# Chapter 1 Structure and Bonding - faculty.swosu.edu

I. Base Theory A. Concepts 1) Arrhenius Concept: base produces OH- in water 2) Bronsted-Lowery Model: base is a H+ acceptor 3) Strong Hydroxide Bases a) Alkali Metal Hydroxides: NaOH, KOH, etc i. Completely Dissociated in Water ii. NaOH Na+ + OH- K = very large b) Alkaline Earth Hydroxides: Ca(OH)2, Mg(OH)2, etc i. Not very soluble in water ii. What does dissolve is completely dissociated iii. Ca(OH)2 Ca2+ + 2 OHK = very large c) Example: pH of 0.05 M NaOH pOH - log(0.05) 1.30 pH 14 - 1.30 12.70 4) H H N Non-Hydroxide Bases (Weak Bases) a) Any atom with a lone pair of electrons can accept a proton = base H b) Ammonia in water: Kb = 1.8 x10-5 OH H N H + + H O H H

acid base c) CH2CH3 CH3CH2 H conjugate acid conjugate base Other amine molecules are also bases N N CH2CH3 pyridine triethylamine 5) The general base equation B + H2 O BH+ + OH- 6) Strong Base: equilibrium lies far to the right ([OH-] [B]0) 7) Weak Base: equilibrium lies far to the left ([OH-] << [B]0) a) [BH ][OH ] Kb [B] Calculations are similar to weak acids b)

Example: pH of 15.0 M NH3 i. Kb = 1.8 x 10-5 We can find [H+] from KW = [H+][OH-] = 1 x 10-14 or pH + pOH = 14 ii. Percent dissociation still means the same thing iii. The 5% rule for approximations: x/[B] x 100% < 5% c) II. Example: pH of 1.0 M methylamine Kb = 4.38 x 10-4 Polyprotic Acids A. Carbonic Acid is a diprotic acid 1) Polyprotic means there are more than one ionizable proton 2) Carbonic acid is the acid that helps maintain body pH H2CO3 H2CO3 H+ + HCO3- [H ][HCO 3 ] K a1 4.3 x 10-7 [H 2 CO 3 ] 2 HCO33) H+ + CO32- K a2 [H ][CO3 ] -11

5 . 6 x 10 [HCO3 ] Ka1 and Ka2 stand for the loss of the first and second protons, respectively 4) B. Usually Ka1 >> Ka2 a) As (-) charge builds up, it is harder to remove the next proton b) H+ from the first ionization forces the second ionization to the left c) We can usually ignore all but the first ionization in calculations Phosphoric Acid is a triprotic acid 1) Ionizations H3PO4 H+ + H2PO4- Ka = 7.5 x 10-3 2) H2PO4- H+ + HPO42- Ka = 6.2 x 10-8 HPO42- H+ + PO43- Ka1 >> Ka2 >> Ka3 Ka = 4.8 x 10-13 K a1 7.5 x 10 -3 5

1 . 2 x 10 K a2 6.2 x 10 -8 3) K a2 6.2 x 10-8 5 1 . 3 x 10 K a3 4.8 x 10-13 Example: pH of 5.0 M H3PO4, and the concentrations of all of the phosphoric acid derived species a) Use Ka1 only to find [H+] and the pH b) Then, [H+] = [H2PO4-] c) [H3PO4] = [H3PO4]0 - [H+] d) Find [HPO42-] and [PO43-] from what is already calculated and Ka2, Ka3 C. Sulfuric Acid is a unique diprotic acid 1) The first ionization of sulfuric acid is a strong acid (Ka1 = large) H2SO4 H+ + HSO4-

2) The second ionization of sulfuric acid is a weak acid HSO4H+ + SO42Ka2 = 1.2 x 10-2 3) Example: pH of 1.0 M H2SO4 a) b) c) d) 4) Assume complete dissociation of the first proton Use the equilibrium calculation on Ka2 i. [H+]0 does not = 0 because of Ka1 ii. [H+]0 = 1.0 M Approximate 1 + x 1 in this case (5% rule says this is ok) When [H2SO4] > 1.0 M, you can ignore Ka2 Example: pH of 0.01 M H2SO4 a) The 5% rule tells us we cant ignore Ka2 in this case b) c) We must use the quadratic equation to solve for x When H2SO4 < 1.0 M, we cant ignore Ka2 III. Acid-Base Properties of Salts A) 1) 2)

a) b) c) 1) 2) 3) Simple Salts Salt = ionic compound = one that completely ionizes in water Some salts have no effect on pH Cations of strong bases have no effect on pH i. Na+, K+, etc ii. These cations have no affinity for OH- in water Anions of strong acids have no effect on on pH i. Cl-, NO3-, etc ii. These anions have no affinity for H+ in water Solutions of these combined ion salts have pH = 7.00 B) Basic Salts Salts containing the conjugate base of a weak acid produce basic solutions The conjugate base must be strong if the acid is weak, so it must have a strong affinity for H+, which will affect the pH of a solution Sodium Acetate Example NaC2H3O2 Na+ + C2H3O2C2H3O2- + H2O [HC 2 H 3O 2 ][OH ] Kb ??? [C 2 H 3O 2 ] HC2H3O2 + OH- 4) How do we find Kb for the conjugate base of a weak acid? [H ][C 2 H 3O 2 - ] [HC2 H 3O 2 ][OH ]

