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Entry Task: Feb 1 st Friday. Question: A 0.55 M solution of a weak acid (HA) has an [H + ] of 2.36 x10 -4 M. What is the K a 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.
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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
K+(aq) + OH(aq) Cu+(aq) + OH(aq) H+(aq) + CN(aq) H+(aq) + C2H3O2(aq) H+(aq) + HSO4(aq) NH4+(aq) + OH(aq) Na+(aq) + OH(aq) H+(aq) + Cl(aq) (aq) + H2O(l) (aq) + H2O(l) (aq) + H2O(l) (aq) + H2O(l) (aq) + H2O(l) (aq) + H2O(l) (aq) + H2O(l) (aq) + H2O(l) 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_________________________________________
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 0.0020 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
[1.66 10−4] [1.66 10−4] [0.30] Ka = 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- • 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-]
[3.98 10−6] [3.98 10−6] [0.10] Ka = 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- • 1.58 x 10-10
[X] [X] [x- 0.10] 1.8 x 10-5 = Ka = 6. Determine the pH of a 0.10 M CH3COOH solution if the Ka = 1.8 x 10-5? • CH3COOH H+ + CH3COO- [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 1.0 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-]
[2.0 10−4] [1.0] Ka = 7. Determine the percent dissociation of a 1.0 M HClO solution if the pH = 3.7. • HClO H+ + ClO- 0.02 % X 100 =
[X] [X] [x- 0.05] 1.8 x 10-5 = Kb = 8. Determine the pH of a 0.05 M NH3 if Kb = 1.8 x 10-5 • NH3 NH4+ + OH - [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.
H3PO3 Name or provide the formula to the following acid Phosphorous acid____________ H2CO3 ____________________ Chlorous acid ______________ HCN______________________ 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. 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= [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 = 1.8 x 10-5? 1.8 x10-5= [x] [x] [0.10M] Ka= [H+][A-] [HA] 1.8 x10-6 = x2 1.8 x10-5 (0.10)= x2 1.34 x10-3 = x log -1.34 x10-3 = pH pH= 2.9
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] 9.0 x10-7 = x2 1.8 x10-5 (0.05)= x2 9.5 x10-4 = x log -1.34 x10-3 = pOH pH= 11.0 pOH= 3.0