Question

consider the titration of 85 mL of 1.55 diethylamine (kb=1.3e-3) with .25 M HCl.

What is the pH after 263.5mL of HCl is added to the diethylamine solution.

Answer 1

mol of amine = M*V = 85*1.55 = 131.75 mmol of amine

then

mmol of acid = MV = 263.5*0.25 = 65.875mmol of acid added

then

this will be in th buffer zone

that is,

conjugate and acid are present

expct a basic buffer

i.e.

pOH = pKb + log(conjugate/base)

pKb = -ñog(Kb) = -log(1.3*10^-3) = 2.886056

then

pOH = 2.886056 + log(conjugate/base)

calculate the conjugate formed due to acid

conjugate = 0 + 65.875

base after addition = 131.75 -65.875 = 65.875

then

pOH = 2.886056 + log(65.875/65.875 )

pOH = 2.886056

pH = 14-pOH = 14-2.886056

pH = 11.113944