CO 3 2-

1. Chapter 11 2- : O O : C . . . . O : : . . Chemical Bonds Step 1: Find the total number of valence e- by adding up the group numbers of all atoms. For ions, adjust the dot count accordingly (subtract e- for cation, add for anion). CO32- Atomic radii increase right to left across the period, and top to bottom down the group. Opposite is true for ionization energy. (4 + 3x6 +2)e- = 24e- total Covalent bonds are made when difference in electronegativity of the elements is less than 1.9, i.e. when ions cannot be formed. Step 2: Assume that the first non-hydrogen atom in the formula of the group is the central atom. Connect atoms with single bonds. O O C O Covalent bonding gives all atoms apparent octet, i.e. the configuration of the closest noble gas. Only valence electrons are involved. [24 - (3x2)]e- = 18e- Step 3: Subtract 2e- for each bond from total #e- to get#e- in lone pairs. in lone pairs Step 4: Put in the remaining electrons, two at a time, as lone pairs. Start with the terminal atoms, and continue with the central atom if there are any electron pairs left. . . . . Octet rule not satisfied for C : O O : C . . . . O : : . . Step 5: Check that each atom has octet satisfied (doublet for H). If not, move electron pair(s) from the adjacent atom to form multiple bonds. . . The ion has three resonance structures. Step 6. Count the number of electron pairs around the central atom, including lone pairs (i.e. the steric number, SN). A multiple bond counts as a single electron group. SN=4, tetrahedral, 109.5o SN=3, trigonal planar, 120o SN=2, linear, 180o Step 7. Pretend the lone pairs are invisible and describe the resulting shape of the molecule. SN = 3, no lone pairs  all atoms in a plane, 120o.

2. Volume A Volume B Moles B Mass A Moles A Mass B Atoms A Atoms B P1 n1 P1 T1 P2 n2 P2 T2 = = 22.4 L 1 mol O2 1 mol KClO3 122.55 g p.c. 3 mol O2 2 mol KClO3 = 2.7 L 10. g p.c.x x x g - mol A mol A - mol B mol - Liters B Chapter 12 Behavior of an ideal gas can be totally described by: Gaseous State PV = nRT Standard temp. and pressure (STP) is 1 atm and 0oC. Pressure (P) – atmospheres Temperature (T) – Kelvin Volume (V) – liters Amount of Gas (n) – moles Conversion factors 1 mol 22.4 L One constant parameter: Volume (L) x If two parameters are constant, the relationship simplifies: P1V1 = P2V2 1 mol molar mass (g) Mass (g) x n1T1 = n2T2 P1V1 n1 P2V2 n2 V1 T1 V2 T2 = 1 mol 6.022 x 1023 atoms = etc. # atoms x Gas Stoichiometry: What volume of O2 at STP can you make from 10. g KClO3 (p.c.)? The total pressure of a mixture of gases is the sum of the partial pressures exerted by each gases: 2 KClO3 2 KCl + 3 O2 Ptotal = PA + PB + PC + … Partial pressure: What is the volume of the dry oxygen if it was collected over water at 23oC and 760. torr in a 500 mL container? Water vapor pressure at 23oC is 21.2 torr. Experimentally determined that 1 mol of ANY gas occupies 22.4 L at STP (Avogadro’s law). Step 1: determine the pressure of dry O2: PO2 = Ptot – PH2O If the mass and volume of the gas at STP are known, molar mass M is: PO2 = 760 torr – 21.2 torr = 739 torr. P1 = 739 torr V1 = 500 mL P2 =760 torr V2 = ? mRT PV M = P1V1 P2 739 torr x 500 mL 760. torr V2 = = 486 mL dry O2. =

3. . . . . H – O : H – O : | | H H evaporation liquid gas . . . . condensation O H H melting solid liquid freezing Chapter 13 Liquids have intermediate properties between solids and gases. Liquids are almost incompressible, have definite volume and assume the shape of the container. Liquid State d- Vapor pressure is the pressure exerted by a liquid at equilibrium with its liquid. d+ Vapor pressure of any gas at the boiling point is equal to the atmospheric pressure. Rate of vaporization is proportional to vapor pressure. Qfusion = (mass) (spec.heat of fusion) Qvaporization = (mass) (spec.heat of vaporization) Qheating = (mass) (spec.heat) (temp.change) Heating curve for a pure substance Qtot = Qfusion + Qheating + Qvaporization Hydrogen bond exists if H is directly bonded to either N, O, or F. Very strong intermolecular forces. H bond . . . . Hydrateis a true compound, water is an integral part of it. CuSO4 5 H2O Copper(II) sulfatepentahydrate. Water is unique because it is liquid at room temp, its solid form (ice) has lower density than the liquid, and is an excellent solvent. Temperature is constant during changes of phase.

