Solids, Liquids, and Solutions

Solids, Liquids, & Solutions 1970 What is meant by the lattice energy of an ionic compound? What quantities need to be determined and how are they used to calculate the lattice energy of an ionic compound. Answer: Lattice energy - quantity of energy released in the formation of one mole of an ionic solid from its separated gaseous ions. The energy quantities needed to be determined: sublimation of solid metal ionization of gaseous atomic metal (ionization energy) dissociation of gaseous non-metal ion formation by gaseous atomic non-metal (electron affinity) They are used with Hess’s Law to determine the combination of gaseous ions. This is the Born-Haber Cycle. page 1 (2.4×10-2 mol/L)(187.8 g/mol)(0.050 L) = 0.22 g AgBr 1971 Molarity and molality are two ways of expressing concentration. (a) Clearly distinguish between them (b) Indicate an experimental situation where expressing concentrations as molarity is particularly appropriate. (c) Indicate an experimental situation where expressing concentration as molality is particularly appropriate. Answer: (a) molarity (M) - molar concentration; composition or concentration of a solution expressed as number of moles of solute per liter of solution. molality (m) - solution concentration expressed as number of moles of solute per kilogram of sol1971 vent. Solve the following problem many possibilities, examples: AgBr(s) → Ag+(aq) + Br-(aq) Ksp = 3.3×10-13 Ag+(aq) + 2 NH3(aq) → Ag(NH3)2+(aq) K = 1.7×10+7 (b) Acid - base titrations (a) How many grams of silver bromide, AgBr, can be (c) Molecular weight determination by freezing point dissolved in 50 milliliters of water? depression change. (b) How many grams of silver bromide can be dissolved in 50 milliliters of 10 molar ammonia solu- 1972 D tion? (a) How many moles of Ba(IO3)2 is contained in 1.0 Answer: liter of a saturated solution of this salt at 25°. Ksp of Ba(IO3)2 = 6.5×10-10 (a) [Ag+][Br-] = Ksp = 3.3×10-13 = X2 X = 5.7×10-7 M = [Ag+] = mol/L AgBr that dissolve (b) When 0.100 liter of 0.060 molar Ba(NO3)2 and 0.150 liter of 0.12 molar KIO3 are mixed at 25°C, 5.7 ∞10− 7 mol AgBr 187 .8 g AgBr ∞ = 0.050 L ∞ how many milligrams of barium ion remains in 1L 1 mol AgBr each milliliter of the solution? Assume that the = 5.4×10-6 g AgBr volumes are additive and that all activity coeffi+ -13 cients are unity. (b) AgBr(s) → Ag (aq) + Br (aq) Ksp = 3.3×10 Ag+(aq) + 2 NH3(aq) → Ag(NH3)2+(aq) K = 1.7×10+7 Answer: (a) Ba(IO3)2 ↔ Ba2+ + 2 IO3-----------------------------------------Ksp = [Ba2+][IO3-]2 = 6.5×10-10 AgBr + 2 NH3 → Ag(NH3)2+ + Br[Ba2+] = X; [IO3-] = 2X; (X)(2X)2 = 6.5×10-10 K = Ksp×K = 5.6×10-6 X = 5.5×10-4 M = mol/L of dissolved Ba(IO3)2 [Ag(NH3)2+] = [Br-] = X M; [NH3] = (10 - 2X) M (b) initial mol Ba2+ = (0.060 mol/L)(0.100L) = 0.0060 [Ag( NH 3 )+ ][ Br − ] X _X − 6 K= = = 5.6∞10 mol (10 − 2X)2 [ NH 3 ]2 initial mol IO3- = (0.150L)(0.120 mol/L) = 0.0180 -2 X = 2.4×10 M = [Br ] = mol/L dissolved AgBr mol Solids, Liquids, & Solutions after reaction, essentially all Ba2+ reacts while IO3= {0.0180 - (2)(0.0060)} mol = 0.0060 mol/0.250 L = 0.024M [IO3-] Ksp 6.5 ∞10−10 [Ba 2+ ] = = =1.1 ∞10− 6 M − 2 2 [ IO 3 ] ( 0.024 ) 1.1∞10− 6 mol 137340 mg Ba 2+ = ∞ 1 mol 100 0 mL = 1.5×10-4 mg / mL Ba2+ 1973 D The molar solubility of silver bromide is diminished by the addition of a small amount of solid potassium bromide to a saturated solution. However, the molar solubility of silver bromide is increased by the addition of solid potassium nitrate, a salt whose ions are not common to those of silver bromide. Explain these experimental observations in terms of the principles involved. Answer: