Acids and Bases Chapter 15 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Acids Have a sour taste. Vinegar owes its taste to acetic acid. Citrus fruits contain citric acid. React with certain metals to produce hydrogen gas. React with carbonates and bicarbonates to produce carbon dioxide gas Have a bitter taste. Feel slippery. Many soaps contain bases. Bases 4.3

There are three definitions (or ways of looking at) acids and bases: Arrhenius Brønsted or Brønsted-Lowry Lewis Key vocabulary from last semester: - proton = H + (remember H atom has 1 proton, 1 electron and no neutrons => remove electron to get positive charge), also H 3 O + - Most acids are compounds containing H bonded to a high electronegative element - monoprotic - one proton; ex: HCl, HNO 3, CH 3 COOH - diprotic - two protons; ex: H 2 SO 4, H 2 CO 3 - triprotic - three protons; ex: H 3 PO 4,

Arrhenius acid is a substance that produces H + (H 3 O + ) in water Arrhenius base is a substance that produces OH - in water 4.3 HA H + + A -

The stronger the acid, the more H + (H 3 O + ) there are in water. The stronger the base, the more OH - there are in water. Strong AcidWeak Acid Full dissociation Partial dissociation

A Brønsted acid is a proton donor A Brønsted base is a proton acceptor acid baseacidbase 15.1 acid conjugate base base conjugate acid proton donor proton acceptor Acid base reaction is a transfer of a proton from an acid to a base.

A Brønsted acid is a proton donor A Brønsted base is a proton acceptor acidbaseacidbase 15.1 acid conjugate base base conjugate acid proton donor proton acceptor

Notice that water can be a proton donor or a proton acceptor => amphoteric Acid donates H + so a base must have non-bonding electrons. Acid / base conjugate pairs relate by one H + HF / F - H 2 O / H 3 O + H 2 O / OH - NH 3 / NH 2 - B + HA HB + A - baseacid conjugate acid of base add proton conjugate base of acid subtract proton

CH 3 COOH + H 2 O CH 3 COO - + H baseacid conjugate acid of base add proton conjugate base of acid subtract proton HF + OH - F - + H 2 0 Always work in pairs. Acids and bases are opposites.

O H H + O H H O H HH O H - + [] + Acid-Base Properties of Water H 2 O (l) H + (aq) + OH - (aq) H 2 O + H 2 O H 3 O + + OH - acid conjugate base base conjugate acid 15.2 autoionization of water

H 2 O (l) H + (aq) + OH - (aq) The Ion Product of Water K c = [H + ][OH - ] [H 2 O] [H 2 O] = constant K c [H 2 O] = K w = [H + ][OH - ] The ion-product constant (K w ) is the product of the molar concentrations of H + and OH - ions at a particular temperature. At 25 0 C K w = [H + ][OH - ] = 1.0 x [H + ] = [OH - ] [H + ] > [OH - ] [H + ] < [OH - ] Solution Is neutral acidic basic 15.2

What is the concentration of OH - ions in a HCl solution whose hydrogen ion concentration is 1.3 M? K w = [H + ][OH - ] = 1.0 x [H + ] = 1.3 M [OH - ] = KwKw [H + ] 1 x = = 7.7 x M 15.2

pH – A Measure of Acidity pH = - log [H + ] [H + ] = [OH - ] [H + ] > [OH - ] [H + ] < [OH - ] Solution Is neutral acidic basic [H + ] = 1 x [H + ] > 1 x [H + ] < 1 x pH = 7 pH < 7 pH > 7 At 25 0 C pH[H + ] 15.3

pOH = -log [OH - ] [H + ][OH - ] = K w = 1.0 x log [H + ] – log [OH - ] = pH + pOH = 14.00

The pH of rainwater collected in a certain region of the northeastern United States on a particular day was What is the H + ion concentration of the rainwater? pH = - log [H + ] [H + ] = 10 -pH = = 1.51 x M The OH - ion concentration of a blood sample is 2.5 x M. What is the pH of the blood? pH + pOH = pOH = -log [OH - ]= -log (2.5 x )= 6.60 pH = – pOH = – 6.60 =