4. Solution Chapter 14 Moles of solute 1 L of solution Mdiluted soln. x Vdiluted soln. Mstock soln. Vstock soln. = Solubility of ions in water …is a homogeneous mixture of two or more substances with the same composition throughout, thus the concentration of the substances is the same. Consists of solvent (present in greatest amount), and solute(s). soluble insoluble all Na+, K+, NH4+ most CO32-, PO43-, OH-, S2-. all NO3-, C2H3O2- (except Na+,K+,NH4+) most SO42-, Cl-, Br-, I- Solubility, or the maximum mass of solute in 100. g of solvent, increases with temperature for most solids. For gases, solubility increases with P and decreases with T. Solution can be saturated,unsaturated, and supersaturated. The latter is unstable and precipitates upon disturbance. Amount of solute Amount of solution Mass% Vol% Mass/vol% % composition = x 100 Molarity is the number of moles Molarity (M) = of solute in 1 L of solution To prepare 1 L of NaCl, measure molar mass of NaCl, add it to a 1L volumetric flask and add water to the mark. To prepare 1 L of NaCl, calculate Vstock soln of NaCl, add it to a 1L volumetric flask and add water to 1L. Molality(m) is the number of moles of solute per kilogram of solvent. ∆tf = m x Kf ∆tb = m x Kb Freezing Point Depression and Boiling Point Elevation depend on molality. Prepare 500. mL of 0.15 M NaCl using solid NaCl. What is the freezing point (F.p.) of solution of 100.0 g ethylene glycol in 200. g H2O? 58.4 g NaCl 1 mol NaCl 0.15 mol NaCl 1 L soln. x = 4.4 g NaCl 0.5 L x 1 mol eg 62.05 g eg 100.0 g eg x = 8.05 m F.p. = -15 oC Prepare above using stock soln of 0.50 M NaCl. 0.15 M x 500 mL 0.5 M M1V1 M2 mol eg kg H2O Vstock = = = oC kg H2O mol eg ∆tf = m x Kf = 8.05 x 1.86 = 15 oC = 150 mL stock soln.  150 mL stock soln. + 350 mL H2O.

5. Strong acids #H3O+ Weak acids Hydrochloric (HCl) 1 1 Hydrofluoric (HF) Hydrobromic (HBr) 1 1 Hypochlorous (HClO) Hydroiodic (HI) 1 1 Acetic (HC2H3O2) Nitric (HNO3) 1 2 Carbonic (H2CO3) Sulfuric (H2SO4) 2 3 Phosphoric (H3PO4) Chapter 15 Electrolytes, Acids and Bases Acid (Latin acidus - sour): sour taste; turns plant dye litmusred; dissolves metals producing H2. Base: bitter taste; turns plant dye litmusblue; aqueous solutions feel slippery to touch. Ionic compounds are usually metal plus nonmetal or group of nonmetals. An electrolyte is a solute that dissolves in water and dissociates into ions, yielding a solution that conducts electricity. Find electrolytes from the list: Exceptions: HX (X-halide) - polar covalent, produce acids in H2O; ammonium salts. Al(NO3)3, (CH3)2O, (NH4)2SO4, CH3OH, HBr H2SO4 + H2O H3O+ + HSO4- Keq > 1.0 x 103 Dissociation constant, Keq, shows how many protons can be obtained from 1 mol of acid. HSO4- + H2O H3O+ + SO42- Keq = 1.2 x 10-2 Base - anything that accepts a proton. Acid - anything that donates a proton. Water auto- ionization. [H3O+] x [OH-] [H2O] x [H2O] Keq = Kw = Keq x [H2O]2= [H3O+] x [OH-] Kw = 1 x 10-14 pH = - log [H3O+] Buffer, a solution of a weak acidand its conjugate base, maintains pH. Addition of a strong acid or base to a buffer changes pH only slightly. [H3O+] = 10-(pH) Ionic Equations HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) Formula eq. Total ionic eq. H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq)  Na+(aq) + Cl-(aq) + H2O(l) Net ionic eq. spectator ions cancel H+(aq) + OH-(aq)  H2O(l)