1	The Chemistry of Acids and Bases Chapter 17
2	Acid and Bases
3	Acid and Bases
4	Acid and Bases
5	Strong and Weak Acids/Bases Generally divide acids and bases into STRONG or WEAK ones. STRONG ACID: HNO₃(aq) + H₂O(liq) ---> H₃O⁺(aq) + NO₃^-(aq) HNO₃ is about 100% dissociated in water.
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7	"Weak acids are much less..." Weak acids are much less than 100% ionized in water. One of the best known is acetic acid = CH₃CO₂H
8	"Strong Base:" Strong Base: 100% dissociated in water. NaOH(aq) ---> Na⁺(aq) + OH^-(aq)
9	"Weak base:" Weak base: less than 100% ionized in water One of the best known weak bases is ammonia NH₃(aq) + H₂O(liq) e NH₄⁺(aq) + OH^-(aq)
10	ACID-BASE THEORIES The most general theory for common aqueous acids and bases is the BRØNSTED - LOWRY theory ACIDS DONATE H⁺ IONS BASES ACCEPT H⁺ IONS
11	ACID-BASE THEORIES The Brønsted definition means NH₃ is a BASE in water — and water is itself an ACID
12	ACID-BASE THEORIES NH₃ is a BASE in water — and water is itself an ACID
13	Conjugate Pairs
14	More About Water H₂O can function as both an ACID and a BASE. In pure water there can be AUTOIONIZATION
15	More About Water K_w = [H₃O⁺] [OH^-] = 1.00 x 10^-14 at 25 ^oC In a neutral solution [H₃O⁺] = [OH^-] so K_w = [H₃O⁺]² = [OH^-]² and so [H₃O⁺] = [OH^-] = 1.00 x 10^-7 M
16	Calculating [H₃O⁺] & [OH^-] You add 0.0010 mol of NaOH to 1.0 L of pure water. Calculate [H₃O⁺] and [OH^-]. Solution 2 H₂O(liq) e H₃O⁺(aq) + OH^-(aq) Le Chatelier predicts equilibrium shifts to the ____________. [H₃O⁺] < 10^-7 at equilibrium. Set up a ICE table.
17	Calculating [H₃O⁺] & [OH^-] You add 0.0010 mol of NaOH to 1.0 L of pure water. Calculate [H₃O⁺] and [OH^-]. Solution
18	Calculating [H₃O⁺] & [OH^-] You add 0.0010 mol of NaOH to 1.0 L of pure water. Calculate [H₃O⁺] and [OH^-]. Solution 2 H₂O(liq) e H₃O⁺(aq) + OH^-(aq) [H₃O⁺] = K_w / 0.0010 = 1.0 x 10^-11 M
19	[H₃O⁺], [OH^-] and pH A common way to express acidity and basicity is with pH pH = log (1/ [H₃O⁺]) = - log [H₃O⁺] In a neutral solution, [H₃O⁺] = [OH^-] = 1.00 x 10^-7 at 25 ^oC pH = -log (1.00 x 10^-7) = - (-7) = 7
20	[H₃O⁺], [OH^-] and pH What is the pH of the 0.0010 M NaOH solution? [H₃O⁺] = 1.0 x 10^-11 M pH = - log (1.0 x 10^-11) = 11.00 General conclusion — Basic solution pH > 7 Neutral pH = 7 Acidic solution pH < 7
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22	[H₃O⁺], [OH^-] and pH If the pH of Coke is 3.12, it is ____________. Because pH = - log [H₃O⁺] then log [H₃O⁺] = - pH Take antilog and get [H₃O⁺] = 10^-pH [H₃O⁺] = 10^-3.12 = 7.6 x 10^-4 M
23	pH of Common Substances
24	Other pX Scales In general pX = -log X and so pOH = - log [OH^-] K_w = [H₃O⁺] [OH^-] = 1.00 x 10^-14 at 25 ^oC Take the log of both sides -log (10^-14) = - log [H₃O⁺] + (-log [OH^-]) pK_w = 14 = pH + pOH
25	Equilibria Involving Weak Acids and Bases Aspirin is a good example of a weak acid, K_a = 3.2 x 10^-4
26	Equilibria Involving Weak Acids and Bases Acid Conjugate Base acetic, CH₃CO₂H CH₃CO₂^-, acetate ammonium, NH₄⁺ NH₃, ammonia bicarbonate, HCO₃^- CO₃^2-, carbonate A weak acid (or base) is one that ionizes to a VERY small extent (< 5%).
