Acids and Bases with Key

Chemistry 12 UNIT 4 ACIDS AND BASES PACKAGE #5 Ka Kb = Kw Proof: H2O + H2O ⇄ H3O+ + OH- Recall, at 25oC: Kw = [H3O+][OH-] [H3O+] = [OH-] = 10-7 M Kw = (10-7) (10-7) = 10-14 Consider the dissociation of any acid: HX + H2O ⇄ H3O+ + XKa = [H3O+][X-] [HX] Consider the reaction of a basic anion with water: X+ H2O ⇄ HX + OHKb = [HX][OH-] [X-] KaKb = [H3O+][X-][HX][OH-] [HX] [X-] KaKb = [H3O+][OH-] = Kw at 25oC KaKb = 10-14 USE KaKb = Kw to solve for Kb given only Ka The acid and base in this equation will be conjugate acid-base pairs example: The Ka for NH4+ is 5.6 x 10-10. What is the Kb for NH3 ? NH3 + Kb (NH3 ) H2O ⇄ = Kw =10 Ka (NH4+ ) 5.6 x 10-10 NH4+ + OH-14 = 1.8 x 10-5 pH of solutions containing an ion which can act as an acid and an ion which can act as a base NH4CH3COO (s) acidic or basic? + - NH4CH3COO (s) ➝ NH4 (aq) + CH3COO (aq) + • Consider NH4 (aq): NH4+(aq) + H2O (l) ⇄ NH3 (aq) + H3O+ (aq) Ka (NH4+ ) = 5.6 x 10-10 - • Consider CH3COO (aq): CH3COO-(aq) + Kb (CH3COO- ) = H2O (l) ⇄ CH3COOH (aq) + OH- (aq) Kw Ka (CH3COOH) = 10 -14 = 5.6 x 10-10 1.8 x 10-5 Since Ka = Kb the pH = 7.0 (neutral) If Ka > Kb then the solution would be acidic pH < 7 If Ka < Kb then the solution would be basic pH > 7 Ka Kb = Kw 1. SAMPLE CALCULATIONS Complete the following chart: ACID Ka BASE Kb HBr HSO4HNO2 H2CO3 NH4+ HS- 2. Consider a solution containing 1.0 M CN - (aq). CN - (aq) 3. + HOH (l) + OH- (aq) Find its pH and [H3O+]. ⇄ HCH3COO(aq) + OH- (aq) Consider a 1.0 M solution with [H3O+] = 2.4 x 10-12. Calculate the Kb. ?? (aq) + HOH (l) 5. HCN (aq) Consider a solution containing 1.0 M CH3COO-(aq). CH3COO-(aq) + HOH (l) 4. ⇄ Find its pH. ⇄ ?? (aq)+ OH- (aq) Consider a 0.25 M solution with [H3O+] = 6.3 x 10-11. Calculate the Kb. ?? (aq) + HOH (l) ⇄ ?? (aq) + OH- (aq) Ka Kb = Kw 1. SAMPLE CALCULATIONS Complete the following chart: Ka ACID BASE Kb ~infinity HBr Br- 0 1.2 x 10-2 HSO4- SO4-2 8.3 x 10-13 4.6 x 10-4 HNO2 NO2- 2.2 x 10-11 4.3 x 10-7 H2CO3 HCO3- 2.3 x 10-8 5.6 x 10-10 NH4+ NH3 1.8 x 10-5 1.3 x 10-13 HS- S-2 7.7 x 10-2 NOTE THAT FROM THE TOP TO THE BOTTOM OF THE LIST THE ACIDS GO FROM STRONGEST TO WEAK (best to least conductor) THE BASES GO FROM WEAK TO STRONGEST (least to best conductor) 2. I R E Consider a solution containing 1.0 M CN - (aq). IGNORE CATION! CN - (aq) + HOH (l) ⇄ HCN (aq) 1.0 M X --------y X +y 1.0-y X y Find its pH. + OH- (aq) -----+y y Kb = [HCN][OH-] = Kw/Ka = 10-14 / 4.9 x 10-10 [CN-] = 2.0 x 10-5 = y2 Assume that y <<1.0 1.0-y 2 y = 2.0 x 10-5 y = 4.5 x 10-3 so the assumption is valid (meaning that subtracting y would NOT have changed our answer in the denominator, above) [OH-] = y = 4.5 x 10-3 pOH = 2.35 (2 sig figs) pH = 11.65 (BASIC!) 