Entry Task: Feb 1st Friday

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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 worksheets Self Check on learned content HW: Ch. 16 sec. 8-11 reading notes

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 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 concentration 3.16 x 10-2M

3. Calculate the [OH‐] and the [H+] of a 0 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 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 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 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 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 = [X]2 = 1.8 x10-6 1.3 x 10-3 -log(1.3 x 10-3) = 2.8

7. Determine the percent dissociation of a 1 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 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 = [X]2 = 9.0 x10-7 9.5 x 10-4 -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 acid H2ClO3 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 pH 2. Calculate the amount of [H+] with a solution with a pH of 3.76 3.76 = -log [H+] Get the negative to the other side -3.76 = log [H+] AntiLog both sides [H+] = -3.76 [H+]= 1.7 x10-4

3. A 0. 30 M solution of a weak acid (HA) has an [H+] of 1. 66 x10-4 M 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 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= [3.98 x 10-6 M ][3.98 x 10-6 M ] [0.10M] Ka= 1.6 x10-10

5. Determine the pH of a 0. 10 M CH3COOH solution if the Ka = 1 5. Determine the pH of a 0.10 M CH3COOH solution if the Ka = 1.8 x 10-5? 1.8 x10-5= [x] [x] [0.10M] Ka= [H+][A-] [HA] 1.8 x10-5 (0.10)= x2 1.8 x10-6 = x2 1.34 x10-3 = x log -1.34 x10-3 = pH pH= 2.9

6. Determine the percent dissociation of a 1 6. Determine the percent dissociation of a 1.0 M HClO solution if the pH = 3.7. pH= -log[H+] 100 x [H+] [HA] -3.7 antillog= 2.0 x10-4 = [H+] 100 x [2.0 x10-4 ] [1.0] 0.02%

7. Determine the pH of a 0.05 M NH3 if Kb = 1.8 x 10-5 1.8 x10-5= [x] [x] [0.05M] Kb= [H+][B-] [HB] 1.8 x10-5 (0.05)= x2 9.0 x10-7 = x2 9.5 x10-4 = x log -1.34 x10-3 = pOH pOH= 3.0 pH= 11.0