1. Acids and Bases

2. Part 1
Arrhenius acids produce ____________ in water.
Arrhenius bases produce ______________ in water.
What is the difference between a Bronstead-Lowry Acid and Base?
Hydrogen Ions
Hydroxide Ions
Acids donate H+ and Bases accept H+

3. Calculate the pH of a 0.045 M solution of KOH.
log M = 1.34 (pOH)
14 – 1.34 = 12.66 OR
[0.045M] [H+] = 1 x 10-14
[H+] = 2.22 x 10 –13
-log 2.22 x 10 –13 = 12.66

4. Write an equation showing HCO3-1 acting as an acid dissolved in water.
HCO3-1 + H2O  CO3-2 + H3O+
Write an equation showing HCO3-1 acting as a base dissolved in water.
HCO3-1 + H2O  H2CO3 + OH-
Vocabulary word to describe HCO3-1
Amphoteric!

5. Which of the following indicate an acidic solution?
[H+] < [OH-]
pOH > 7
[H+] > 1 x 10-7 M
Basic
Acidic
Acidic

6. If 26.5 mL of a 0.2 M solution of NaOH is required to titrate 50 mL of H2SO4, what is the concentration of H2SO4?
H2SO4 + 2NaOH  Na2SO4 + 2H2O
0.2 M = x moles = moles NaOH L
moles NaOH  1:2 ratio  M M = moles = M mol H2SO L

7. Write the conjugate acid for each:
HSO4- SO3-2
Write the conjugate base for each:
HCO3- HCl
H2SO4
HSO3-
CO3-2
Cl-

8. If the pOH of a solutions is 3.96, calculate:
pH:
Acid or Base:
= 9.12 x M
= 1.1 x 10-4 M
14 – 3.96 = 10.04
Base

9. What is the difference between a strong and a weak acid
What is the difference between a strong and a weak acid? Give an example of each.
Strong acids completely disassociate or ionize in water. HCl
Weak acids do not completely disassociate in water. HC2H3O2.

10. 1x10-14 = 9.54x10-8 * [OH- 9.54x10-8 9.54x10-8 [OH-] = 1.05x10-7
You have two solutions, one has a [OH-] = 3.21 x M and one has a [H+] = 9.54 x 10-8 M. Which solution is more basic? Explain why
1x = 9.54x10-8 * [OH- 9.54x x10-8
[OH-] = 1.05x10-7
3.21x10-10 < 1.05x10-7
The second one, has a greater amount of OH-

11. Given a solution with a [H+] of 4.73 x 10-6 M calculate:
pH:
pOH:
[OH-]:
Acid or Base:
-log 4.73 x 10-6 M = 5.33
14 – 5.33 = 8.67
= 2 x 10-9 M
Acid

12. What is the pH of the solution when 26 mL of 0
What is the pH of the solution when 26 mL of 0.2 M NaOH is added to 50 mL of M HNO3? Show work.
0.2M NaOH = x mol/0.026 L
x = moles NaOH
0.104M HNO3 = x mol/0.050 L
x = moles HNO3
The pH is 7, the # of moles of each is equal.

13. What must a buffered solution contain? Give an example.
A weak acid and a conjugate base. HC2H3O2 and C2H3O2-

14. Calculate the pOH of a solution with a [H+] = 4.96 x 10-5 M.
log 4.96 x 10-5 M = 4.3 (pH)
14 – 4.3 = 9.7