Question

A 25mL sample of 0.10M C2H5NH2 (ethylamine) is titrated with 0.150M HCl. What is the pH of the solution after 9mL of acid have been added to the amine?

[Kb(C2H5NH2) = 6.5 X 10^-4]

Answer 1

Given that Kb(C2H5NH2) = 6.5 X 10^-4

pKb = - log Kb

= - 6.5 X 10^-4

= 3.18

We know that ; number of mole s= molarity * volume in L

Then;

Moles Acid HCl = 0.150 mol/L x .00900L

= 0.00135 mol

Moles of C2H5NH2 = 0.10 mol/L x0 .0250L

= 0.0025 mol

Hence
Acid will be used up; 0.00135 mol C2H5NH3+ will be produced.

Therefore;
Base ; C2H5NH2 is in excess:

0.0025 mol - 0.00135 mol = 0 .00115 mol.

Total volume = 25 ml ++9 ml =3 4 ml

= 0.034 L

Then calculate molalrity of C2H5NH2 as follows:

Molarity = number of mole s / Total volume in L

= 0 .00115 mol /0.034 L

= 0.0338 M

[B] = moles B / new volume = .00115mol/.0340L = .0338mol/L
and

[BH+] = moles BH+/new volume = .00135mol/.034L = .0397mol/L

[OH-] = x (unknown)
Kb = [0.0397][x]/[.0338]
x = (6.5x10-4)[.0338]/[.0397]
x = [OH] = [.00055]
pOH = 3.26, pH = 14 - pOH = 10.74