Unit 2 Acids and Bases 1
Topics Properties / Operational Definitions Acid-Base Theories ph & poh calculations Equilibria (Kw, K a, K b ) Indicators Titrations STSE: Acids Around Us 2
Operational Definitions An Operational Definition is a list of properties, or operations that can be performed, to identify a substance. See p. 550 for operational definitions of acids and bases 3
Operational Definitions (Properties see p. 550) Acids ph < 7 taste sour react with active metals (Mg, Zn) to produce hydrogen gas Bases ph > 7 taste bitter no reaction with active metals feel slippery 4
Operational Definitions Acids blue litmus turns red react with carbonates to produce CO gas 2 Bases red litmus turns blue no reaction with carbonates 5
Operational Definitions Acids conduct electric current neutralize bases to produce water and a salt any ionic compound Bases conduct electric current neutralize acids to produce water and a salt any ionic compound 6
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Acid-Base Theories 1. Arrhenius Theory (p. 549 ) acid any substance that dissociates or IONIZES in water to produce H + ions ie. an acid must contain H + ions and some negative ion. 9
Arrhenius Theory eg. HCl (aq) H + (aq) + Cl- (aq) H 2 SO 4(aq) H + (aq) + HSO 4 - (aq) HSO 4 - (aq) H + (aq) + SO 4 2- (aq) 10
Arrhenius Theory base any substance that dissociates in water to produce OH - ions ie. a base must contain OH - ions and some positive ion 11
Arrhenius Theory eg. NaOH (aq) Na + (aq) + OH- (aq) Ca(OH) 2(aq) Ca + (aq) + 2 OH- (aq) 12
Arrhenius Theory Which is an Arrhenius acid? Which is an Arrhenius base? a) KOH c) CH 4 b) HCN d) CH 3 OH 13
Limitations of Arrhenius theory (p.551) 1. H + cannot exist as an ion in water. The positive H + ions are attracted to the polar water molecules forming HYDRONIUM ions or H 3 O + (aq) H + (aq) + H 2 O (l) H 3 O + (aq) 14
Limitations of Arrhenius theory 2. CO dissolves in water to produce an 2 acid. NH dissolves in water to produce a 3 base. Neither of these observations can be explained by Arrhenius theory 15
Limitations of Arrhenius theory 3. Some acid-base reactions can occur in solvents other than water. Arrhenius theory can explain only aqueous acids or bases. 16
Limitations of Arrhenius theory 4. Arrhenius theory is not able to predict whether certain species are acids or bases. eg. NaHSO H PO - HCO - 4 2 4 3 Arrhenius theory needs some work 17
Acid-Base Theories To be used when Arrhenius is inadequate 2. Modified Arrhenius Theory (p. 552) acid any substance that reacts with water to produce H O + ions 3 eg. HCl + H O H O + (aq) 2 (l) 3 (aq) + Cl- (aq) 18
Modified Arrhenius Theory base any substance that reacts with water to produce OH - ions eg. NH + H O NH + + OH - 3(aq) 2 (l) ) 4 (aq) (aq) pg. 559 # s 3, 8, & 9 19
Acid-Base Theories 3. Brønsted-Lowry Theory (p. 553) acid any substance from which a proton (H + ) may be removed ie. an acid is a substance that loses a proton (H + ) 20
Brønsted-Lowry Theory base any substance that can remove a proton (H + ) from an acid. ie. a base is a substance that gains a proton (H + ) In BLT, an acid-base reaction requires the transfer of a proton (H + ) from an acid to a base. 21
