1 1 Chapter 10 Acids, Bases, and Salts
2 2 Ch 10.1 Arrhenius Acid-Base Theory (also in Chapter Medley) Arrhenius Acids produce Arrhenius Bases produce H + in water OH - in water HCl hydrochloric acid KOH HNO 3 nitric acid Ba(OH) 2 HClO 4 perchloric acid H 2 SO 4 sulfuric acid H 3 PO 4 phosphoric acid
3 3 Arrhenius Acids Sour taste Change blue litmus paper to Corrosive Arrhenius Bases Bitter taste Change red litmus paper to Slippery (soapy) to the touch
4 4 Ch 10.2 Brønsted Lowry Acid-Base Theory Brønsted Lowry acid = proton (H + ion) donor Brønsted Lowry base = proton (H + ion) acceptor In aqueous solution, a proton is bonded to water through a coordinate covalent bond: Ion
5 5 Base Acid H + acceptor H + donor H 2 O(l) + HCl(g) H 3 O + (aq) + Cl - (aq) Base Acid H + acceptor H + donor NH 3 (g) + HCl(g) [NH + 4 Cl - ] NH 4 Cl(s) Figure 10.3 White cloud of solid formed from gaseous HCl and NH 3.
6 6 Conjugate Acid-Base pairs differ by a single Acid Base Acid Base H + donor H + acceptor H + donor H + acceptor HCl(g) + H 2 O(l) H 3 O + (aq) + Cl - (aq) Acid Base Conjugate Conjugate Acid
7 7 Drill Problem. Write the chemical formula for: 1. The conjugate base of H 2 PO 4-2. The conjugate acid of H 2 PO 4-3. The conjugate base of H 2 O 4. The conjugate acid of H 2 O Take note that H 2 PO 4 - and H 2 O can act as H + donor or H + acceptor. Substances that can either donate or accept a H + are called amphiprotic.
8 8 Ch 10.3 Mono-, Di-, and Triprotic Acids Monoprotic acids can transfer 1 H + to H 2 O or base Examples: HCl and HNO 3 Diprotic acids can transfer 2 H + Example: H 2 SO 4 + H 2 O H 3 O + + HSO 4 - HSO H 2 O H 3 O + + SO 4 2- Triprotic acids can transfer Example: H 3 PO 4 + H 2 O H 2 PO 4 + H 2 O 2 HPO 4 + H 2 O H 3 O + + H 2 PO 4 H 3 O HPO 4 H 3 O PO 4 A polyprotic acid supplies
9 9 Only acidic H atoms are donated: Acetic acid is protic Acidic CH 3 CO 2 H(l) + H 2 O(l) CH 3 CO 2 - (aq) + H 3 O + (aq) acetate ion
10 Ch 10.4 Strengths of Acids and Bases A strong acid donates all or nearly 100% of its H + to 10 Table 10.1 Learn the names and formulas of these commonly encountered strong acids, and then assume that all other acids you encounter are weak, unless you are told otherwise. These acids are strong even in solution because in water they are all or mostly ionized.
11 A weak acid does not ionize completely. Acetic acid is a weak acid; less than of its molecules are ionized: CH 3 CO 2 H(l) + H 2 O(l) CH 3 CO 2 - (aq) + H 3 O + (aq) Figure 10.5 Comparison of ionized species for a strong and a weak acid. 11 demo
12 12 Strong bases are limited to the hydroxides of Group IA and IIA listed in Table Ammonia gas (NH 3 ) is the most common base. In water, it produces only small amounts of OH - ions: NH 3 (g) + H 2 O(l) NH 4 + (aq) + OH - (aq) Less than of ammonia is ionized. demo
13 13 Ch 10.5 Ionization Constants for Acids and Bases HA(aq) + H 2 O(l) H 3 O + (aq) + A - (aq) acid ionization constant K a = [H 3 O + ][A - ] [HA] Recall from Ch 9 that the concentrations of liquids and solids are into the equilibrium constants. Table 10.3 K a and % ionization values for selected weak acids (1.0 M) at 25 o C Acid strength increases with K a values. Note that for polyprotic acids, each successive step of proton transfer occurs to a lesser extent than the previous one.