AgBr(s) ↔ Ag+(aq) + Br-(aq); As KBr dissolves, the concentration of Br- ions increase and force the equilibrium to shift to the left (LeChatelier’s principle) where the concentrations of the ions in solution decrease and less can dissolve. The diverse (“uncommon”) ion effect – “the salt effect”. As the total ionic concentration of a solution increases, interionic attractions become more important. Activities become smaller than the stoichiometric or measured concentrations. For the ions involved in the solution process this means that a higher concentration must appear in solution before equilibrium is established. - the solubility must increase. page 2 Description of process (rates of vaporization and condensation) 1975 D Alcohol dissolves in water to give a solution that boils at a lower temperature than pure water. Salt dissolves in water to give a solution that boils at a higher temperature than pure water. Explain these facts from the standpoint of vapor pressure. Answer: An alcohol-water solution has a higher than normal (pure water) vapor pressure because alcohol is a volatile solute and contributes substantially to the vapor of the solution. The higher the vapor pressure, the lower the boiling point. A salt-water solution has a lower than normal vapor because salt is a non-volatile solute and solute-solvent interaction decrease the vapor of the solution, the lower the vapor pressure, the higher the boiling point. 1976 B (a) Calculate the molality of a 20.0 percent by weight aqueous solution of NH4Cl. (Molecular weight: NH4Cl = 53.5) (b) If this NH4Cl solution is assumed to be ideal and is completely dissociated into ions, calculate the pressure of this solution at 29.0°C. (c) Actually a solution of NH4Cl of this concentration is not ideal. Calculate the apparent degree of dissociation of the NH4Cl if the freezing point of this solution is -15.3°C? (Molal freezing point constant = 1.86°C) Answer: 20.0 ∞1000 = 250 g NH 4 Cl in 1000g H 2O (a) 80.0 1974 D Two beakers, one containing 100 milliliters of a 0.10 1 mol NH4 Cl 250 g = 4.67 mol 1 kg = 4.67molal ∞ molal solution of sucrose (a nonvolatile nonelectrolyte) the O 53.5 g k g H 1 2 other containing 100 milliliters of pure water, are placed side by side in a closed system, such as under a (b) P1 = (P°)(X1) bell jar. Explain in terms of the principles involved mol ions = (2)(4.67 mol) = 9.34 mol what changes, if any, occur to bring the system to equi1 kg water = 55.6 mol water librium. 55.6 Answer: = 0.856 X1 = (1) Volume of sugar solution increases; (2) volume of 55.6 + 9.34 pure water decreases; (3) water beaker finally empty. P1= (29.8 mm Hg)(0.856) = 25.5 mm Hg Raoult’s Law, vapor pressure and volatility (c) Assume no dissociation. Solids, Liquids, & Solutions ∆T = kfm = (1.86)(4.67) = 8.69°C i = 15.3 / 8.69 = 1.76 (1.76 - 1.00)(100) = 76% dissociated page 3 of fused salt, etc. 1978 D The freezing point and electrical conductivities of three aqueous solutions are given below. Solution Freezing Electrical (0.010 molal) Point Conductivity sucrose -0.0186°C almost zero formic acid -0.0213°C low sodium formate -0.0361°C high Explain the relationship between the freezing point and electrical conductivity for each of the solutions above. Account for the differences in the freezing points among the three solutions. Answer: sucrose a non electrolyte 1977 D The solubility of Zn(OH)2 is not the same in the following solutions as it is in pure water. In each case state whether the solubility is greater or less than that in water and briefly account for the change in solubility. (a) 1-molar HCl (c) 1-molar NaOH (b) 1-molar Zn(NO3)2 (d) 1-molar NH3 