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?
Section 19.2 Assessment
14. Identify the acid-base pairs:
Section 19.2 Assessment
14. Identify the acid-base pairs:
HCOOH + H2O ↔
+
HCOO + H3O
Section 19.2 Assessment
14. Identify the acid-base pairs:
HCOOH + H2O ↔
+
HCOO + H3O
Section 19.2 Assessment
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
-10
4.3 x 10 . Explain
what this tells you
about aniline.
Section 19.2 Assessment
+
[HX ]
[OH ]
Kb = ------------------[X]
Section 19.2 Assessment
+
[HX ]
4.3 x
-10
10
[OH ]
= ------------------[X]
Section 19.2 Assessment
+
[HX ]
4.3 x
-10
10
[OH ]
= ------------------[X]
This number is very small
Section 19.2 Assessment
+
[HX ]
4.3 x
-10
10
[OH ]
= ------------------[X]
This number is very small.
Very little of X has been able to
+
grab any H
Section 19.2 Assessment
+
[HX ]
4.3 x
-10
10
[OH ]
= ------------------[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
+
[HX ]
4.3 x
-10
10
[OH ]
= ------------------[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
16. Why is a strong base such as sodium
hydroxide generally not considered to
have a conjugate acid?
NaOH ↔
+
Na
+
OH
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
HF
-8
10
Ka = 4.0 x
Ka = 6.3 x 10-4
Section 19.2 Assessment
Bigger Ka = stronger acid
+
= more H ions in solution
= more electrical conductivity
HCLO
HF
Ka = 4.0 x 10-8
-4
Ka = 6.3 x 10
Section 19.2 Assessment
Bigger Ka = stronger acid
+
= more H ions in solution
= more electrical conductivity
HCLO
HF
Ka = 4.0 x 10-8
-4
Ka = 6.3 x 10