Strong Electrolyte – 100% dissociation NaCl (s) Na + (aq) + Cl - (aq) H2OH2O Weak Electrolyte – not completely dissociated CH 3 COOH CH 3 COO - (aq) + H + (aq) Strong Acids are strong electrolytes HCl (aq) + H 2 O (l) H 3 O + (aq) + Cl - (aq) HNO 3 (aq) + H 2 O (l) H 3 O + (aq) + NO 3 - (aq) HClO 4 (aq) + H 2 O (l) H 3 O + (aq) + ClO 4 - (aq) H 2 SO4 (aq) + H 2 O (l) H 3 O + (aq) + HSO 4 - (aq) 15.4

HF (aq) + H 2 O (l) H 3 O + (aq) + F - (aq) Weak Acids are weak electrolytes HNO 2 (aq) + H 2 O (l) H 3 O + (aq) + NO 2 - (aq) HSO 4 - (aq) + H 2 O (l) H 3 O + (aq) + SO 4 2- (aq) H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH - (aq) Strong Bases are strong electrolytes NaOH (s) Na + (aq) + OH - (aq) H2OH2O KOH (s) K + (aq) + OH - (aq) H2OH2O Ba(OH) 2 (s) Ba 2+ (aq) + 2OH - (aq) H2OH2O 15.4

F - (aq) + H 2 O (l) OH - (aq) + HF (aq) Weak Bases are weak electrolytes NO 2 - (aq) + H 2 O (l) OH - (aq) + HNO 2 (aq) Conjugate acid-base pairs: The stronger the acid the weaker the conjugate base. The stronger the base the weaker the conjugate acid. The conjugate base of a strong acid has no measurable strength. H 3 O + is the strongest acid that can exist in aqueous solution. The OH - ion is the strongest base that can exist in aqeous solution. 15.4

Acid-base reactions : Proton transfer reaction. Neutralization reaction of two opposites. Competition of two bases for a proton, H +. => Who wants the proton more (Think back to last semester’s discussions of electronegativities and who wanted the electrons more.) 15.4

Right side favored CH 3 COOH + H 2 O CH 3 COO - + H weak base weak acid strong acid strong base HF + OH - F - + H 2 0 Equilibrium favors the weak side of the reaction. The stronger the acid the weaker the conjugate base. The stronger the base the weaker the conjugate acid. weak base weak acid strong acid strong base Compare the pairs, HF is a weak acid, but it is a stronger acid than water. Left side favored

Strong AcidWeak Acid 15.4

What is the pH of a 2 x M HNO 3 solution? HNO 3 is a strong acid – 100% dissociation. HNO 3 (aq) + H 2 O (l) H 3 O + (aq) + NO 3 - (aq) pH = -log [H + ] = -log [H 3 O + ] = -log(0.002) = 2.70 Start End M 0.0 M What is the pH of a 1.8 x M Ba(OH) 2 solution? Ba(OH) 2 is a strong base – 100% dissociation. Ba(OH) 2 (s) Ba 2+ (aq) + 2OH - (aq) Start End M M0.0 M pH = – pOH = log(0.036) =

Right side favored HCN + HCOO - CN - + HCOOH weak base weak acid strong acid strong base HF + OH - F - + H 2 0 Equilibrium favors the weak side of the reaction. weak base weak acid strong acid strong base Left side favored K = [H 2 O][F - ] [HF][OH - ] = 7.2 x K = [HCOOH][CN - ] [HCN][HCOO - ] = 3.4 x 10 -6

HA (aq) + H 2 O (l) H 3 O + (aq) + A - (aq) Weak Acids (HA) and Acid Ionization Constants HA (aq) H + (aq) + A - (aq) K a = [H + ][A - ] [HA] K a is the acid ionization constant KaKa weak acid strength 15.5 K >> 1 K << 1 Lie to the rightFavor products Lie to the leftFavor reactants Remember from last chapter that: Be careful when looking at negative exponents.

