Question

Consider the titration of 70.0 mL of 0.0300 M (CH3)3N (a weak base; Kb = 6.40e-05) with 0.100 M HBr. Calculate the pH after the following volumes of titrant have been added:

(a) 0.0 mL

pH = _____

(b) 5.3 mL

pH = _____

(c) 10.5 mL

pH = _____

(d) 15.8 mL

pH = _____

(e) 21.0 mL

pH = _____

(f) 27.3 mL

pH = _____

Answer 1

a)

The initial pH:

consider the reaction:

You need to do and ICE table, to calculate the disociation of the base

Initial	0.03M	---------	0	0
Change	-X	--------	+X	+X
Equilibrium	0.03-X	------------	X	X

Using the mass action equation:

and the concentrations at equilibrium:

Solving for X you will get the concentrations

----->using the quadratic formula:

--->

b)

You added a strong acid, thus, all the H+ willl react to neutralize the base and form the conjugate acid

you need to calculate the moles of base an acid initially, using the concentration at equilibrium from above.

moles using the 70mL:

for the conjugate acid is the same

moles :

The moles of acid added are:

After the reaction the moles of are going to decrease:

and the moles of conjugate acid are going to increase

Using the henderson hasselbach equation you can calculate the pOH:

------>since the volume for both species is the same the equation can fit as:

c)

as above:

moles of acid added 10.5mL

inital moles of base- moles of acid added, and initial moles of acid + acid added:

d)

as above:

moles of acid added 15.8mL

inital moles of base- moles of acid added, and initial moles of acid + acid added:

e)

moles of acid added ar 21mL

at this volume you have added the same moles of acid as initial weak base:

thus the pH is going to be given by the the acid formed, you are formiing the same moles as base added:

thus :

by this point the volume is 70+21mL=91mL and the concentration of acid is:

you need to do another ICE table to calculate the pH, in this case for the acid disociation:

Initial	0.023M		0	0
Change	-X		+X	+X
Equilibrium	0.023-X		X	X

Using the mass action equation:

and the concentrations at equilibrium:

Solving for X you will get the concentrations

----->using the quadratic formula:

--->

f)

By this point you have converted all the base into its conjugate acid, so the pH is going to be given by the excess of H+ in the solution

the excess moles of acid is the moles added in the last step:

<---this moles added with the moles in the last step:

from the last step->>>

<----moles of H+

the new moles of H+ are:

and the concentration of H+ in the 97.3mL is: