A 20.0 ml sample of 0.150 m ethylamine is titrated with 0.050 m hcl. you may want to reference (page 805) section 17.4 while completing this problem. part a what is the ph after the addition of 5.0 ml of hcl? for ethylamine, pkb = 3.25.

First, we need to get moles of OH-:

moles of ethylamine = molarity * volume

= 0.15 M * 0.02 L

= 0.003 mol

moles of H + = molarity * volume

= 0.05 M * 0.005 L

= 0.00025 mol

when the total volume = 0.02 L + 0.005L

= 0.025 L

[H+] = moles / total volume

= 0.00025 / 0.025 = 0.01 M

[ethaylamine] = moles / total volume

= 0.003 / 0.025 = 0.12 M

when Pkb = 3.25 so we can get Pka from this formula:

Pka = 14 - Pkb

= 14 - 3.25

= 10.75

by using H-H equation:

PH = Pka + ㏒[salt]/[acid]

= 10.75 + ㏒[0.12/0.01]

= 11.8