Answer: (a) greater: Zn(OH)2(s) + 2 H+ → Zn2+ + 2 H2O (b) lower: increased [Zn2+] decreases [OH-] and decreases the amount of Zn(OH)2 in solution. Ksp =[Zn2+][OH-]2 (c) greater: Zn(OH)2(s) + 2 OH- → Zn(OH)42- (Can al0.010 mol/kg lowers freezing point 0.0186°C so site common ion effect) formic acid (d) greater: Zn(OH)2(s) + 4 NH3 → Zn(NH3)42+ + OHa weak electrolyte; low conductance as a result of low ion concentration 1977 D The state of aggregation of solids can be described as mtotal > 0.010 molal due to partial ionization and belonging to the following four types: ∆Tf somewhat greater than 0.0186°C (1) ionic (3) covalent network sodium formate (2) metallic (4) molecular a salt and strong electrolyte For each of these types of solids, indicate the kinds of approximately 100% dissociation into ions, mtotal particles that occupy the lattice points and identify approaching 0.02 molal forces among these particles. How could each type of solid be identified in the laboratory? Answer: 1979 A Particles Binding Experimental A saturated solution of lead iodate in pure water has a Forces Identification lead ion concentration of 4.0×10-5 mole per liter at ionic + & - ions electrostatic conductivity 20°C. attraction of fused salt (a) Calculate the value for the solubility-product conmetallic + ions electrostatic conductivity stant of Pb(IO3)2 at 25°C. attraction of the solid (b) Calculate the molar solubility of Pb(IO3)2 in a 0.10 between molar Pb(NO3)2 solution at 25°C. ions and (c) To 333 milliliters of a 0.120-molar Pb(NO3)2 soluelectrons tion, 667 milliliters of 0.435-molar KIO3 is added. covalent atoms covalent high melting Calculate the concentrations of Pb2+ and IO3- in the network bonds pt., extreme solution at equilibrium at 25°C. hardness, etc. Answer: molecular molecules van der low melting Waals pt., non(a) Pb(IO3)2(s) ↔ Pb2+ + 2 IO3conductivity Ksp = [Pb2+][IO3-]2 = (4.0×10-5)(8.0×10-5)2 = 2.6×10-13 Solids, Liquids, & Solutions (b) [Pb2+] = (0.10 + X) M ≈ 0.10 M [IO3-]2 = (2X) M Ksp = [Pb2+][IO3-]2 = 2.6×10-13 = (0.10)(2X)2 X = 8.1×10-7M = mol Pb(IO3)2 dissolved per L solution. (c) initial mol Pb2+ = 0.0400 mol initial mol IO3- = 0.290 mol ppt must occur; after ppt, mole IO3- in solution = (0.290 - 0.080) mol = 0.210 mol [IO3-] = 0.210 M since the volume is 1.00 L Ksp = [Pb2+][IO3-]2 = 2.6×10-13 = [Pb2+](0.210)2 [Pb2+] = 5.8×10-12 M page 4 62g 0.5156 g O 1 mol O 2.0 mol O ∞ ∞ = 1 mol cmpd 1g cmpd 16 g O 1 mol cmpd = C2H6O2 (c) 2 C2H6O2 + 5 O2 → 6 H2O(g) + 4 CO2(g) 1 mol C 2H6O2 10 mol gas = ∞ 1.05 g C 2H6O2 ∞ 62 g C 2H6O2 2 mol C 2H6O2 = 0.0847 mol gas P = (nRT) / V (0.0847 mol )( 0.08205 L_atm )( 400 .K) mol _K P= = 0.926 atm. 3.00L 1980 D Account for the differences in solubility described in each of the following experimental observations: (a) BaCO3, BaSO3, and BaSO4 are only slightly soluble in water, but the first two dissolve in HCl solution whereas BaSO4 does not. (b) CuS cannot be dissolved by warm dilute HCl but it does dissolve in warm dilute HNO3. (c) AgCl, Hg2Cl2 and PbCl2 are only slightly soluble in water, but AgCl does dissolve in ammonia solution whereas the other two do not. (d) Fe(OH)3 and Al(OH)3 are only slightly soluble in water, but Al(OH)3 dissolves in concentrated NaOH whereas Fe(OH)3 does not. Answer: (a) BaCO3(s) ↔ Ba2+ + CO32BaSO3(s) ↔ Ba2+ + SO32BaSO4(s) ↔ Ba2+ + SO42Dissolving takes place if equilibrium is shifted to the right. CO32- + H+ → HCO3- + H+ → H2O + CO2(g) SO32- + H+ → HSO3- + H+ → H2O + SO2(g) In these two cases, equilibrium is shifted to the right by the production of a removed product (a gas). SO42- + H+ do not react since SO42- is a weak Bronsted base. 1980 B (a) A solution containing 3.23 grams of an unknown compound dissolved in 100.0 grams of water freezes at -0.97°C. The solution does not conduct electricity. Calculate the molecular weight of the compound. (The molal freezing point depression constant for water is 1.86°C kg mole-1) (b) Elemental analysis of this unknown compound yields the following percentages by weight H=9.74%; C=38.70%; O=51.56%. Determine the molecular formula for the compound. (c) Complete combustion of a 1.05 gram sample of the compound with the stoichiometric amount of oxygen gas produces a mixture of H2O(g) and CO2(g). What is the pressure of this gas mixture when it is contained in a 3.00 liter flask at 127°C? Answer: (a) ∆Tf = kfm; 0.97°C = (1.86°Cm-1)(m) m = 0.52 mol solute/kg solvent In this solution, 3.23 g solute in 100.0 g water or 32.3 g solute in 1 kg of water 1 kg solvent 32.3 g = 62 g mol ∞ mol .wt. = 0.52 mol 1 kg solvent 0.0974 g H 1 mol H 6.0 mol H 62g 2∞ ∞ = 1g cmpd 1 mol cmpd 1g H 1 mol cmpd (b) Warm dilute HNO3 oxidizes S to S° (or higher). (b) This reaction shifts the equilibrium between 0.3870 g C 1 mol C 2.0 mol C 62g CuS(s) and its ions toward the ions. = ∞ ∞ 1g cmpd 12 g C 1 mol cmpd (c) AgCl(s) + 2 NH → [Ag(NH ) ]+ + Cl1 mol cmpd 3 3 2 Solids, Liquids, & Solutions page 5 silver ions complex with ammonia to form the (a) The hydrocarbon is found to contain 93.46 percent soluble [Ag(NH3)2]+, neither Hg22+ nor Pb2+ form carbon and 6.54 percent hydrogen. Calculate the such complexes. empirical formula of the unknown hydrocarbon. (d) Al(OH)3(s) + OH- → Al(OH)4- (or similar) (b) A solution is prepared by dissolving 2.53 grams of p-dichlorobenzene (molecular weight 147.0) in Al(OH)3 is amphoteric. The product is a hydroxo25.86 grams of naphthalene (molecular weight aluminate ion, Fe(OH)3 is not amphoteric. 128.2). Calculate the molality of the pdichlorobenzene solution. 1984 C Give a scientific explanation for the following obser- (c) The freezing point of pure naphthalene is determined to be 80.2°C. The solution prepared in (b) vations. Use equations or diagrams if they are relevant. is found to have an initial freezing point of (a) It takes longer to cook an egg until it is hard75.7°C. Calculate the molal freezing-point depresboiled in Denver (altitude 1 mile above sea level) sion constant of naphthalene. than it does in New York City (near sea level). (b) Burn coal containing a significant amount of sul- (d) A solution of 2.42 grams of the unknown hydrocarbon dissolved in 26.7 grams of naphthalene is fur leads to ôacid rain.ö found to freeze initially at 76.2°C. Calculate the (c) Perspiring is a mechanism for cooling the body. apparent molecular weight of the unknown hydro(d) The addition of antifreeze to water in a radiator carbon on the basis of the freezing-point depresdecreases the likelihood that the liquid in the radision experiment above. ator will either freeze or boil. (e) What is the molecular formula of the unknown Answer: hydrocarbon? (a) Water boils at a lower temperature in Denver than Answer: in NYC because the atmospheric pressure is less at high altitudes. At a lower temperature, the (a) Assume 100. g sample of the hydrocarbon 1 mol C cooking process is slower, so the time to prepare a = 7.782 mol C 93.46 g C ∞ 12.01 g C hard-boiled egg is longer. 1 mol H (b) S + O2 → SO2 (as coal is burned) = 6.49 mol H 6.54 g H ∞ 1.008 g H SO2 + H2O → H2SO3 (in the atmosphere) 7.782 mol C 1.20 H2SO3 is sulfurous acid. ; C 1.20 H1.00 = C6H5 = 6.49 mol H 1.00 (c) Vaporization or evaporation of sweat from the skin is an endothermic process and takes heat (b) 1 mol from the body and so cool the skin. 2.53 g ∞ 147.0 g mol solute (d) Colligative properties, which depend on the numm= = = 0.666 mol al ber of particles present, are involved. 