Chapter 17 Acids and Bases

Arrhenius Definition Acids produce hydrogen ions (H+1) in aqueous solution (HCl → H1+ + Cl1-) Bases produce hydroxide ions (OH1-) when dissolved in water. (NaOH → Na1+ + OH1-) Limited to aqueous solutions. Only one kind of base (hydroxides) NH3 (ammonia) could not be an Arrhenius base: no OH1- produced.

Brønsted-Lowry A broader definition than Arrhenius Acid is hydrogen-ion donor (H+ or proton); base is hydrogen-ion acceptor. Acids and bases always come in pairs. HCl is an acid. When it dissolves in water, it gives it’s proton to water. HCl(g) + H2O(l) ↔ H3O+(aq) + Cl-(aq) Water is a base; makes hydronium ion.

Monoprotic Acids Compounds that donate one hydrogen atom HNO3 (nitric acid) HF (hydrofluoric acid) HCl (hydrochloric acid) CH3COOH (acetic acid)

Polyprotic Acids Compounds have more than one hydrogen to donate H2SO4 (sulfuric acid) - diprotic - 2 H+ H3PO4 (phosphoric acid) - triprotic - 3 H+ H2CO3 (carbonic acid) - diprotic - 2 H+ Having more than one ionizable hydrogen does not mean stronger!

Amphoteric – a substance that can act as both an acid and base- as water shows HCl + H2O H3O+ + Cl- acid base NH3 + H2O NH4+ + OH- base acid

Amphoteric – a substance that can act as both an acid and base- as water shows H2PO4- + H2O HPO4-2 + H3O+ acid base H2PO4- + H2O H3PO4 + OH- base acid

Conjugate Acid-Base Pairs A “conjugate base” is the remainder of the original acid, after it donates it’s hydrogen ion A “conjugate acid” is the particle formed when the original base gains a hydrogen ion Thus, a conjugate acid-base pair is related by the loss or gain of a single hydrogen ion.

Hydrogen Ions from Water Water ionizes, or falls apart into ions: 2H2O H3O+ + OH- Called the “self ionization” of water Occurs to a very small extent: [H+1] = [OH-1] = 1 x 10-7 M Since they are equal, a neutral solution results from water Kw = [H+1] x [OH-1] = 1 x 10-14 Kw is called the “ionization constant” for water

Based on writing equilibrium constants, we do not include water in the expression. Kw is just an equilibrium constant used for water. The low number for concentrations (1 x 10-7) means that very few water molecules ionize.

2H2O ↔ H3O+ + OH- Kw is constant in every aqueous solution: [H+] x [OH-] = 1 x 10-14 If [H+] > 10-7 then [OH-] < 10-7 If [H+] < 10-7 then [OH-] > 10-7 If we know one, other can be determined If [H+] > 10-7 , the solution is acidic If [H+] < 10-7 , the solution is basic

pH

Calculating pH definition: pH = -log[H+] in neutral pH = -log(1 x 10-7) = 7 in acidic solution [H+] > 1 x 10-7 pH < 7 (from 0 to 7 is the acid range) pH > 7 (7 to 14 is base range)

Calculating pOH pOH = -log [OH-] [H+] x [OH-] = 1 x 10-14 pH + pOH = 14

Calculating [H+] from pH [H+] = 10-pH Example If the pH of a solution is 3.12, what is the hydronium ion concentration? [H+] = 10-3.12 = 7.6 x 10-4 M

Calculating [OH-] from pOH [OH-] = 10-pOH Example If the pH of a solution is 3.12, what is the hydroxide ion concentration? pH + pOH = 14, so the pOH is 10.88 [OH-] = 10-10.88 = 1.3 x 10-11 M

Problems, page 763 What is the pH of a 0.0012 M NaOH solution. The pH of diet soda is 4.32. What is the hydronium ion concentration? If pH of a solution containing the strong base NaOH is 10.46, what is the concentration of NaOH?