15.5

CH 3 COOH CH 3 COO - + H + Arrhenius Model Acid strength can be quantified with equilibrium constant, K a. K a = [CH 3 COOH] [H + ][CH 3 COO - ] = 1.8 x CH 3 COOH + H 2 O CH 3 COO - + H 3 O + Bronsted Model [CH 3 COOH] [H 2 O] [H 3 O + ][CH 3 COO - ] = [H 3 O + ] = [H + ] HOCl OCl - + H + K a = [HOCl] [H + ][OCl - ] = 3.5 x K a,CH3COOH = 1.8 x > K a,HOCl = 3.5 x CH 3 COOH is a stronger acid than HOCl.

What is the pH of a 0.5 M HF solution (at 25 0 C)? HF (aq) H + (aq) + F - (aq) K a = [H + ][F - ] [HF] = 7.1 x HF (aq) H + (aq) + F - (aq) Initial (M) Change (M) Equilibrium (M) x-x+x+x x x+x xx K a = x2x x = 7.1 x Ka  Ka  x2x = 7.1 x – x  0.50 K a << 1 x 2 = 3.55 x x = M [H + ] = [F - ] = M pH = -log [H + ] = 1.72 [HF] = 0.50 – x = 0.48 M pH if completely dissociated = 0.301

When can I use the approximation? 0.50 – x  0.50 K a << 1 When x is less than 5% of the value from which it is subtracted. x = M 0.50 M x 100% = 3.8% Less than 5% Approximation ok. What is the pH of a 0.05 M HF solution (at 25 0 C)? Ka  Ka  x2x = 7.1 x x = M M 0.05 M x 100% = 12% More than 5% Approximation not ok. Must solve for x exactly using quadratic equation or method of successive approximation. 15.5

Solving weak acid ionization problems: 1.Identify the major species that can affect the pH. In most cases, you can ignore the autoionization of water. Ignore [OH - ] because it is determined by [H + ]. 2.Use ICE to express the equilibrium concentrations in terms of single unknown x. 3.Write K a in terms of equilibrium concentrations. Solve for x by the approximate method. If approximation is not valid, solve for x exactly. 4.Calculate concentration of all species and/or pH of the solution. 15.5

What is the pH of a M monoprotic acid whose K a is 5.7 x ? HA (aq) H + (aq) + A - (aq) Initial (M) Change (M) Equilibrium (M) x-x+x+x x x+x xx K a = x2x x = 5.7 x Ka  Ka  x2x = 5.7 x – x  K a << 1 x 2 = 6.95 x x = M M M x 100% = 6.8% More than 5% Approximation not ok. 15.5

K a = x2x x = 5.7 x x x – 6.95 x = 0 ax 2 + bx + c =0 -b ± b 2 – 4ac  2a2a x = x = x = HA (aq) H + (aq) + A - (aq) Initial (M) Change (M) Equilibrium (M) x-x+x+x x x+x xx [H + ] = x = M pH = -log[H + ] =

percent ionization = Ionized acid concentration at equilibrium Initial concentration of acid x 100% For a monoprotic acid HA Percent ionization = [H + ] [HA] 0 x 100% [HA] 0 = initial concentration 15.5

What is the pH of a 5.3 x M HCl solution? HCl (aq) H + (aq) + Cl - (aq) HCl is a strong acid so it will fully dissociate. [H + ] HCl = 5.3 x M pH = 9.3 But, [H + ] pure water = 1.0 x M Since 5.3 x M << 1.0 x M, the acid does not significantly change the [H + ] pH = 7 neutral

NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) Weak Bases and Base Ionization Constants K b = [NH 4 + ][OH - ] [NH 3 ] K b is the base ionization constant KbKb weak base strength 15.6 Solve weak base problems like weak acids except solve for [OH-] instead of [H + ].