27	Equilibria Involving Weak Acids and Bases Consider acetic acid, CH₃CO₂H (HOAc) HOAc + H₂O e H₃O⁺ + OAc^- Acid Conj. base
28	Equilibrium Constants for Weak Acids
29	Equilibrium Constants for Weak Bases
30	Ionization Constants for Acids/Bases
31	Relation of K_a, K_b, [H₃O⁺] and pH
32	K and Acid-Base Reactions
33	K and Acid-Base Reactions A strong acid is 100% dissociated. Therefore, a STRONG ACID—a good H⁺ donor—must have a WEAK CONJUGATE BASE—a poor H⁺ acceptor. HNO₃(aq) + H₂O(liq) e H₃O⁺(aq) + NO₃^-(aq) STRONG A base acid weak B
34	K and Acid-Base Reactions We know from experiment that HNO₃ is a strong acid. 1. It is a stronger acid than H₃O⁺ 2. H₂O is a stronger base than NO₃^- 3. K for this reaction is large
35	K and Acid-Base Reactions Acetic acid is only 0.42% ionized when [HOAc] = 1.0 M. It is a WEAK ACID HOAc + H₂O e H₃O⁺ + OAc^- WEAK A base acid STRONG B Because [H₃O⁺] is small, this must mean 1. H₃O⁺ is a stronger acid than HOAc 2. OAc^- is a stronger base than H₂O 3. K for this reaction is small
36	Types of Acid/Base Reactions Strong acid + Strong base H⁺ + Cl^- + Na⁺ + OH^- e H₂O + Na⁺ + Cl^- Net ionic equation H⁺(aq) + OH^-(aq) e H₂O(liq) K = 1/K_w = 1 x 10¹⁴
37	Types of Acid/Base Reactions Weak acid + Strong base CH₃CO₂H + OH^- e H₂O + CH₃CO₂^- This is the reverse of the reaction of CH₃CO₂^- (conjugate base) with H₂O. OH^- stronger base than CH₃CO₂^- K = 1/K_b = 5.6 x 10⁴
38	Types of Acid/Base Reactions Strong acid + Weak base H₃O⁺ + NH₃ e H₂O + NH₄⁺ This is the reverse of the reaction of NH₄⁺ (conjugate acid of NH₃) with H₂O. H₃O⁺ stronger acid than NH₄⁺ K = 1/K_a = 5.6 x 10⁴
39	Types of Acid/Base Reactions Weak acid + Weak base
40	Types of Acid/Base Reactions: Summary
41	Calculations with Equilibrium Constants pH of an acetic acid solution. What are your observations?
42	Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H₃O⁺, OAc^-, and the pH. Step 1. Define equilibrium concs. in ICE table. [HOAc] [H₃O⁺] [OAc^-] initial change equilib
43	Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H₃O⁺, OAc^-, and the pH. Step 1. Define equilibrium concs. in ICE table. [HOAc] [H₃O⁺] [OAc^-] initial 1.00 0 0 change -x +x +x equilib 1.00-x x x Note that we neglect [H₃O⁺] from H₂O.