14 14 To compare bases, we look at their tendency to accept protons from water and generate ions: B(aq) + H 2 O(l) BH + (aq) + OH - (aq) base ionization constant K b = [BH + ][OH - ] [B] NH 3 (g) + H 2 O(l) K b = 1.8 x 10-5 NH 4 + (aq) + OH - (aq) 0.42% is ionized
15 15 Acid-Base Neutralization Reactions (Ch 10.1, 10.6 & 10.7 covered in Chapter Medley) Acid + Base Salt + Water HX + BOH BX + HOH Double-replacement reaction ionic equation: H + + OH - H 2 O
16 16 Ch 10.8 Self-Ionization of Water H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH - (aq) K eq = [H 3 O + ][OH - ] [H 2 O] = M [H 2 O] 2 memorize this # K eq [H 2 O] 2 = K w = [H 3 O + ][OH - ] = 1.00 x K w = Ion product constant for pure H 2 O at 25 o C In pure H 2 O at 25 o C [H 3 O + ] = [OH - ] = 1.00 x 10-7 M
17 17 What happens to [H 3 O + ] & [OH - ] when an acid is added to H 2 O? Example. When HCl is added to produce a M solution: M HCl M H 3 O + acidic solution [H 3 O + ] = 1.0 x ,000 times more than in pure H 2 O In acidic solution [H 3 O + ] > And what happens to [OH - ]? Le Châtelier 2 H 2 O H 3 O + + OH - large # added with HCl pushes to left some must [OH - ] decreases K w = [H 3 O + ][OH - ] = 1.00 x [OH - ] = K w /[H 3 O + ] = 1.00 x /1.0 x 10-2 [OH - ] = In acidic solution [H 3 O + ] > [OH - ]
18 18 What happens to [H 3 O + ] & [OH - ] when a base is added to H 2 O? Example. When NaOH is added to produce a M solution: [OH - ] = 1.0 x M NaOH M OH - basic solution In basic solution [OH - ] > And what happens to [H 3 O + ]? 10,000 times more than in pure H 2 O Le Châtelier 2 H 2 O H 3 O + + OH - [H 3 O + ] decreases large # added pushes left K w = [H 3 O + ][OH - ] = 1.00 x [H 3 O + ] = K w /[OH - ] = 1.00 x /1.0 x 10-3 [H 3 O + ] = In basic solution [OH - ] > [H 3 O + ]
19 19 Figure 10.9 Summary of relationship between [H 3 O + ] and [OH - ] I n c r e a s e Correction note: T = 25 o C for values throughout this chapter Table 10.4 Summary
20 20 Ch 10.9 The ph concept The ph scale offers an easier method of expressing acidity ph is the negative logarithm of the of the hydronium ion: ph = -log[h 3 O + ] when [H 3 O + ] = 1 x 10 -x then ph = x [H O + ] = 1.0 x 10-9 then ph = sig fig 2 digits If the coefficient in the exponential expression is not 1, then the ph value will be : [H O + ] = 6.3 x ph = sig fig 2 digits enter 6.3 x 10-5 in your calculator, press the log key, then switch the
21 21 Figure Relationships among ph values, [H 3 O + ], and [OH - ] at 25 o C Higher [H 3 O + ] = lower in 1 ph unit = 10-fold in [ ]
22 22 Fig Most fruits and vegetables are acidic (tart or sour taste) Fig ph values of common liquids
23 23 Measurements of ph Impregnated strips of paper, ph test paper, are still widely in use. HInd + H 2 O H 3 O + + Ind - Phenolphthalein is an acid-base indicator added to solutions. Figure The ph meter measures the voltage across a special membrane that passes only H + when immersed in the solution.
24 24 Ch The pk a Method for Expressing Acid Strength pk a = -logk a example: K a acetic acid = 1.8 x 10-5 pk a = 4.74 Acid strength increases with increasing K a (Ch 10.5), while it increases with pk a. Drill question: Compare acetic acid with HF. Which is the stronger acid? HF K a = 6.8 x 10-4 pk a 3.17
25 25 Ch The ph of Aqueous Salt Solutions Acid + Base Water + When a strong acid reacts with a strong base, a neutral salt is formed: HCl + NaOH H 2 O + NaCl ph = When a weak acid and/or base is involved, the resulting salts react with water (hydrolysis reaction) to produce hydronium ions or hydroxide ions or both. To understand this phenomenon, let s have another look at the ionization of acids and bases in water: HA + H 2 O H 3 O + + A - weak acid strong conjugate base B + H 2 O BH + + OH - weak base strong conjugate acid Strong conjugate acids and bases react with!
