1. Section 19.2 Assessment
12. An acid is highly ionized in aqueous solution. Is the acid strong or weak? Explain your reasoning.

2. Section 19.2 Assessment
13. How is the strength of a weak acid related to the strength of its conjugate base?

3. 14. Identify the acid-base pairs:
Section 19.2 Assessment
14. Identify the acid-base pairs:

4. 14. Identify the acid-base pairs: HCOOH + H2O ↔ HCOO- + H3O+
Section 19.2 Assessment
14. Identify the acid-base pairs:
HCOOH + H2O ↔ HCOO- + H3O+

5. 14. Identify the acid-base pairs: HCOOH + H2O ↔ HCOO- + H3O+
Section 19.2 Assessment
14. Identify the acid-base pairs:
HCOOH + H2O ↔ HCOO- + H3O+

6. 14. Identify the acid-base pairs: NH3 + H2O ↔ NH4+ + OH-
Section 19.2 Assessment
14. Identify the acid-base pairs:
NH3 + H2O ↔
NH4+ + OH-

7. 14. Identify the acid-base pairs: NH3 + H2O ↔ NH4+ + OH-
Section 19.2 Assessment
14. Identify the acid-base pairs:
NH3 + H2O ↔
NH4+ + OH-

8. Section 19.2 Assessment
15. Kb for aniline is
4.3 x Explain
what this tells you
about aniline.

9. [HX+ ] [OH-] Kb = ------------------- [X]
Section 19.2 Assessment
[HX+ ] [OH-]
Kb =
[X]

10. [HX+ ] [OH-] 4.3 x 10-10 = ------------------- [X]
Section 19.2 Assessment
[HX+ ] [OH-]
4.3 x =
[X]

11. 4.3 x 10-10 = ------------------- [X]
Section 19.2 Assessment
[HX+ ] [OH-]
4.3 x =
[X]
This number is very small

12. 4.3 x 10-10 = ------------------- [X]
Section 19.2 Assessment
[HX+ ] [OH-]
4.3 x =
[X]
This number is very small.
Very little of X has been able to grab any H+

13. 4.3 x 10-10 = ------------------- [X]
Section 19.2 Assessment
[HX+ ] [OH-]
4.3 x =
[X]
This number is very small.
Very little of X has been able to grab any H+. It is a weak base.

14. 4.3 x 10-10 = ------------------- [X]
Section 19.2 Assessment
[HX+ ] [OH-]
4.3 x =
[X]
This number is very small.
Very little of X has been able to grab any H+. It is a weak base.

15. Section 19.2 Assessment
16. Why is a strong base such as sodium hydroxide generally not considered to have a conjugate acid?

16. Section 19.2 Assessment NaOH ↔ Na+ + OH-
16. Why is a strong base such as sodium hydroxide generally not considered to have a conjugate acid?
NaOH ↔ Na+ + OH-

17. It dissociates completely.
Section 19.2 Assessment
16. Why is a strong base such as sodium hydroxide generally not considered to have a conjugate acid?
NaOH ↔ Na+ + OH-
It dissociates completely.
NaOH → Na+ + OH-

18. Section 19.2 Assessment
17. Which 0.1 M solution would have the greater electrical conductivity?
HCLO Ka = 4.0 x 10-8
HF Ka = 6.3 x 10-4

19. Section 19.2 Assessment Bigger Ka = stronger acid
= more H+ ions in solution
= more electrical conductivity
HCLO Ka = 4.0 x 10-8
HF Ka = 6.3 x 10-4

20. Section 19.2 Assessment Bigger Ka = stronger acid
= more H+ ions in solution
= more electrical conductivity
HCLO Ka = 4.0 x 10-8
HF Ka = 6.3 x 10-4