Entry Task: Feb 1st Friday

Question:A 0.55 M solution of a weak acid (HA) has an [H+] of 2.36 x10-4 M. What is the Ka of

this weak acid? You have 5 minutes!!

Agenda:Discuss Ka and Kb worksheetsSelf Check on learned contentHW: Ch. 16 sec. 8-11 reading notes

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1. Write chemical equations which represent the dissociation of each of these acids or bases in aqueous solution. Use a single arrow in the case of a strong acid or base, and a double arrow to represent the equilibrium condition that exists in the solution of a weak acid or base. a. HCl___________________________________________b. NaOH__________________________________________c. H2SO4_________________________________________d. KOH___________________________________________e. HC2H3O2_______________________________________f. HCN___________________________________________g. Cu(OH)2_______________________________________h. NH4OH_________________________________________

(aq) + H2O(l) H+(aq) + Cl(aq)

(aq) + H2O(l) Na+(aq) + OH(aq)

(aq) + H2O(l) H+(aq) + HSO4(aq)

(aq) + H2O(l) K+(aq) + OH(aq)

(aq) + H2O(l) H+(aq) + C2H3O2(aq)

(aq) + H2O(l) H+(aq) + CN(aq)

(aq) + H2O(l) Cu+(aq) + OH(aq)(aq) + H2O(l) NH4

+(aq) + OH(aq)

2. Calculate the [H+] and [OH ] of a 1.0 x 10‐ 3‐ M solution of HCl, a strong acid.

1.0 x 10-3M = [H+][OH-]

1.0 x 10-3M = x2

3.16 x 10-2= x

BOTH are going to have the same concentration3.16 x 10-2M

3. Calculate the [OH ] and the [H+] of a 0.0020 M ‐solution of NaOH, a strong base.

0.0020= [H+][OH-]

2.0 x 10-3M = x2

4.47 x 10-2= x

BOTH are going to have the same concentration 4.47 x 10-2 M

4. A 0.30 M solution of a weak acid (HA) has an [H+] of 1.66 x10-4 M. What is the Ka of this weak

acid?

• HA H+ + A-

[1.66 10−4] [1.66 10−4][0.30]

Ka =

• 9.2 x 10-8

5. The pH of a 0.10 M solution of a weak acid is 5.40. What is the Ka of the acid?

• HA H+ + A-

pH = −log [H+]5.40 = −log [H+]−5.40 = log [H+]

10−5.40 = 10log [H+] = [H+]3.98 10−6 = [H+] = [A-]

5. The pH of a 0.10 M solution of a weak acid is 5.40. What is the Ka of the acid?

• HA H+ + A-

[3.98 10−6] [3.98 10−6][0.10]

Ka =

• 1.58 x 10-10

6. Determine the pH of a 0.10 M CH3COOH solution if the Ka = 1.8 x 10-5?

• CH3COOH H+ + CH3COO-

[X] [X][x- 0.10]1.8 x 10-5 = Ka =

1.3 x 10-3

[X]2 = 1.8 x10-6

-log(1.3 x 10-3) = 2.8

7. Determine the percent dissociation of a 1.0 M HClO solution if the pH = 3.7.

• HClO H+ + ClO -

pH = −log [H+]3.7 = −log [H+]−3.7 = log [H+]

10−3.7 = 10log [H+] = [H+]2.0 10−4 = [H+] = [A-]

7. Determine the percent dissociation of a 1.0 M HClO solution if the pH = 3.7.

• HClO H+ + ClO -

[2.0 10−4] [1.0]

Ka = X 100 = 0.02 %

8. Determine the pH of a 0.05 M NH3 if Kb = 1.8 x 10-5

• NH3 NH4+ + OH -

[X] [X][x- 0.05]1.8 x 10-5 = Kb =

9.5 x 10-4

[X]2 = 9.0 x10-7

-log(9.5 x 10-4) = 3.0

3.0 - 14 = 11

I can…

• Distinguish an acid from a base by its properties.• Explain the difference theories of acid and base behavior of

an Arrhenius A/B, Bronsted-Lowry A/B and Lewis A/B. • Use the concentrations of acids and bases to calculate pH

and pOH.• Explain how strong acid and strong base ionize in water and

how weak acids and weak based dissociate.• Use Ka and concentrations weak acids to find pH and

likewise with bases (Kb).• Explain how the weak acid/base concentrations affect the

magnitude of Ka or Kb.

Name or provide the formula to the following acid

Phosphorous acid____________H2CO3 ____________________

Chlorous acid ______________HCN______________________

H3PO3

Carbonic acidH2ClO3

Hydrocyanic acid

Practice on this concept: For the following, label the acid, base, conjugate acid and conjugate base.

NH4+ + OH- NH3 + H2O

HBr + H2O H3O+ + Br-

Acid Base C-Acid C-Base

Acid Base C-Acid C-Base

1. What is the pH of a solution which has [H+] of 5.0 x10-6?

pH = -log (5.0 x10-6) =

pH = -(-5.30) = 5.30

Calculate concentration from pH2. Calculate the amount of [H+] with a solution with a pH of 3.76

3.76 = -log [H+]

AntiLog both sides

[H+]= 1.7 x10-4

[H+] = -3.76

-3.76 = log [H+] Get the negative to the other side

3. A 0.30 M solution of a weak acid (HA) has an [H+] of 1.66 x10-4 M. What is the Ka of this weak acid?

Ka= [H+][A-] [HA]

Ka= [1.66 x10-4 M][1.66 x10-4 M] [0.30M]

Ka= 9.2 x10-8

4. The pH of a 0.10 M solution of a weak acid is 5.40. What is the Ka of the acid?

Ka= [H+][A-] [HA]

The pH is 5.40 which means that the [H+] is 3.98 x 10-6 M

Ka= 1.6 x10-10

Ka= [3.98 x 10-6 M ][3.98 x 10-6 M ] [0.10M]

5. Determine the pH of a 0.10 M CH3COOH solution if the Ka = 1.8 x 10-5?

Ka= [H+][A-] [HA]

pH= 2.9

1.8 x10-5= [x] [x] [0.10M]

1.8 x10-5 (0.10)= x2 1.8 x10-6 = x2

1.34 x10-3 = x

log -1.34 x10-3 = pH

6. Determine the percent dissociation of a 1.0 M HClO solution if the pH = 3.7.

100 x [H+] [HA]

0.02%

-3.7 antillog=

2.0 x10-4 = [H+]

pH= -log[H+]

100 x [2.0 x10-4 ] [1.0]

7. Determine the pH of a 0.05 M NH3 if Kb = 1.8 x 10-5

Kb= [H+][B-] [HB]

pOH= 3.0

1.8 x10-5= [x] [x] [0.05M]

1.8 x10-5 (0.05)= x2 9.0 x10-7 = x2

9.5 x10-4 = x

log -1.34 x10-3 = pOH

pH= 11.0

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