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

PKa = 14 - pKb

pKa = (14 - 3.25) = 10.75

pH = pKa + log (remaining base/added acid)

--> base and acid in moles

.15 M x. 02 L =.003 moles Et;.005 M x. 0981 = 4.905 x 10^-4 moles HCl;.003 - 4.905 x 10^-4 =.0025095 remaining moles of Et

pH = 10.75 + (log (((.0025095 moles Et) / (4.095 x 10^-4 moles HCl))) =

pH = 10.75 +.70895 = 11.46 pH units