3. I R E Consider a solution containing 1.0 M CH3COO-(aq). Find its pH and [H3O+]. IGNORE CATION! CH3COO-(aq) + HOH (l) ⇄ CH3COOH (aq) + OH- (aq) 1.0 M X -------------y X +y +y 1.0-y X y y Kb = [HCH3COO][OH-] [CH3COO-] = Kw/Ka = 5.6 x 10-10 = Assume that y <<1.0 y2 1.0-y y2= 5.6 x 10-10 y = 2.4 x 10-5 = [OH-] pOH = 4.62 (2 sig figs) pH = 9.38 (BASIC!) 4. = 10-14 / 1.8 x 10-5 assumption valid Consider a 1.0 M solution with [H3O+] = 2.4 x 10-12. Calculate the Kb. So pH = 11.62 (BASIC!) pOH = 2.38 [OH-] = 4.2 x 10-3 I have represented this value as “y” in the table below I R E ?? (aq) + HOH (l) 1.0 M X -y X 1.0-y X Kb = ⇄ ?? (aq) -------+y y + OH- (aq) -----+y y (4.2 x 10-3)2 1.0 - 4.2 x 10-3 = 1.8 x 10-5 If we wanted to figure out which base it was… Kb = Kw 1.8 x 10-5 = 1.0 x 10-14 Ka Ka Ka = 5.6 x 10 -10 = NH4+ So the base was NH3 !!!!! 5. Consider a 0.25 M solution with [H3O+] = 6.3 x 10-11. Calculate the Kb. [H3O+] = 10-10.20 = 6.3 x 10-11 pH = 10.20 (BASIC!) pOH = 3.80 (2 sig figs) [OH-] = 1.6 x 10-4 ?? (aq) + HOH (l) I R E 0.25 M -y 0.25-y ⇄ X X X ?? (aq) + OH- (aq) -------+y y Where Y represents [OH-] = 1.6 x 10-4 Kb = y2 0.25-y (1.6 x 10-4)2 0.25 – 1.6 x 10-4 = 1.0 x 10-7 = Ka Kb = Kw If Kb = 1.0 x 10-7 Then Ka = 1.0 x 10-7 The identity of the acid with Ka = 1.0 x 10-7 is HSO3So therefore the base with Kb = 1.0 x 10-7 is SO3-2 -----+y y HYDROLYSIS -a reaction between an ion and water which alters the [OH-] and [H+] in the resulting solution. example: CN- (aq) + HOH (l) ⇄ HCN (aq) + OH- (aq) Note: CN- (aq) hydrolysis produces a basic solution Typical anion hydrolysis: Kh = Kb = [HCN][OH-] [CN-] example: NH4+ (aq) + HOH (l) ⇄ H3O+ (aq) + NH3 (aq) Note: NH4+ (aq) hydrolysis produces an acidic solution Typical cation hydrolysis: Kh = Ka = [H3O+][NH3] [NH4+] NOT ALL IONS HYDROLYZE!!! example: Cl- (aq) Kb + HOH (l) ⇄ HCl (aq) + OH- (aq) = Kw / Ka = 10-14 / infinity =0 K <<1 therefore reactants are preferred This hydrolysis reaction does not proceed. HCl is a strong acid and will dissociate 100%. Anions incapable of hydrolysis: Cl-, HSO4-, I-, Br-, NO3-. ClO4- Cations incapable of hydrolysis: Na+, K+, Li +, Cs+, Fr+, Rb+, Ca+2, Ba+2 HYDROLYSIS OF SALTS: SALT PARENT NaF NaOH HF pH of salt strong base weak acid HYDROLYSIS of ions basic pH >7 KCl NH4NO3 NH4IO3 NH4F NH4CH3COO SUMMARY: 1) 2) 3) 4) HYDROLYSIS OF SALTS: Neither cation or anion hydrolyze: the salt solution will be neutral Only the cation hydrolyzes: the solution