26 1. Basic Hydrolysis: Example. When sodium nitrite dissolves in water, it dissociates completely: NaNO 2 Na + (aq) + NO 2 - (aq) strong conjugate base of a weak acid (HNO 2 ) has strong attraction for H + NO H 2 O HNO 2 + OH - NaNO 2 = salt! solution has higher [OH - ] than pure H 2 O, solution is Anions of strong acids do not hydrolyze because the weak conjugate base cannot take H + from H 2 O. 26
27 2. Acidic Hydrolysis: Example. When ammonium chloride dissolves in water, it dissociates completely: NH 4 Cl Cl - (aq) + NH 4 + (aq) strong conjugate acid of a weak base (NH 3 ) has strong tendency to donate H + NH H 2 O NH 3 + H 3 O + NH 4 Cl = salt! solution has higher [H 3 O + ] than pure H 2 O, solution is 27
28 28 Examples of neutral, acidic, and basic ions Ions of Neutral Salts Cations Na + K + Rb + Cs + Mg 2+ Ca 2+ Sr 2+ Ba 2+ Anions Cl - Br - I -, ClO - 4 BrO - 4 ClO NO 3 NH 4 + Acidic Ions Al 3+ Pb 2+ Sn 2+ Transition metal ions HSO H 2 PO 4 Basic Ions F - C 2 H 3 O - 2 NO HCO 3 CN - 2- CO 3 S 2-2- SO 4 HPO PO 4 Table 10.6 Examples of neutral, acidic, and basic salts. Table 10.7 Approximate ph of selected 0.1 M salt solutions.
29 29 Ch Buffers demo A buffer is a solution capable of absorbing small amounts of acid or base without dramatic changes in solution Buffered aspirin has a buffering agent, such as MgO, that will maintain the ph of the aspirin as it passes through the stomach of the patient. Many hair shampoos are ph-controlled.
30 30 Typically, a buffer is composed of a conjugate acid-base pair: a weak acid and its conjugate base - CH 3 CO 2 H/CH 3 CO 2 H 2 PO /HPO 4 - H 2 CO 3 /HCO 3 (buffer in human blood) Added base is absorbed by the - OH + H 2 CO 3 HCO3 - + H2O Added acid is absorbed by the conjugate H 3 O HCO 3 H 2 CO 3 + H 2 O H 2 CO 3 is unstable and decomposes; excess CO 2 is expelled through the lungs.
31 31 Typical graphs depicting how ph is buffered. (Note that Figure in your text is in error; the labeling of the two lines must be switched.)
32 32 Ch Electrolytes Solutions in which ions are present are conductors of electricity (recall demo comparing conductivity of ionic and covalent compounds in Ch 4). Strong Electrolyte (NaCl) Nonelectrolyte (sugar) Figure Weak electrolytes, such as weak acids and bases, incompletely ionize/dissociate into ions in aqueous solutions.
33 Electrolytes are important components of body fluids and essential to the proper functioning of the human body. 33
34 34 Drill Problem. Classify each of the following compounds as a strong electrolyte, weak electrolyte, or nonelectrolyte: H 3 PO 4 HCl Cl 2 HF KBr CH 3 CH 2 -OH CH 3 COOH
35 35 Ch Equivalents and Milliequivalents of Electrolytes For solutions containing multiple types of ions, concentrations are often given in equivalent units. An equivalent (Eq) of an ion is the molar amount of that ion needed to supply one mole of positive or negative charge. 1 meq = Eq 1 mole Na + 1 mole Ca mole PO 4 = 1 equivalent = 2 equivalents = equivalents Table 10.8 Concentrations of Major Electrolytes in Blood Plasma
36 36 Sample Calculation. Human blood plasma contains 2.4 mg Mg 2+ per dl. How many Eq or meq are in 1.0 L of plasma? Strategy: dl L mg g mol Eq meq 1.0 L x 2.4 mg Mg 2+ x 10 dl x. 1 g. x 1 mol x Mg 2+ 1 dl 1 L 10 3 mg 24.3 g mol Mg 2+ = 2.0 x 10-3 E q Mg 2+ or 2.0 meq Mg 2+
37 37 Ch Acid-Base Titrations An acid-base titration is a neutralization reaction in which a measured volume of acid or base of known concentration is completely reacted with a measured volume of a base or acid of unknown concentration. The concentration of the unknown solution is then calculated. To detect the endpoint or neutralization point, an acid-base indicator, such as phenolphthalein, is added. HInd + H 2 O H 3 O + + Ind -
38 38 Sample calculation. In an acid-base titration, ml of M H 3 PO 4 is needed to neutralize 25.0 ml of KOH of unknown concentration. Calculate the molarity of the KOH. H 3 PO 4 (aq) + 3 KOH(aq) K 3 PO 4 (aq) + 3 H 2 O(l) ml x mol H 3 PO 4 x mole KOH = mole KOH 1000 ml 1 mol H 3 PO 4 M = mol/l = mole KOH/0.025 L = M KOH