Æ Acid / Base Titrations
Æ pH Indicators
• And much, much more …..
Æ Next Week: Dr. Murphy and Nuclear
Chemistry
Æ Next Friday: “Biotechnology:
Connecting Science, Society, and
Ethics”, Ms.
Michelle Serapiglia
Ph.D. – Student, EFB/FCH
,
What’s a Titration ?
9 An experiment where you add a measured amount of one reactant (for example, an acid or base solution in water) to another reactant (like a solution of acid or base of unknown concentration ) to determine the point of equivalence (where equal moles of each reactant are consumed or
“neutralized”).
9 Useful for determining the Molarity of solutions of unknown concentration.
Titrations: Mixing Acids and
Bases - Neutralization
¾ An acid in solution contributes a proton, a base provides an OH .
Acid: HCl (aq) → H + (aq) + Cl (aq)
Base: NaOH (aq) → Na + (aq) + OH (aq)
Neutralization: H + (aq) + OH (aq) → H
2
O(l) plus NaCl…
(or, H
3
O + + OH → 2 H
2
O (l))
Another Example: 3HCl(aq) + Al(OH)
3
Overall: Acid + Base
(s) → AlCl
3
(aq) + 3 H
2
→ Salt + Water
O (l)
1

Titrations: Neutralizing a Strong
Acid with a Strong Base
9 H
2
SO
4
(aq) + 2NaOH (aq) → Na
2
SO
4
+ 2H
9 What if the molar concentration (M) of H
2
2
0 (l)
SO was unknown and you needed to find out what it
4 is ?
9 First, make up a solution of known NaOH concentration (add solid NaOH of known weight to a specific volume of water; M is known).
SO
4
9 Then add NaOH solution to the H
2 solution and measure pH. When it gets to 7.0 (neutral), record the volume of NaOH added.
9 Volume x Molarity = Volume x Molarity
1000 ml
Volumetric Flask
1 mole of
Copper Sulfate
(160g)
Fill almost full and swirl
Photo from Brown, LeMay, Bursten
Fill to
1000ml line
Neutralization of Sulfuric Acid with NaOH
¾ Assume: Molar NaOH (aq) concentration is known... Let’s say it’s 2.0M
¾ Now, let’s determine the concentration of a known volume of H
2
SO
4
, say 40 ml = 0.04 L
7.0
Phenolphthalein:
Colorless to pink
2

0.04 L of ?? M H
2
SO 05_32.jpg
20 ml = 0.02 L
2.0 M
0.04 mol
2 moles of [H+] from H
2
SO
4
0.02 mol
0.04 L
0.5 M
In General, For Titrations….
Assuming “Monoprotic” Acids
Problem: Molarity is unknown for the acid or base …
Volume x Molarity = Volume x Molarity
Acid Base
Volume x Molarity = L x mol / L = moles moles of acid = moles of base
For diprotic acids: 0.5 mole acid can neutralize one mole of base
3

Titration of a Strong and Weak Base
Original Solutions:
20 ml of 0.10M Base
Titration of a Strong and Weak Acid
Original Solutions:
20 ml of 0.10M Acid
Equivalent Points….
• Equivalent Point: Equal moles of acid and base have reacted.
• Sufficient standard solution of known concentration has been added to react with all of the unknown solute.
• pH = pK a versa)
+ log [base] / [conjugate acid] (or vice
• If [base] = [conjugate acid], then [base] / [acid] =
1 and pH = pK a
4

Fun with Titrations….
‰ How many moles of a diprotic acid like sulfuric acid (H
2
SO
4
) are required to neutralize one mole of a strong base like NaOH ?
; Answer: 0.5 moles
‰ How many ml of 0.1M HCl are required to neutralize 5 ml of 1.2M KOH ?
; Use: Volume x Molarity = Volume x Molarity
; Answer = 60 ml
“Basic” Carbonate Minerals –
Two Equivalence Points
(1) H
3
O + (aq) + CO
3
2-
(aq)
→
HCO
3
(aq) + H
2
O (l)
(2) H
3
O + (aq) + HCO
3
→
H
2
CO
3
+ H
2
O (l)
(aq) pH Buffers and Buffering…..
• Definition of a Buffer : A solution of acidic and basic solutes that resists changes in its pH when acids and bases are added to it.
• Critically important in biological systems.
Blood pH needs to remain near 7.4 and surface seawater is buffered to be 8.1-8.3.
• Buffering Capacity: The amount of acid or base that can be added to a buffer before pH changes significantly.
More on Buffers Later in Course
5