1. Acid-Base Titration and pH
CHAPTER 15
Acid-Base Titration and pH

2. 15.1 Aqueous Solutions and the Concept of pH

3. Ionization of Water
Self-ionization of water: two water molecules produce a hydronium ion and hydroxide ion by transfer of a proton
Concentrations are represented by the molecule’s name enclosed in brackets.
Example: [H3O+]
Concentrations of hydronium and hydroxyl ions are inversely proportional - as one increases, the other decreases.

4. Ionization Constant Ionization constant of water is Kw
It’s equal to the concentration of hydronium ions times hydroxyl ions, which equals 1.0 x
[H3O+] x [OH-] = 1.0 x 10-14
Ionization of water increases as temperature increases.

5. Solutions Neutral Solution [H3O+] = [OH-] Acidic Solution
Basic Solution
[H3O+] < [OH-]
Calculating Hydronium and Hydroxyl ion concentrations
Do dissociation problem to determine number of moles of specific ion made via one mole of solute
Use strong acids and bases; they completely ionize

6. Strong Acids and Bases

7. Example
A 1.0 x 10-4 M solution of nitric acid has been prepared in a lab. Calculate the hydronium ion concentration and the hydroxyl ion concentration.

8. Solution HNO3 + H2O  H30+ + NO3- 1.0 x 10-4 M HNO3 =
(1.0 x 10-4 mol / 1 L) x (1 mol H30+ / 1 mol HNO3) = 1.0 x 10-4 M H30+
[H3O+] x [OH-] = 1.0 x 10-14
[1.0 x 10-4] x [OH-] = 1.0 x 10-14
[OH-] = 1.0 x M

9. Example
Barium Hydroxide has a hydronium ion concentration of 1.0 x What is the hydroxyl concentration? What is the molarity of solution?

10. Solution Ba(OH)2 –> Ba2+ + 2OH- [H3O+] x [OH-] = 1.0 x 10-14
[1.0 x 10-11] x [OH-] = 1.0 x 10-14
[OH-] = 1.0 x 10-3 M
1.0 x 10-3 M OH = 1.0 x 10-3 mol / 1L
1.0 x 10-3 mol OH x (1mol Ba(OH)2 / 2 mol OH) = 5.0 x 10-4 M Ba(OH)2

11. pH Scale

12. pH Scale pH: negative of the logarithm of hydronium ion concentration
pH = -log[H30+]
Example:
[H30+] = 1x10-7 M
pH = -log[1x10-7] = 7
pOH: negative logarithm of hydroxide ion concentration
pOH = -log[OH-]

13. pH and pOH Recall that Kw = 1.0 x 10-14 M pH of solutions
Therefore, pH + pOH = 14
pH of solutions
Neutral: pH = 7.0; pH = pOH
Basic: pH > 7.0; pH > pOH
Acidic: pH < 7.0; pH < pOH

14. Calculating pH
If starting from hydronium concentration, just take negative log
If starting from hydroxyl concentration, find hydronium concentration, then take negative log
Calculate pOH similarly
take negative log for hydroxyl concentrations
for hydronium concentrations, find hydroxyl concentration and then take negative log

15. Examples What is the pH of a 1.0 x 10-3 M NaOH solution?
[OH-] = 1.0 x and [H30+] = 1.0 x 10-11
pH = -log(1.0 x 10-11) = 11
What is the pOH of a 1.0 x 10-8 M NaOH solution?
pOH = -log(1.0 x 10-8) = 8
What is the pH of a solution if the [H30+] is 2.7 x 10-3 M?
pH = -log(27 x 10-3) = 2.6

16. Common pH Ranges

17. Calculating Concentrations
Find concentrations from pH in reverse order
Hydronium concentration = 10-pH
Example
Determine the hydronium concentration of an aqueous solution that has a pH of 4.0
[H30+] = 10-pH = 10-4 = 1.0 x 10-4 M

18. Another Example
A shampoo has a pH of What are the hydronium and hydroxyl ion concentrations?
[H30+] = 10-pH = = 2.0 x 10-9 M
[H30+][OH-] = 1.0 x 10-14
[OH-] = 1.0 x / 2.0 x = 5.0 x 10-6 M

19. 15.2 Determining pH and Titrations

20. Indicators
Acid-Base Indicator: compounds whose colors are sensitive to pH
Indicators change colors because they are either weak acids or weak bases
Indicators come in different colors and work over a variety of ranges

21. Different Color Ranges of Various Indicators

22. How Indicators Work Equilibrium indicator eq: HIn <–> H+ + In-
An indicator’s colors result from the fact that HIn and In- are different colors.
Acidic solutions - In- acts as a base and accepts acid protons. Indicator is then present in largely unionized form, Hin.
Basic solutions - H+ ions combine with the base’s OH- ions. The indicator further ionizes since H+ ions have been lost. Indicator is then largely present in the form of In- .

23. Indicators cont’d and pH Meters
Transition Interval: pH range over which an indicator changes color
The lower the pH that an indicator changes colors means the stronger the acid of the indicator.
pH Meter: determines pH of solution by measuring the voltage between two electrodes placed in the solution

24. Titrations
Definition: controlled addition and measurement of amount of solution of known concentration required to react completely with measured amount of solution of unknown concentration

25. More Titration
Equivalence point: point at which two solutions used in titration are present in chemically equivalent amounts
End point: point in titration where indicator changes color
If we know the concentration of one solution, we can find the concentration of the other in a titration from the chemically equivalent volumes.

26. Titration Equivalence Points

27. Titration Solutions
Standard solution: solution that contains a precisely known concentration of a solute
Compare our known solution concentrations with a solution of a primary standard
Primary standard: highly purified solid compound used to check concentration of known solution in titration

28. Steps to Solve a Titration Problem
Start with a balanced equation for neutralization reaction and determine chemically equivalent amounts of acid and base
Determine moles of acid or base from known solution used during titration
Determine moles of solute of unknown solution used during titration
Determine molarity of unknown solution

29. Example
In a titration, 27.4 mL of M Ba(OH)2 are added to a 20.0 mL sample of HCl solution of unknown concentration until the equivalence point is reached. What is the molarity of the acid solution?

30. Solution Ba(OH)2 + 2HCl  2H2O + BaCl2 Ba(OH)2 : 27.4 mL, 0.0154 M
HCl: 20.0 mL, ?M
M = ? Mol Ba(OH)2 / L
? = 4.22 x 10-4 mol Ba(OH)2
4.22 x 10-4 mol Ba(OH)2 x (2HCl / 1 Ba(OH)2) = 8.44 x 10-4 mol HCl
8.44 x 10-4 mol HCl / L = M

31. Example
You have a vinegar solution you believe to be 0.83 M. You are going to titrate mL of it with a NaOH solution that you know to be M. At what volume of added NaOH solution would you expect to see an end point?

32. Solution HC2H3O2 + NaOH  H2O + NaC2H3O2 HC2H3O2: 0.83 M, 0.02000L
NaOH: M, ? L
0.83 M HC2H3O2 = ? Mol / L
? = mol HC2H3O2
mol HC2H3O2 x (1 mol NaOH / 1 mol HC2H3O2) = mol NaOH
0.519 M NaOH = mol / ? L
? = L = 32 mL

33. Created by: Savannah Sisk
THE END
Created by:
Savannah Sisk