eg. Brønsted-Lowry Theory base conjugate acid HCN (aq) + NH 3(aq) CN - (aq) + NH 4 + (aq) acid conjugate base 22
Brønsted-Lowry Theory What is a conjugate acid-base pair?? (p. 554) Two particles (molecules or ions) that differ by one proton are called a conjugate acid-base pair. The conjugate base forms when an acid loses a proton (H + ). The conjugate acid forms when a base gains a proton (H + ). 23
Brønsted-Lowry Theory base conjugate acid acid conjugate base 24
eg. Brønsted-Lowry Theory base conjugate acid NH + H O NH + 3(aq) 2 (l) + OH - 4 (aq) (aq) acid conjugate base 25
Brønsted-Lowry Theory - an amphoteric substance can be either an acid or a base - usually these are negative ions that contain at least one hydrogen atom, OR water eg. H 2 O, HCO 3 - (aq), H 2 PO 4 - (aq) 26
Brønsted-Lowry Theory p.557-558 # s 1 9 27
Strength of Acids and Bases A strong acid is an acid that ionizes or dissociates 100% in water eg. HCl (aq) H + (aq) + Cl- (aq) Strong acids react 100% with water (BLT) eg. HCl + H O H O + (aq) 2 (l) 3 (aq) + Cl- (aq) 28
Strong acids produce more H + ions OR more H 3 O + ions than weak acids with the same molar concentration
Strength of Acids and Bases NOTE: The equilibrium symbol is NOT used for strong acids because there is NO REVERSE REACTION. 30
Strength of Acids and Bases A weak acid is an acid that ionizes or dissociates LESS THAN 100% eg. HF (aq) Weak acids react less than 100% with water eg. HF (aq) + H 2 O (l) F - (aq) + H 3 O+ (aq) 31
Strength of Acids and Bases For weak acids, an equilibrium is established between the original acid molecule and the ions formed. DO NOT confuse the terms strong and weak with concentrated and dilute. 32
Strength of Acids and Bases eg. Classify the following acids: 0.00100 mol/l HCl(aq) strong and dilute 12.4 mol/l HCl(aq) strong and concentrated 10.5 mol/l CH3 COOH (aq) weak and concentrated 33
Strength of Acids and Bases monoprotic acids that contain or lose one proton diprotic acids that contain or lose two protons polyprotic more than one proton acid 34
Strength of Acids and Bases A strong base is a base that dissociates 100% in water, or reacts 100% with water, to produce OH - ion. The only strong bases are hydroxide compounds of most Group 1 and Group 2 elements eg. NaOH (s) Ca(OH) 2(s) 35
Strength of Acids and Bases A weak base is a base that reacts less than 100% in water to produce OH - ion. eg. S 2- (aq) + H 2 O (l) HS - (aq) + OH- (aq) 36
Writing Acid-Base Equations (BLT) Step 1: List all the molecules/ions present in the solution ionic compounds dissociate strong acids exist as hydronium ion and the conjugate base for weak acids use full formula of the compound (i.e. un-ionized molecule) always include water in the list. 37
Writing Acid-Base Equations (BLT) Step 2: Identify the STRONGEST ACID and the STRONGEST BASE from Step 1. Step 3: Write the equation for the reaction by transferring a proton from the strongest acid to the strongest base. 38