will be acidic Only the anion hydrolyzes: the solution will be basic If both the cation and anion hydrolyze: the solution will be: acidic if Ka > Kb neutral if Ka = Kb basic if Ka < Kb Sample Calculation: DETERMINE THE pH of a 0.06 M NaCH3COO SOLUTION. Na+ F- - NO - yes HYDROLYSIS OF SALTS: SALT PARENT NaF NaOH HF strong base weak acid basic pH >7 Na+ F- - NO - yes KCl HCl KOH strong acid strong base neutral pH = 7 K+ Cl- - NO - NO NH4NO3 HNO3 strong acid acidic NH4OH weak base NH4+ - yes NO3- - NO HIO3 weak acid NH4OH weak base NH4IO3 pH of salt pH < 7 **acidic** HYDROLYSIS of ions NH4+ - yes IO3- - yes NH4 + + HOH ⇄ H3O+ + NH3 This Bronsted-Lowry equilibrium equation can also be written as: NH4 + + HOH ⇄ H+ + NH4OH You should be OK with this second way to write the same reaction! Ka = 5.6 x 10-10 HOH ⇄ HIO3 + OHKb = Kw / Ka = 5.9 x 10-14 IO3 - + Ka > Kb **Therefore Acidic!!!!!!!!! ** NH4F HF weak acid NH4OH weak base HOH ⇄ Ka = 5.6 x 10-10 NH4 + + **acidic** H3O+ + NH3 HOH ⇄ HF Kb = Kw / Ka = 2.8 x 10-11 F- + Ka > Kb **Therefore Acidic!!!!!!!!! ** + OH- NH4+ - yes F- yes SALT PARENT NH4CH3COO HCH3COO NH4OH pH of salt weak acid **neutral** weak base HOH ⇄ Ka = 5.6 x 10-10 NH4 + + NH4+ - yes CH3COO- - yes H3O+ + NH3 HOH ⇄ Kb = Kw / Ka = 5.6 x 10-10 CH3COO - HYDROLYSIS of ions + CH3COOH + OH- Ka = Kb **Therefore Neutral!!!!!!!!! ** SUMMARY: HYDROLYSIS OF SALTS: 1) Neither cation or anion hydrolyze: the salt solution will be neutral 2) Only the cation hydrolyzes: the solution will be acidic 3) Only the anion hydrolyzes: the solution will be basic 4) If both the cation and anion hydrolyze: the solution will be acidic if Ka > Kb neutral if Ka = Kb basic if Ka < Kb Sample Calculation: DETERMINE THE pH OF A 0.06 M NaCH3COO SOLUTION. I R E HCH3COO weak acid NaOH strong base CH3COO 0.06 M -y 0.06 -y + HOH ⇄ XXX XXX XXX basic CH3COOH --+y y Kb = Kw / Ka = 5.6 x 10-10 5.6 x 10-10 = y2 / (0.06-y) assume y <<0.06 3.4 x 10-11 = y2 y = 5.8 x 10-6 M= [OH-] pOH = 5.24 pH = 8.76 Basic!!! Na+ - NO CH3COO- - yes + OH--+y y HYDROLYSIS REVIEW EXERCISE: 1 Find the pH of: a) b) c) d) e) f) g) h) 0.50 M C6H5OH 0.50 M KOH 0.50 M H2SO4 0.50 M HBr 0.50 M H2CO3 0.50 M NaCN 0.50 M NH4I 0.50 M K2SO3 2 Give numerical proof to support your prediction of acidic, basic, or neutral character for the following compounds: a) NH4F b) NH4HS c) (NH4)3PO4 3 Will HPO4-2 behave as an acid or a base in water at RTP? Show your proof. HINTS for the HYDROLYSIS WORKSHEET: THIS IS A DECEIVING WORKSHEET. YOU NEED TO KNOW A LOT TO COMPLETE THIS WORKSHEET. SUGGESTION FOR SUCCESS: Make up flash