1. Tutorial 4

2. Objectives: Define titration and pH curves.
Construct titration curves of weak bases with strong Acids.
Apply.

3. pH curve or a titration curve.

4. 1- Consider the titration of 60. 0 ml of 0. 2M NH3 (Kb 1
1- Consider the titration of 60.0 ml of 0.2M NH3 (Kb 1.8x10^-5) with 0.2 M HCl
Before doing anything we have to calculate the volume of HCl required for complete neutralization!
(MV) NH3 = (MV)HCl
(60x0.2) NH3 = (0.2xV)HCl
VHCl= 60 ml

5. Stages of Titration: A) At the beginning [No Hcl added]
B) Before the equivalence point
C) At half way point
D) At the equivalence point
E) After the equivalence point

6. When No HCl has been Added:
A) At the beginning :
When No HCl has been Added:
The only species in the flask is NH3.
NH3  weak base.
Calculate the pH using the equation for weak base:
pOH=-log(√Kb [Base] )
pOH=-log(√ 1.8x10^-5 [0.2] ) = PH= =11.28

7. B) Before the equivalence point:
When 15 mls of 0.2M HCl are added to the flask
NH3 +
HCl
NH4Cl
Before Rx
0.2M x 60ml
0.2M x 15ml
12 mmol
3 mmol
0 mmol
“Limiting reactant”
After Rx
12-3=
9 mmol
3- 3=
Now what we have in the flask is mainly NH3 / NH4Cl Basic buffer

8. C) Half way point to equivalence point:
When 30 mls (total) of 0.2M HCl are added to the flask
NH3 +
HCl
NH4Cl
Before Rx
0.2M x 60ml
0.2M x 30ml
12 mmol
6 mmol
0 mmol
“Limiting reactant”
After Rx
12-6= 6 mmol
6- 6= 0 mmol
Now what we have in the flask is mainly NH3 / NH4Cl of ratio = 1
pH= pka
pH= -log (5.56x10^-10) =9.26

9. D) At the equivalence point:
When 60 mls (total) of 0.2M HCl are added to the flask
Before Rx
0.2M x 60ml
12 mmol
0 mmol
After Rx
12-12=
12- 12=
NH3 +
HCl
NH4Cl
Now what we have in the flask is mainly NH4Cl
Acidic salt

10. D) At the equivalence point:
Calculate the pH using the equation for basic salt.
Total volume
pH = -log √ Ka[Acidic]
pH= -log √ ((5.55x ) (12/(60+60)) pH=5.12
Ka = Kw/Kb
= 1x / 1.8x10^ = 5.55x 10-10

11. E) After the equivalence point:
When 80 mls (total) of 0.2M HCl are added to the flask
NH3 +
HCl
NH4Cl
Before Rx
0.2M x 60ml
0.2M x 80ml
12 mmol
16 mmol
0 mmol
“Limiting reactant”
After Rx
12-12=
16- 12=
4 mmol
Now what we have in the flask is mainly NH4Cl
and HCl
both are sources for H+ in the medium.
However due to the weak acid character of the salt, it contributes with very small amount of H+ compared to the strong acid HCl so it could be ignored.
Consider that we have only the strong Acid HCl in the flask.

12. E) After the equivalence point:
Calculate the pH using the equation for strong Acid
PH= -log [H+]
[H+] = 4 mmol/ ( ml) = 0.03 M
pH = -log [H+] = 1.54

13. Titration curve of weak base with a strong acid

14. Acidic salt + Strong Acid pH= (-log [H+])
Before the End point
Basic buffer
pH= pka+ log[base]/[acid]
At the beginning
Half way point
After the E.P
E.P
Acidic salt only
pH= -log √ Ka[Salt])
Weak Base only
pH=-log(√Kb [Base)
Acidic salt + Strong Acid
pH= (-log [H+])
(ignore the salt and only work on strong Acid )
[base]/[salt] =1
pH= pka

15. Notice the following:
Comparing the titrations of NH3 Kb 1.8x10^-5 and that of pyridine C5H5N Kb 1.9 x 10-9., we found the following
1. The same amount of 0.1 M HCl is required to reach the equivalence point in both cases. (the fact that pyridine is a much weaker base than NH 3has no effect on the amount of Acid required)
2. The pH value at the equivalence point is affected by the base strength . The weaker the base , the lower the pH value at the equivalence point.

16. In General,
The amount of the baseand not its strength that determines the equivalence point.
The weaker the base the lower the pH value at the equivalence point.