44	Equilibria Involving A Weak Acid Step 2. Write K_a expression
45	Equilibria Involving A Weak Acid Step 3. Solve K_a expression
46	Equilibria Involving A Weak Acid Step 3. Solve K_a approximate expression
47	Equilibria Involving A Weak Acid Consider the approximate expression
48	Equilibria Involving A Weak Acid Calculate the pH of a 0.0010 M solution of formic acid, HCO₂H. HCO₂H + H₂O e HCO₂^- + H₃O⁺ K_a = 1.8 x 10^-4 Approximate solution [H₃O⁺] = [K_a • C_o]^1/2 = 4.2 x 10^-4 M, pH = 3.37 Exact Solution [H₃O⁺] = [HCO₂^-] = 3.4 x 10^-4 M [HCO₂H] = 0.0010 - 3.4 x 10^-4 = 0.0007 M pH = 3.47
49	Weak Bases
50	Equilibria Involving A Weak Base You have 0.010 M NH₃. Calc. the pH. NH₃ + H₂O e NH₄⁺ + OH^- K_b = 1.8 x 10^-5 Step 1. Define equilibrium concs. in ICE table [NH₃] [NH₄⁺] [OH^-] initial change equilib
51	Equilibria Involving A Weak Base You have 0.010 M NH₃. Calc. the pH. NH₃ + H₂O e NH₄⁺ + OH^- K_b = 1.8 x 10^-5 Step 1. Define equilibrium concs. in ICE table [NH₃] [NH₄⁺] [OH^-] initial 0.010 0 0 change -x +x +x equilib 0.010 - x x x
52	Equilibria Involving A Weak Base You have 0.010 M NH₃. Calc. the pH. NH₃ + H₂O e NH₄⁺ + OH^- K_b = 1.8 x 10^-5 Step 2. Solve the equilibrium expression
53	Equilibria Involving A Weak Base You have 0.010 M NH₃. Calc. the pH. NH₃ + H₂O e NH₄⁺ + OH^- K_b = 1.8 x 10^-5 Step 3. Calculate pH [OH^-] = 4.2 x 10^-4 M so pOH = - log [OH^-] = 3.37 Because pH + pOH = 14, pH = 10.63
54	"MX" MX + H₂O ----> acidic or basic solution? Consider NH₄Cl NH₄Cl(aq) ----> NH₄⁺(aq) + Cl^-(aq) (a) Reaction of Cl^- with H₂O Cl^- + H₂O ----> HCl + OH^- base acid acid base Cl^- ion is a VERY weak base because its conjugate acid is strong. Therefore, Cl^- ----> neutral solution
55	"NH₄Cl(aq)" NH₄Cl(aq) ----> NH₄⁺(aq) + Cl^-(aq) (b) Reaction of NH₄⁺ with H₂O NH₄⁺ + H₂O ----> NH₃ + H₃O⁺ acid base base acid NH₄⁺ ion is a moderate acid because its conjugate base is weak. Therefore, NH₄⁺ ----> acidic solution See TABLE 17.4 for a summary of acid-base properties of ions.
56	Acid-Base Properties of Salts
57	"Calculate the pH of a..." Calculate the pH of a 0.10 M solution of Na₂CO₃. Na⁺ + H₂O ---> neutral CO₃^2- + H₂O e HCO₃^- + OH^- base acid acid base K_b = 2.1 x 10^-4 Step 1. Set up concentration table [CO₃^2-] [HCO₃^-] [OH^-] initial change equilib
58	"Calculate the pH of a..." Calculate the pH of a 0.10 M solution of Na₂CO₃. Na⁺ + H₂O ---> neutral CO₃^2- + H₂O e HCO₃^- + OH^- base acid acid base K_b = 2.1 x 10^-4 Step 1. Set up ICE table [CO₃^2-] [HCO₃^-] [OH^-] initial 0.10 0 0 change -x +x +x equilib 0.10 - x x x
59	"Calculate the pH of a..." Calculate the pH of a 0.10 M solution of Na₂CO₃. Na⁺ + H₂O ---> neutral CO₃^2- + H₂O e HCO₃^- + OH^- base acid acid base K_b = 2.1 x 10^-4
60	"Calculate the pH of a..." Calculate the pH of a 0.10 M solution of Na₂CO₃. Na⁺ + H₂O ---> neutral CO₃^2- + H₂O e HCO₃^- + OH^- base acid acid base K_b = 2.1 x 10^-4