Writing Acid-Base Equations (BLT) Step 4: Determine the type of reaction arrow to use in the equation. Stoichiometric (100%) reactions occur between: Hydronium (H 3 O + ) and bases stronger than nitrite (NO 2- ) hydroxide (OH - )and acids stronger than hypochlorous acid (HOCl) 39
Writing Acid-Base Equations (BLT) Step 5: Determine the position of the equilibrium by comparing the strengths of both acids in the equation. The favoured side is the side with the weaker acid! 40
Writing Acid-Base Equations (BLT) Sample problems: Write the net ionic equation for the acid-base reaction between: - aqueous sodium hydroxide (NaOH (aq) ) and hydrochloric acid (HCl (aq) ). 41
species present Na + (aq) OH - (aq) H 3 O + (aq) Cl - (aq) H 2 O(l) strongest base strongest acid H 3 O + (aq) + OH - (aq) H 2 O(l) + H 2 O(l) OR H 3 O + (aq) + OH - (aq) 2 H 2 O(l) 42
Writing Acid-Base Equations (BLT) Sample problems: Write an equation for the acid-base reaction between nitrous acid (HNO 2(aq) ) and aqueous sodium sulfite (Na 2 SO 3(aq) ). 43
species present HNO 2 (aq) Na + (aq) SO 3 2- (aq) H 2 O(l) SA SB HNO 2 (aq) + SO 3 2- (aq) NO 2 - (aq) + HSO3 - (aq) Stronger Acid Weaker Acid Products favored 44
Write the Net Ionic Equation for each aqueous reaction below: 1. Na CO and CH COOH 2 3(aq) 3 (aq) 2. NH and HNO 3(aq) 2(aq) 3. HNO and RbOH 3(aq) 4. H SO and K PO 2 4(aq) 3 4(aq) 5. HF and NH CH COO (aq) 4 3 (aq) 6. CaCl and PbSO 2(aq) 4(aq) p. 564 # s 10 &11 45
species present Na + (aq) CO 3 2- (aq) CH 3 COOH(aq) H 2 O(l) SB SA CH 3 COOH (aq) + CO 3 2- (aq) CH 3 COO - (aq) + HCO 3 - (aq) Stronger Acid Weaker Acid Products Favoured 46
species present NH 3 (aq) HNO 2(aq) H 2 O(l) SB SA HNO 2 (aq) + NH 3(aq) NO 2 - (aq) + NH4 + (aq) Stronger Acid Weaker Acid Products favored 47
species present H 3 O + NO 3 - (aq) Rb + (aq) OH - (aq) H 2 O(l) (aq) strongest acid strongest base H 3 O + (aq) + OH - (aq) H 2 O(l) + H 2 O(l) OR H 3 O + (aq) + OH - (aq) 2 H 2 O(l) 48
species present H 3 O + (aq) HSO - K + PO 3-4 (aq) (aq) 4 (aq) H O 2 (l) SA SB H 3 O + (aq) + PO 4 3- (aq) H 2 O (l) + HPO 4 2- (aq) 49
species present HF(aq) NH 4 + (aq) CH 3 COO - (aq) H 2 O(l) SA SB HF (aq) + CH 3 COO - (aq) F - (aq) + CH 3 COOH (aq) Stronger Acid Weaker Acid Products favored 50
species present strongest acid Ca 2+ (aq) Cl - (aq) H 2 O(l) Pb 2+ (aq) SO 4 2- (aq) strongest base H 2 O (l) + SO 4 2- (aq) HSO 4 - (aq) + OH - (l) Weaker Acid Reactants favored Stronger Acid 51
Products are NOT always favoured Try these: CH 3 COOH (aq) + NH 4 F (aq) HCN (aq) + NaHS (aq) 52
Acid-Base Calculations K w K a K b [H 3 O + ] [OH - ] ph poh 53
K w (Ionization Constant for water) With very sensitive conductivity testers, pure water shows slight electrical conductivity. PURE WATER MUST HAVE A SMALL CONCENTRATION OF DISSOLVED IONS 54
K w Auto-Ionization of water H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH- (aq) K = [H O+] [OH - ] w 3 [H 2 O] [H 2 O] 55
K w In pure water at 25 C; [H 3 O+] = 1.00 x 10-7 mol/l [OH - ] = 1.00 x 10-7 mol/l Calculate K w at 25 C. 56