Problems, page 763 What is the pH of a 0.0012 M NaOH solution. Answer is 11.08 The pH of diet soda is 4.32. What is the hydronium ion concentration? 4.8 x 10-5 If pH of a solution containing the strong base NaOH is 10.46, what is the concentration of NaOH? 2.9 x 10-4

Ionization of Acids and Bases Strong acid: HCl H+ + Cl- Weak acid: CH3COOH CH3COO- + H+

Ionization of Acids and Bases Strong acid: [HA] = [H+] Weak acid: [HA] >>> [H+]

Ionization of Acids and Bases Strong base: NaOH Na+ + OH- Weak base: NH3 NH4+ + OH- H2O

Ionization of Acids and Bases Strong base: [B] = [OH-] in H2O solution Weak base: [B] >>> [OH-] in H2O solution

Increasing Acid Strength HCO3- HClO HF Ka = 4.8 x 10-11 3.5 x 10-8 7.2 x 10-4 Increasing acid strength

pKa and Ka pKa = - log Ka The higher the Ka value, the lower the pKa value. Lower pKa value – the stronger the acid Common in Organic and Biochemistry This is similar to pH = -log[H+]

Increasing Acid Strength CH3CH2COOH CH3COOH HCOOH Ka = 1.3 x 10-5 1.8 x 10-5 1.8 x 10-4 pKa = 4.89 4.74 3.74 Increasing acid strength

Direction of Acid/Base Reactions All acid/base reactions proceed from the stronger acid and base, to the weaker acid and base

Direction of Acid/Base Reactions All acid/base reactions proceed from the stronger acid and base, to the weaker acid and base

Using a table of acid strength (Ka values), you can predict the direction of an acid/base reaction. Is the reaction product favored or reactant favored.

Example 17.3, page 773 Write a balanced, net ionic equation for the reaction that occurs between acetic acid (CH3COOH) and sodium bicarbonate (NaHCO3). Decide which direction does the equilibrium lie.

CH3COOH + NaHCO3 CH3COO-Na+ + H2CO3 CH3COOH + HCO3- CH3COO- + H2CO3 Ka = 1.8 x 10-5 4.7 x 10-7 All acid/base reactions proceed from the stronger acid and base, to the weaker acid and base

Types of Acid-Base Reactions

Calculations with Equilibrium Constants Values of Ka and Kb are determined through experiments. Ka and Kb values are determined in a similar fashion to other K values. To determine Ka and Kb values, you use ICE tables.

What is the value of Ka for lactic acid? Example 17.4, page 776 A 0.10 M aqueous solution of lactic acid, CH3CHOHCOOH, has a pH of 2.43. What is the value of Ka for lactic acid? CH3CHOHCOOH + H2O CH3CHOHCOO- + H3O+

To determine Ka, you need to know the equilibrium concentrations of each species. The pH allows you to calculate the [H3O+]. pH = 2.43 [H3O+] = 10-pH [H3O+] = 10-2.43 [H3O+] = 3.7 x 10-3 M

CH3CHOHCOOH + H2O CH3CHOHCOO- + H3O+ (I) 0.10 0 0 (C) - x + x +x (E) 0.1 –x x 3.7 x 10-3 x = 3.7 x 10-3

[CH3CHOHCOO-] [H3O+] Ka = [CH3CHOHCOOH] (3.7 x 10-3) (3.7 x 10-3) Ka = (0.1- 3.7 x 10-3) (3.7 x 10-3) (3.7 x 10-3) Ka = (0.096) Ka = 1.4 x 10-4

Knowing the values of equilibrium constants for weak acids and bases enables you to calculate the pH of a solution.

Example 17.5, page 778 Calculate the pH of a 0.020 M aqueous solution of benzoic acid (C6H5CO2H) given that the value of Ka = 6.3 x 10-5. C6H5COOH + H2O C6H5COO- + H3O+

C6H5COOH + H2O C6H5COO- + H3O+ (I) 0.02 0 0 (C) - x + x +x (E) 0.02 –x x x

[C6H5COO-] [H3O+] Ka = [C6H5COOH] (x) (x) 6.3 x 10-5 = (0.02 - x) (x)2 6.3 x 10-5 = (0.02) x = 0.001

Using the value of x calculated, return to the ICE table and substitute x into the equilibrium line. X = [H3O+] [H3O+] = 0.001M pH = -log(0.001) pH = 2.96

Example 17.7, page 781 Calculate the pH of a 0.015 M aqueous solution of sodium acetate (Na+CH3CO2-) given that the value of Kb = 5.6 x 10-10. CH3COO- Na+ + H2O CH3COOH + OH-

