# A 20.0 mL sample of 0.150 M ethylamine is titrated

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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Jonathan Burroughs

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 $3.0×{10}^{-3}$ 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 .

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.

$=3.25-0.786$
$=2.46$
The pH of the solution is calculated as shown below. the pH of the solution is 11.54.
$pH+pOH=145$
$pH=14-pOH$
$=14-2.46$
$=11.54$