1. Acids, Bases & pH

2. Types of solutes Non-electrolyte -
no conductivity
Non-electrolyte -
No dissociation into ions, all molecules in solution
sugar

3. Types of solutes Electrolyte - Dissociation into ions in solution.
conductivity
Electrolyte -
Dissociation into
ions in solution. H+
Cl-

4. Electrolytes Acids, bases, and salts are electrolytes.
HNO3, NaOH, KBr.

5. Acid-Base Theory - Arrhenius
Arrhenius defined an acid as follows:
An acid is a substance that produces H+ ions when dissolved in water (now described as hydronium rather than H+).

6. Hydronium Ion (H3O+) or (H+(aq))

7. Acid-Base Theory - Arrhenius
Arrhenius defined a base as follows:
A base is a substance that produces OH- ions when dissolved in water

8. Acid-Base Theory - Arrhenius
We can demonstrate these definitions as follows:
H2O
HNO3(l) H+(aq) + NO3-(aq)
H2O
Ca(OH)2(s) Ca2+(aq) + 2OH-(aq)

9. pH scale
The pH scale is used to indicate how acidic or basic a solution is by measuring the relative amount of H+ and OH- in a solution.

10. The pH scale

11. pH and pOH scales
Acidic Neutral Basic
pH + pOH = 14.00

12. pH and pOH formulas Remember that H+ = H3O+ pH = -log[H+]
pOH = -log[OH-]
[H+] x [OH-] = 1.0 x 10-14
pH + pOH = 14.00
[H+] = antilog(-pH)
[OH-] = antilog(-pOH)

13. Find the [OH-] in a solution where [H3O+] = 6.80 x 10-10.
pH = -log[H+]
pOH = -log[OH-]
[H+] x [OH-] = 1.0 x 10-14
pH + pOH = 14.00
[H+] = antilog(-pH)
[OH-] = antilog(-pOH)

14. Find the [OH-] in a solution where [H3O+] = 6.80 x 10-10.
[H+] x [OH-] = 1.0 x 10-14
[6.80 x 10-10] x [OH-] = 1.0 x 10-14
1.47 x 10-5 M

15. Find the pH of a solution where [H3O+] = 7.01 x 10-6.
pH = -log[H+]
pOH = -log[OH-]
[H+] x [OH-] = 1.0 x 10-14
pH + pOH = 14.00
[H+] = antilog(-pH)
[OH-] = antilog(-pOH)

16. Find the pH of a solution where [H3O+] = 7.01 x 10-6.
pH = -log[H+]
pH = -log[7.01 x 10-6]
5.15

17. Find the [OH-] in a solution with a pOH of 4.976.
pH = -log[H+]
pOH = -log[OH-]
[H+] x [OH-] = 1.0 x 10-14
pH + pOH = 14.00
[H+] = antilog(-pH)
[OH-] = antilog(-pOH)

18. Find the [OH-] in a solution with a pOH of 4.976.
[OH-] = antilog(-pOH)
[OH-] = antilog(-4.976)
1.06 x 10-5 M

19. What is the pH of a solution with a [OH-] = 9.80 x 10-9.
pH = -log[H+]
pOH = -log[OH-]
[H+] x [OH-] = 1.0 x 10-14
pH + pOH = 14.00
[H+] = antilog(-pH)
[OH-] = antilog(-pOH)

20. What is the pH of a solution with a [OH-] = 9.80 x 10-9.
pOH = -log[OH-]
pOH = -log[9.80 x 10-9]
pOH = 8.01
pH + pOH = 14.00
14.00 – 8.01 =
5.99

21. Acid Base Neutralization
HCl + NaOH 
HCl + NaOH  NaCl + HOH
Acid + Base  Salt + Water
salt water

22. Acid-Base Neutralization Reactions
HCl + NaOH  NaCl + H2O
The products of an acid-base neutralization reaction are a salt made up of the positive ion of the base (Na+) and the negative ion (Cl-) of the acid, and water.

23. HCl + NaOH  NaCl + H2O
Remember, in writing the correct formula for the salt, the total positive charge of the ions in the salt must equal the total negative charge of the ions in the salt.
Make sure you write the correct formula for the products, salt and water first, then balance the equation.

24. Write the balanced equation for the acid-base neutralization of perchloric acid, HClO4 and potassium hydroxide, KOH
HClO4 + KOH  KClO4 + H2O

25. Ca(OH)2 + 2HNO3  Ca(NO3)2 + 2H2O
Write the balanced equation for the acid-base neutralization of nitric acid, HNO3 and calcium hydroxide Ca(OH)2
Ca(OH)2 + 2HNO3  Ca(NO3)2 + 2H2O

26. MA x VA = MB x VB
Acids and bases often react together in what is called a neutralization reaction. The amount (volume) of acid or base and the strength (molarity) of acid and base needed to neutralize each other can “sometimes” be found by using the formula above.

27. Formula cannot be used: Ca(OH)2 + 2HNO3  Ca(NO3)2 + 2H2O
MA x VA = MB x VB
We can use this formula when the acid and base react together in a 1:1 ratio.
Formula can be used:
HClO4 + KOH  KClO4 + H2O
Formula cannot be used:
Ca(OH)2 + 2HNO3  Ca(NO3)2 + 2H2O

28. MA x VA = MB x VB
15.6ml of acid neutralize 25.8ml of 0.50M base. What is the molarity of the acid?
0.83M

29. MA x VA = MB x VB
32.6 ml of 1.25M acid neutralize how many ml of 5.60M base?
7.28 mL

30. Homework Summarize the Lab: “Acid – Base Titration”
Worksheet: Acids, Bases and pH