + Cl-(aq)



Step 1: Calculate the initial amount (in moles) of each species:

Given:

20.0 mL of 0.150 M C2H5NH2
5.0 mL of 0.0981 M HCl


molarity (M)=molL

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: Construct an ICF chart to determine the final amount of each species.

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Problem Details

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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