K a x K b [H ][OH ] [C2 H 3O 2 ] [HC2 H 3O 2 ] [H ][OH ] K W 1 x 10-14 5) For any weak acid and its conjugate base, Ka x Kb = KW Kb = KW / Ka = 1 x 10-14 / 1.8 x 10-5 = 5.6 x 10-10 for acetate 6) C. Example: pH of 0.30 M NaF Ka for HF = 7.2 x 10-4 Base Strength in Water 1) Any base in water must compete with the hydroxide anion for protons 2) Hydrocyanic acid example: HCN + H2O H3O+ + CN- Ka = 6.2 x 10-10 weak acid (compared to OH-) CN- + H2O weak base strong base (compared to H2O) HCN + OHstrong acid Kb = KW / Ka = 1.6 x 10-5 OH- > CN- > H2O D.

Acidic Salts 1) Salts having the conjugate acid of a weak base produce acidic solutions 2) Ammonium chloride = NH4Cl NH4+ + ClNH4+ NH3 + H+ 3) Since ammonia is a weak base, ammonium is strong acid and will effect pH 4) Example: pH of 0.1 M NH4Cl Kb = 1.8 x 10-5 a) Find Ka from KW b) Make sure the conjugate acid is stronger than water or it will have no effect 5) Highly charged metal ions can also be acidic a) AlCl3 + 6 H2O Al(H2O)63+ + 3 Clb) Al(H2O)63+ + H2O Al(H2O)5(OH)2+ + H3O+ c) The higher the metals charge the more acidic Example: pH of 0.01 M AlCl3 Ka = 1.4 x 10-5 E. Salts containing both acidic and basic components 1) NH4C2H3O2 NH4+ + C2H3O22) 3) 4) The calculations for these compounds are complex We can, however, at least decide if the solution is acidic or basic a) If Ka > Kb, the solution will be acidic b) If Kb > Ka, the solution will be basic c)

If Ka = Kb, the solution will be neutral Example: Will the following solutions be acidic or basic? NH4C2H3O2 NH4CN Al2(SO4)3 IV. Acid-Base Properties and Molecular Structure A. Polarity 1) Not all H containing molecules are acidic a) CHCl3 H+ + CCl3b) Strong, nonpolar bonds dont dissociate easily 2) Polar XH bonds are easily dissociated (acidic) a) HCl is a polar bond b) HOH is a polar bond 3) Bond strength also plays a part in acidity a) Hydrohalide Polarity: HF > HCl > HBr > HI b) Bond Strength (kJ.mol) 565 427 363 295 c) Acidity weak strong strong strong 4) Oxyacids: the more O on the central atom, the stronger the acid in a series a) Electronegative Oxygens remove electrons from the center atom b) This polarizes and weakens the OH bond even more c) HClO4 > HClO3 > HClO2 > HClO d)

H2SO4 > H2SO3 5) HOX Molecules a) The more electronegative X is, the more acidic the molecule is b) Electronegative X removes electrons from the HO bond c) Acidity: HOCl > HOBr > HOI > HOCH3 d) B. Electronegativity: 3.0 2.8 2.5 2.3 Acid-Base Properties of Oxides 1) A generic oxide can be represented as XO 2) If XO is a strong and covalent bond, the oxide will be acidic in water a) HOX H+ + -OX b) If X is electronegative, like O, it should form a strong covalent OX bond 3) If XO is weak and ionic, the oxide will be basic in water a) HOX X+ + OHb) NaOH Na+ + OHc) The X atom in these oxides is usually not electronegative (Na +) 4) Examples of Acidic Oxides a) SO2 + H2O H2SO3 b)

CO2 + H2O H2CO3 H+ + HSO3H+ + HCO3- 5) Examples of Basic Oxides a) CaO + H2O b) V. K2O + H2O Ca(OH)2 KOH Ca2+ + 2 OHK+ + OH- Lewis Acid-Base Definition A. Definitions 1) Lewis Acid = an electron pair acceptor 2) Lewis Base = an electron pair donor 3) Example: H+ + :NH3 NH4+ Lewis acid Lewis Base Lewis Acid-Base complex 4) This model includes the other acid-base concepts 5) This model accounts for many other chemical reactions that the others dont a) BF3 + :NH3 H3N:BF3

Lewis acid Lewis Base Lewis Acid-Base complex b) AlCl3 + 6 H2O c) Example: Identify Lewis Acid and Lewis Base i. Ni2+ + 6 NH3 Ni(NH3)62+ ii. H+ + H2O Al(H2O)63+ + 3 Cl- H3O+ Summary of Acid-Base Problem Solving 1. List major species in solution 2. Look for reactions that go to completion a. concentration of product b. major species left 3. Identify acids and bases 4. Solve the equilibrium problem, check the approximation, find pH

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