A 20.0 mL sample of 0.150 M ethylamine is titrated

elvishwitchxyp 2021-12-15 Answered
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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Expert Answer

Jonathan Burroughs
Answered 2021-12-16 Author has 37 answers

The volume of 1 L is equal to 1000 mL. The number of mole (n) of ethylamine is calculated as shown below where 20.0 mL is the volume of the solution, 0.150 M is the given concentration of ethylamine. The number of mole (n) of ethylamine is 3.0×103 mol.
n=0.150 mol1000 mL×20.0 mL
=3.0×103 mol
The number of mole (n’) of HCl is calculated as shown below where 5 mL is the volume of the concentration of 0.0981 M of HCl. The number of mole (n’) of HCl is 4.91×104 mol.
n=0.0981 mol1000 mL×5.0 mL
=4.91×104 mol
The pOH of the solution after the addition of 5.0 mL of 0.0981 M of HCl in 20.0 mL of 0.150 M of ethylamine is calculated as shown below where 25 mL is the total volume of the solution. On substitution, the pOH of the solution after the addition of 5.0 mL of 0.0981 M of HCl in 20.0 mL of 0.150 M of ethylamine is 2.46.
pOH=pKb+log(n/25 mLn/25 mL)
=3.25+log(4.91×104 mol25 mL3.0×103 mol25 mL)
=3.250.786
=2.46
The pH of the solution is calculated as shown below. the pH of the solution is 11.54.
pH+pOH=145
pH=14pOH
=142.46
=11.54

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