1. 10.3
Acid-Base
Reactions
Titrations

2. “Acids And Bases Have Two Different Faces” Coolest Song Ever!

3. Neutralization Reaction
Neutralization Rxn-
A double displacement reaction in which an acid and a base combine to form water and a salt.
Salt-
An ionic compound that is made from the anion from an acid and a cation from a base.

4. ACID-BASE REACTIONS Titrations
H2C2O4(aq) + 2 NaOH(aq) Na2C2O4(aq)+ 2H2O(l)
acid base salt water
Base cation = Na+
Acid anion = C2O Na2C2O4(aq)
Carry out this reaction using a TITRATION.

5. Adding Acid and Bases
When a strong acid reacts with a strong base in the same mole ratio from the balanced --- a neutral aqueous solution of a salt and water is formed.
Why is this important?
pH =7

6. What is an Acid-Base Titration?
is a chemical analysis involving the addition of a known concentration of titrant to a known volume but unknown concentration of sample.
Chemical Indicator-
is used to determine the end point of the reaction.

7. Chemical Indicators Most indicators are weak monoprotic acids
H-indicator (aq) H+ (aq) + indicator - (aq)
Colour #1=acid Colour #2=base

8. Titration Between a Strong Acid and Strong Base
The equivalence point will occur at a pH of 7.
Want an indicator that will change around 7.
Titration Between a Weak Acid and Strong Base
The equivalence point will occur at a pH above 7.
Want an indicator that will change above 7.
Titration Between a Strong Acid and Weak Base
The equivalence point will occur at a pH below 7.
Want an indicator that will change below7.

9. Titration Set up Burette Funnel Pipette Bulb Acid or base (Titrant)
Burette Clamp
Graduated Pipette
Acid or base
(Analyte)
Acid or base
Retort Stand

10. Acid-Base Titration

11. Setup for titrating an acid with a base

12. Sample Problem #1
What is the concentration of a HCl solution when mL of M NaOH is needed to neutralize 20 mL of the acid?
HCl(aq) + NaOH(aq)  NaCl (aq) + H2O(l)
V=20 mL
C= M
V= mL (you will titrate this!)
C= ???? M

13. Solve the problem 1st write the equation for the reaction:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
2nd solve for the amount of moles of the titrant used.
nNaOH mol = CxV= mol/L x L
= 3.85 x 10-3 mol NaOH
Volume found in titration experiment
3rd using stoichiometry, solve for the concentration of HCl , knowing it is a 1:1 mole ratio
C= n = x 10-3 mol= M
V L

14. Sample Problem #2
What volume of mol/L of Mg(OH)2 is needed to react completely with 43.5 mL of mol/L of HCl?
Mg(OH)2 + 2HCl  2H2O (l) + MgCl2 (aq)
C=0.400 mol/L
V= ?
C=0.781 mol/L
V= 43.5 mL 2. 1.
Mole Ratio!
Mg(OH)2 : HCl
1:2
X: mol / 2mol
n Mg(OH)2 = mol
Determine number of
moles of HCl:
nHCl= CxV
= (0.781 mol/L) ( L)
= mol

15. 3. FIX THIS SLIDE! VHCl= n = 0.0166 mol = 0.0414 L C 0.400 mol/L
You now have the number of moles of Mg(OH)2
You can now find the
volume of the analyte.
VHCl= n = mol = L
C mol/L
L x 1000 ml = 41.4 mL 1L

16. Titration Steps
“Titration Technique Using a Buret”

17. Step 1:
The NaOH, the titrant, is placed in the buret. The titrant is the solution of known concentration that is added from the buret into the flask with the unknown called the analyte.

18. Step 2:
The HCl is placed in the Erlenmeyer flask along with approximately mL of distilled water and 2-3 drops of phenolphthalein indicator. Since the solution in the flask is acidic, phenolphthalein is colourless.
Phenolphthalein indicator pH range:

19. Step 3:
NaOH is added to the HCl in the flask. When the NaOH comes in contact with the solution in the flask, it turns pink and then the pink colour quickly disappears. This is because the OH- from the NaOH interact with the phenolphthalein to change the phenolphthalein from colourless to pink.

20. The solution becomes clear again as the hydronium ions from the hydrochloric acid neutralize the added hydroxide ions. As more NaOH is added, it takes longer for the pink colour to disappear.
As it starts taking longer for the pink colour to disappear, the sodium hydroxide is added a drop at a time.

21. The equivalence point of the titration is reached when equal numbers of moles of hydronium and hydroxide ions have been reacted.
When this happens in this titration, the pH of the solution in the flask is 7.0 and the phenolphthalein indicator is colourless.
This would be a good time to stop, however the indicator is still colourless, so must keep going.

22. Titration Curve

23. Step 4:
Add as little excess NaOH as possible. We want to add a single drop of NaOH to the colourless solution in the flask and have the solution in the flask turn pink and stay pink while the contents of the flask are swirled.
This permanent colour change in the indicator is known as the endpoint of the titration and the titration is over.

24. Homework
SECTION 10.3 PRACTICE PROBLEMS
Page 398, #12, 13.