CHCOO- + H2O CH3COOH + OH- (I) 0.015 0 0 (C) - x + x +x (E) 0.015 –x x x

[CH3COOH] [OH-] Kb = [CH3COO-] (x) (x) 5.6 x 10-10 = (0.015 - x) (x)2 5.6 x 10-10 = (0.015) x = 2.9 x 10-6

Using the value of x calculated, return to the ICE table and substitute x into the equilibrium line to solve for [OH]. [OH-] = 2.9 x 10-6 pOH = -log(2.9 x 10-6) pOH = 5.54 pH + pOH = 14 pH = 14 – 5.54 = 8.46

Example 17.8, page 782 Skip What is the pH of a solution that results from mixing 25 mL of 0.016 M aqueous NH3 with 25 mL of 0.016 M aqueous HCl? NH3 + HCl + H2O NH4+ + H2O + Cl- Since you are solving for pH, you need the [H3O+]. Rewrite the equation ignoring Cl: NH3 + H3O+ NH4+ + H2O

  +

To solve this type of problem between a strong acid and weak base, reverse the reaction. NH4+ + H2O NH3 + H3O+ (I) 0.008 0 0 (C) - x + x +x (E) 0.008 –x x x

[NH3] [H3O+] Ka = [NH4+] (x) (x) 5.6 x 10-10 = (0.008 - x) (x)2 5.6 x 10-10 = (0.008) x = 2.1 x 10-6

[H3O+] = 2.1 x 10-6 M pH = -log(2.1 x 10-6 ) pH = 5.67 The final solution is acidic, which is predicted when mixing a strong acid with a weak base.

Ionization of Strong Acids and Bases

Strong acids and bases are completely ionized in aqueous solutions: HCl H+ + Cl- NaOH Na+ + OH-

Therefore: NaOH Na+ + OH- [1 x 10-2 M] [1 x 10-2 M] of Na+ and OH-

Problem 17.1, page 762 What are the hydroxide and hydronium ion concentrations in a 0.0012 M aqueous solution of NaOH? NaOH is a strong base and is completely ionized: NaOH Na+ + OH-

NaOH Na+ + OH- [OH-] = 0.0012 M Kw = [H+1] x [OH-1] = 1 x 10-14 [H+1] x [0.0012] = 1 x 10-14 [H+1] = 1 x 10-14 M2 / 0.0012 = 8.3 x 10-12 M

Calculate the hydroxide ion concentration for a 2.5 M Ca(OH)2 solution. Ca(OH)2 is a strong base and is completely ionized: Ca(OH)2 Ca+2 + 2 OH- [Ca(OH)2] = [Ca+2] + 2 [OH-] 2.5 M 2.5 M 5.0 M

Oxyacids Oxyacids are acids containing an oxygen atom, hydrogen atom and another non-metal atom (N, S, P). Examples HNO2, HNO3, H2SO4, H3PO4, HOCl

HClO4 > HClO3 > HClO2 > HClO In a series of related compounds, acid strength increases as the number of oxygen atoms bonded to the central atom increases. HNO3 > HNO2 Stronger weaker HClO4 > HClO3 > HClO2 > HClO Stronger weaker

After the H+ is removed from the molecule, you are left with a negative charge. The more electronegative atoms in the molecule (O atoms), the better able to accommodate the negative charge.

Carboxylic Acids Organic acids represented by the structure Acetic Acid

Acetic Acid

Resonance Structures

Electronegative Groups in the Molecule

Hydrated Metal Cations Hydrated metal cations can act as acids. This occurs with transition elements with a +2 or +3 charge. [Fe(H2O)6]+3 [Ni(H2O)6]+2 [Fe(H2O)6]+2 [Cu(H2O)6]+3

Examples [Fe(H2O)6]+3 + H2O [Fe(H2O)5OH]+2 + H3O+ [Cu(H2O)6]+2 + H2O [Cu(H2O)5OH]+1 + H3O+

Lewis Acids and Bases Lewis Acid – electron pair acceptor Lewis Base – electron pair donor A + B: B-A acid base

Typical Lewis Acids 1) Molecules with a + charge 2) Molecules that do not meet the octet rule (BH3) Typical Lewis Bases Molecules with a – charge Molecules with lone pairs of electrons (H2O, NH3)

Fe+2 + 6 H2O [Fe(H2O)6]+2 Cu+2 + 4 NH3 [Cu(NH3)4]+2 acid base

Lewis Acid – Base Examples

Lewis Acid – Base Examples BH3 AlCl3