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

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

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

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

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

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

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

Section 19.2 Assessment 15. Kb for aniline is 4.3 x 10-10. Explain what this tells you about aniline.

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

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

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

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

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

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

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

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-

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-

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

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

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