15.6

What is the pH of a M C 5 H 5 N (pyridine) whose K b is 1.7 x ? C 5 H 5 N ( aq) + H 2 O C 5 H 6 N + (aq) + OH - (aq) Initial (M) Change (M) Equilibrium (M) x-x+x+x x x+x xx K b = x2x x = 1.7 x Kb  Kb  x2x = 1.7 x – x  K b << 1 x = 4.1 x M = [OH - ] pH = 14 + log[OH - ] = 14 + log (4.1 x ) = 8.65 basic Base hydrolysis N : pH if completely dissociated = 12.0

How much of this pyridine is hydrolized? 4.1 x M M x 100% = 0.41% Less than 5% Approximation is ok percent hydrolized = Hydrolized concentration at equilibrium Initial concentration x 100% Always check to make sure that the assumption made was correct !!

15.7 Ionization Constants of Conjugate Acid-Base Pairs HA (aq) H + (aq) + A - (aq) A - (aq) + H 2 O (l) OH - (aq) + HA (aq) KaKa KbKb H 2 O (l) H + (aq) + OH - (aq) KwKw K a K b = K w Weak Acid and Its Conjugate Base Ka =Ka = KwKw KbKb Kb =Kb = KwKw KaKa

15.7 Ionization Constants of Conjugate Acid-Base Pairs NH 4 + (aq) H + (aq) + NH 3 (aq) NH 3 (aq) + H 2 O (l) OH - (aq) + NH 4 + (aq) K a = 5.6 x K b = 1.8 x H 2 O (l) H + (aq) + OH - (aq) K w = 1.0 x K a K b = K w (5.6 x )(1.8 x ) = 1.0 x If you know K a, you know K b !!

Polyprotic Acids => Each proton dissociates separately 1 st Ionization H 3 PO 4 (aq) H + (aq) + H 2 PO 4 - (aq) K a1 = 7.5 x nd Ionization H 2 PO 4 - (aq) H + (aq) + HPO 4 -2 (aq) K a2 = 6.2 x rd Ionization HPO 4 -2 (aq) H + (aq) + PO 4 -3 (aq) K a3 = 4.8 x K a1 > K a2 > K a2 It is more and more difficult to remove a proton.

15.8

Molecular Structure and Acid Strength H X H + + X - The stronger the bond The weaker the acid HF << HCl < HBr < HI 15.9

Molecular Structure and Acid Strength Z O HZ O-O- + H + -- ++ The O-H bond will be more polar and easier to break if: Z is very electronegative or Z is in a high oxidation state 15.9

Molecular Structure and Acid Strength 1. Oxoacids having different central atoms (Z) that are from the same group and that have the same oxidation number. Acid strength increases with increasing electronegativity of Z H O Cl O O H O Br O O Cl is more electronegative than Br HClO 3 > HBrO

Molecular Structure and Acid Strength 2. Oxoacids having the same central atom (Z) but different numbers of attached groups. Acid strength increases as the oxidation number of Z increases. HClO 4 > HClO 3 > HClO 2 > HClO 15.9

Acid-Base Properties of Salts A - (aq) + B + (aq) BA (aq) Acid + Base Salt + Water H + (aq) + OH - (aq) H 2 O (l) hydrolysis Salt + H 2 O ? pH How do we know the pH of a salt solution ?

Acid-Base Properties of Salts Neutral Solutions: Salts containing an alkali metal or alkaline earth metal ion (except Be 2+ ) and the conjugate base of a strong acid (e.g. Cl -, Br -, and NO 3 - ). NaCl (s) Na + (aq) + Cl - (aq) H2OH2O Cannot donate or accept H + Conjugate base of a strong acid HCL => Cannot accept H + => pH = 7

Basic Solutions: Salts derived from a strong base and a weak acid. CH 3 COONa (s) Na + (aq) + CH 3 COO - (aq) H2OH2O CH 3 COO - (aq) + H 2 O (l) CH 3 COOH (aq) + OH - (aq) Cannot donate or accept H + hydrolysis K b = [CH 3 COOH][OH - ] [CH 3 COO - ] = 5.6 x => generates OH -

CH 3 COO - (aq) + H 2 O (l) CH 3 COOH (aq) + OH - (aq) What is the pH of a 0.50 M CH 3 COONa solution ? Initial (M) Change (M) Equilibrium (M) x-x+x+x x “0.00” +x+x xx K b = x2x x = 5.6 x Kb  Kb  x2x = 5.6 x – x  0.50 K b << 1 x = 1.7 x M = [OH - ] pH = 14 + log[OH - ] = 14 + log (1.7 x ) = 9.23 => basic

