CHM 2046C    Sample Module 12   Name: __Answer_____

 

Part F.  Acid-Base/pH Titration Calculations     10 points

 

 

You titrate 25.0 mL of 0.10 M NH3 with 0.10 M HCl. Kb = 1.8 x 10-5

 

a.  What is the pH of the solution Before acid has been added?

 

First: before acid has been added, the solution contains only the weak acid weak base and  the problem involves the Kb, and the weak base’s

Initial Concentration:

 

NH4OH  +  HOH  ß à NH4+1   +   OH-1

Or

NH3  +  2 HOH  ß à NH4+1   +  OH-1

 

 

 

           [NH41+] [OH1-]

Kb = --------------------

                [NH3]

 

 

 

                          [x] [x-]

1.8 x 10-5 = ------------- 

                       [0.10 - x]

 

Assume x is small in 0.10 – x so it can be neglected:

 

Therefore: 1.8 x 10-6 = x2

 

[OH1-] = x = 1.34 x 10-3

 

pOH = -Log[OH1-] insert the value for x then calculate the pH

pOH = -Log[1.34 x 10-3]

pOH = 2.87

pH = 14.00 – 2.87 = 11.13

 

b.  What is the pH at the equivalence point?

 

At the equivalence point, the solution contains the salt of the weak acid which hydrolyzes to produce a basic solution. The pH can be calculated using only the hydrolysis reaction.

 

At the equivalence point only water and ammonium chloride salt is present. The chloride ion is the conjugate base of a strong acid so it will NOT hydrolyze. But the ammonium ion will hydrolyze by the following reaction:

 

NH41+ (aq) + HOH    ßà    NH4OH    +     H3O1+

 

              Kw           [NH4OH] [H3O1+]

Ka =   ------  =   ----------------------

               Kb               [NH41+]

 

You must remember to adjust the concentration of the salt for the total volume in the flask (acid + base = new volume)

 

25.00 mL of 0.10M NH3  +  25.00 mL of HCl = 50.00 mL total solution

 

25.00 mL 0.10M NH3 X (0.10 molNH3/1000 ml0.10M NH3) = 0.0025 molNH3

 

For every mole of NH3 in the beginning, there will be one mole of ammonium ions formed from the stoichiometry of the reaction at the equivalence point. Therefore at equivalence point there is 0.0025 molNH4+X

 

0.0025 molammonium / 50.00 mLsolution =  0.050 Mammonium chloride formed

 

 

          1.0 x 10-14           [x] [x]

Ka =  ------------  =   ---------------

          1.8 x 10-5           [0.050 - x]

 

Assume x in 0.050 – x to be so small it can be neglected:

 

                             [x]2

5.6 x 10-10 =  ------------ 

                        [0.050 - x]

 

 

2.8 x 10-11 = [x]2

 

[H3O1+] = x = 5.3 x 10-6

 

pH =  - Log [H3O1+] = - log 5.3 x 10-6 =  5.27

 

 

 

c. What is the pH at the halfway point of the titration?

 

Exactly halfway to the equivalence point, the concentrations of the weak base and its conjugate acid are equal and the pOH equals the pKb  :

[B1+ ]/[BOH] = 1    therefore    the log (1) = 0

 

pOH =  pKb + 0 =  pKb

 

Take the – Log of the pKb to Calculate the pOH

 

pOH = - log 1.8 x 10-5   =  4.74

 

therefore after 25.00 mL of HCl the pH = 14.00 – 4.74 = 9.36

 

so there are three points known in the titration curve:

 

Before any HCl the pH = 11.13 at the start

 

After 25.00 mL of HCl the pH = 9.36

 

At equivalence point the pH =  5.27

 

 

 

If you can do Part (d) successfully it will be worth a 10 point bonus:

 

 

Bonus: d.  Calculate the pH of the solution after adding 5.00, 15.0, 20.0, 22.0, and 30.0 mL of acid. Combine this information with that in parts a., b., and c. and plot the titration curve..