H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH- LCP: What happens if we add OH - ions (NaOH ) to water? (aq) shift to the left [H O + ] decreases 3 [OH- ] spikes, decreases, ends up higher than at the start Does K w change? NOPE!! 57
K w = [H 3 O + ] [OH - ] 1.00 x 10-14 = [H 3 O + ] [OH - ] [H 3 O + ] [OH - ] 0.00357 M 2.04 x 10-12 2.80 x 10-12 4.89 x 10-3 mol/l 12.5 M 8.00 x 10-16 1.50 mol/l 6.67 x 10-15 58
Calculations with K w (p. 564 566) For strong acids and strong bases, the [H O + ] and [OH - ] may be calculated using 3 the solute concentration. eg. What is the [H O + ] in a 2.00 mol/l 3 solution of HNO 3(aq)? Ans: 2.00 mol/l [Strong acid] = [H 3 O + ] [OH - ] =??? (5.00 x10-15 mol/l) 59
Calculations with K w eg. What is the [OH - ] in a 2.00 mol/l solution of NaOH? (aq) Ans: 2.00 mol/l eg. What is the [OH - ] in a 2.00 mol/l solution of Ca(OH)? 2(aq) Ans: 4.00 mol/l [H O + ] =??? 3 60
Calculations with K w eg. What molar concentration of Al(OH) is needed to obtain a [OH - ] = 3(aq) 0.450 mol/l? Ans: 0.150 mol/l 61
What is the [H 3 O + ] and [OH - ] in: [H 3 O + ] mol/l [OH- ] mol/l 1.0 x 10-8 1.0 x 10-6 5.00 x 10-14 0.200 1.50 6.67 x 10-15 1.0 x 10-2 1.0 x 10-12 62
solute [H 3 O + ] [OH - ] 0.680 mol/l HCl (aq) 1.50 mol/l NaOH 0.680 6.67 x 10-15 1.47 x 10-14 1.50 0.0500 mol/l Ca(OH) 2(aq) 1.00 x 10-13 0.450 mol/l HClO 4(aq) 0.450 M 0.100 2.22 x 10-14 0.250 mol/l Mg(OH) 2(aq) 2.00 x 10-14 0.500 mol/l p. 566 # s 12 15 (answers p. 581) 63
By what factor does the [H O + ] change when the 3 ph value changes by 1? by 2?
ph and poh (See p. 568) 65
The [H 3 O + ] changes by a factor of 10 (10X) for each ph changes of 1.
ph and poh (See p. 568) 67
ph and poh FORMULAS ph = -log [H 3 O + ] poh = -log [OH - ] [H 3 O + ] = 10 -ph [OH - ] = 10 -poh ph + poh = 14.00 68
ph and poh eg. What is the ph of a 0.0250 mol/l solution of HCl (aq)? [H 3 O + ] = 0.0250 mol/l ph = 1.602 What is the poh of a 0.00087 mol/l solution of NaOH (aq)? [OH - ] = 0.00087 mol/l poh = 3.06 What is the ph of a 1.25 mol/l solution of KOH (aq)? [OH - ] = 1.25 mol/l [H 3 O + ] = 8.00 x 10-15 mol/l ph = 14.097 69
Significant digits in ph values? The number of significant digits in a concentration should be the same as the number of digits to the right of the decimal point in the ph value. eg. In a sample of OJ the [H O + ] = 2.5 10 4 mol/l 3 ph = 3.60 (See p. 568) 70
[H 3 O + ] [OH - ] ph poh 0.0035 1.2 x 10-5 4.68 8.33 x 10-15 -1.10 9.15 71
[H 3 O + ] [OH - ] ph poh 0.0035 2.9 x 10-12 2.46 11.5 8.3 x 10-10 1.2 x 10-5 9.08 4 4.92 2.1 x 10-5 4.8 x 10-10 4.68 9.32 1.4 x 10-5 7.1 x 10-10 4.85 9.15 1.20 8.33 x 10-15 -0.079 14.079-1.10 13 7.9 x 10-16 15.10 72
ph, poh and K w p. 572 # s 20 25 73
Dilutions When a solution is diluted the number of moles does not change. OR n initial = n final C i V i = C f V f 74
eg. 400.0 ml of water was added to 25.0 ml of HCl(aq) that had a ph of 3.563. Calculate the ph of the resulting solution. calculate [H3O + ] dilution formula calculate ph 75
Before dilution: [H3O + ] = 10-3.563 = 2.735 x 10-4 After dilution: (2.735 x 10-4 ) (25.0 ml) = (C f )(425.0 ml) [H3O + ] = 1.609 x 10-5 ph = -log (1.609 x 10-5 ) = 4.793 p.574 # s 26-29 76