Acid-Base Properties of Salts Acid Solutions: Salts derived from a strong acid and a weak base. NH 4 Cl (s) NH 4 + (aq) + Cl - (aq) H2OH2O NH 4 + (aq) NH 3 (aq) + H + (aq) Salts with small, highly charged metal cations (e.g. Al 3+, Cr 3+, and Be 2+ ) and the conjugate base of a strong acid. Al(H 2 O) 6 (aq) Al(OH)(H 2 O) 5 (aq) + H + (aq) Conjugate base of a strong acid HCL => Cannot accept H + K a = [NH 3 ][H + ] [NH 4 + ] = 5.6 x

NH 4 + (aq) NH 3 (aq) + H + (aq) What is the pH of a 0.33 M NH 4 Cl solution ? Initial (M) Change (M) Equilibrium (M) x-x +x+x x “0.00” +x+x xx K a = x2x x = 5.6 x Ka  Ka  x2x = 5.6 x – x  0.33 K a << 1 x = 1.4 x M = [H + ] pH = - log[H + ] = 4.87 => acidic

Acid Hydrolysis of Al Salts with small, highly charged metal cations (e.g. Al 3+, Cr 3+, and Be 2+ ) and the conjugate base of a strong acid.

Acid-Base Properties of Salts Solutions in which both the cation and the anion hydrolyze: K b for the anion > K a for the cation, solution will be basic K b for the anion < K a for the cation, solution will be acidic K b for the anion  K a for the cation, solution will be neutral What about a salt made from a weak acid and a weak base ?

CH 3 COONH 4 (s) NH 4 + (aq) + CH 3 COO - (aq) H2OH2O CH 3 COO - (aq) + H 2 O (l) CH 3 COOH (aq) + OH - (aq) K b = [CH 3 COOH][OH - ] [CH 3 COO - ] = 5.6 x NH 4 + (aq) NH 3 (aq) + H + (aq) K a = [NH 3 ][H + ] [NH 4 + ] = 5.6 x K a = K b => neutral solution

NH 4 NO 2 (s) NH 4 + (aq) + NO 2 - (aq) H2OH2O NO 2 - (aq) + H 2 O (l) HNO 2 (aq) + OH - (aq) K b = [HNO 2 ][OH - ] [NO 2 - ] = 2.2 x NH 4 + (aq) NH 3 (aq) + H + (aq) K a = [NH 3 ][H + ] [NH 4 + ] = 5.6 x K a > K b => acidic solution

NH 4 CN (s) NH 4 + (aq) + CN - (aq) H2OH2O CN - (aq) + H 2 O (l) HCN (aq) + OH - (aq) K b = [HCN][OH - ] [CN - ] = 2.0 x NH 4 + (aq) NH 3 (aq) + H + (aq) K a = [NH 3 ][H + ] [NH 4 + ] = 5.6 x K a < K b => basic solution

Oxides of the Representative Elements In Their Highest Oxidation States Remember discussion of metals and non-metals ? Metals generate basic oxides Non-metals generate acidic oxides

Arrhenius acid is a substance that produces H + (H 3 O + ) in water A Brønsted acid is a proton donor A Lewis acid is a substance that can accept a pair of electrons A Lewis base is a substance that can donate a pair of electrons Definition of An Acid H+H+ H O H + OH - acidbase N H H H H+H+ + N H H H H acidbase 15.12

Lewis Acids and Bases N H H H acidbase F B F F + F F N H H H No protons donated or accepted! Definition is more general than Bronsted sp 2 hybridized with empty 2p orbital

Lewis Acids and Bases O H H O B O H Boric Acid B(OH) 3 (aq) + H 2 O (l) B(OH) 4 - (aq) + H + (aq) Boric acid does not ionize in water.

One Ingredient Problems: One aqueous substance that you need to determine information about.