cards of these questions and on the back of the flashcard write: whether it is a strong/weak acid/base, the dissocation equation and whether you will be calculating a Ka or a Kb. Also keep track of how many significant figures you need to have throughout each question. FRONT OF FLASHCARD: a) Find the pH of 0.50 M C6H5OH b) Find the pH of 0.50 M KOH c) Find the pH of 0.50 M H2SO4 d) Find the pH of 0.50 M HBr e) Find the pH of 0.50 M H2CO3 f) Find the pH of 0.50 M NaCN g) Find the pH of 0.50 M NH4I h) Find the pH of 0.50 M K2SO3 BACK OF FLASHCARD: weak acid So use Ka to solve for [H+] C6H5OH ⇄ C6H5O- + H+ strong base therefore 100% dissoc. So no Ka or Kb needed [KOH] = [OH-] strong DIPROTIC acid 100% dissoc [H+] is not appreciably >K2 therefore a 2 stage dissoc IS necessary add [H+]1 + [H+]2 to find final [H+] and then pH strong monoprotic acid therefore one step calc: 100% dissoc. weak DIprotic acid [H+]> K2 therefore ignore second stage use Ka1 to solve for [H+] parent base NaOH: strong base so: Na+ will not hydrolyze parent acid HCN: weak acid so: SINGLE HYDROYSIS: CN- only the ANION solution will be basic; use Kb to solve CN + HOH <==> HCN + OH HI: strong acid parent (I- does not hydrolyze) NH4OH or NH3: weak base parent SINGLE HYDROYSIS: NH4+ only the CATION solution will be acidic; use Ka to solve NH4+ + HOH <==>H+ + NH3 KOH: strong base base (K+ does not hydrolyze) H2SO3: weak acid parent (you could also look at the HSO3- conj. and come to same conclusion) SINGLE HYDROYSIS: SO3-2 only the ANION solution will be basic Question 2: DETERMINE SINGLE HYDROLYSIS (cation or anion) OR DOUBLE HYDROLYSIS (BOTH cation AND anion). Then use Ka versus Kb to determine acidic / basic character. Question 3: Use Ka of HPO4-2 vs Kb of HPO4-2 to determine if this amphiprotic ion is going to act as acid or base. Note that in question 2b you have to do something similar in addressing the fact that HS- could be an acid OR a base! And also consider the NH4+ ion acting as an acid as well. Highest K value determines it. HYDROLYSIS REVIEW EXERCISE 1 KEY Find the pH of: a) b) c) d) e) f) g) h) 0.50 M C6H5OH 0.50 M KOH 0.50 M H2SO4 5.09 or 5.10 13.70 0.29 0.50 M HBr 0.50 M H2CO3 0.50 M NaCN 0.50 M NH4I 0.50 M K2SO3 0.30 3.33 11.50 or 11.51 4.77 or 4.78 10.35 Acidic Basic Acidic (final [H+] = .512 M) Acidic Acidic Basic Acidic Basic 2 Give numerical proof to support your prediction of acidic, basic, or neutral character for the following compounds: a) NH4F Acidic b) NH4HS Basic c) (NH4)3PO4 Basic 3 Will HPO4-2 behave as an acid or a base in water at RTP? Show your proof. Base