0.02586 kg 1.0 kg solvent Solute (the antifreeze) causes the lowering of the (c) ∆Tf = (80.2 - 75.7)°C = 4.5°C vapor pressure of the solvent. When the vapor kf = ∆Tf /m = 4.5°C / 0.666 molal = 6.8°C/molal pressure of the solvent is lowered, the freezing point is lowered and the boiling point is raised. (d) ∆Tf = (80.2 - 76.2)°C = 4.0°C 2.43 g solute 6.8_C _kg solvent 1 ∞ = ∞ 1985 B 0.0267 kg solvent 1 mol solute 4.0_C The formula and the molecular weight of an unknown =154 g/mol hydrocarbon compound are to be determined by elemental analysis and the freezing-point depression (e) C6H5 = 77 method. # empirical units/mol = 154/77 = 2 molecular formula = (C6H5)2 = C12H10 Solids, Liquids, & Solutions 1986 D Give a scientific explanation for each of the following observations. Use equations or diagrams if they seem relevant. (a) Graphite is used to make electrodes, while diamond, another allotrope of carbon, is a very poor conductor of electricity. (b) Putting rock salt on an icy driveway melts the ice even when the air temperature is -10°C. (c) [question and answer in the GASES section] (d) Carbon dioxide, rather than a stream of water, should be used to extinguish an oil fire. Answer: (a) Distinction or correctly implied distinction between the structures of graphite and diamond. Freedom of movement of electrons in graphite resulting from the structure. (b) The rock salt forms a concentrated solution with very little water from the ice. The solution now has a freezing point lower than the temperature of the ice, therefore, the ice melts. (c) [question and answer in the GASES section] (d) Carbon dioxide is more dense than air and so pushes the air away from the fire. Water is more dense than the oil and so ends up below the oil, leaving the oil still in contact with the air; or the hot burning oil quickly vaporizes the water creating steam that spatters the oil into the air. 1987 D In 1884 the Swedish chemist Svante Arrhenius proposed that salts dissociate into two or more separate, independent, ionic fragments when they dissolve in water. (a) Give one piece of experimental evidence that more than 1 mole of particles is formed when 1 mole of a salt dissolves in water. (b) Give one piece of experimental evidence that the particles formed when a salt dissolves in water are charged. (c) Explain why the heat of neutralization is always the same when 1 mole of any monoprotic strong page 6 acid reacts with enough strong base to form a neutral solution. (d) Explain why hydrogen chloride, HCl, dissociated when it dissolves in water but not when it dissolves in benzene. Answer: (a) The freezing point depression (or any colligative effect) that occurs when a mole of a salt is dissolved is greater than when a mole of a nondissociated substance is dissolved. (The greater the number of solute particles the greater the colligative effect.) (b) The solution of a salt conducts electricity. (c) Every neutralization between a strong acid and a strong base involves the same reaction: H+(aq) + OH-(aq) → H2O since both the strong acid and the strong base are completely dissociated. Spectator ions have no appreciable effect. (d) Because of the polar nature of water, it is capable of solvating the ions that result from the dissociation, whereas the nonpolar benzene interacts very weakly with these ions. OR Because of the greater dielectric constant of water, it is better able to separate the ions. 1988 D The normal boiling and freezing points of argon are 87.3 K and 84.0 K, respectively. The triple point is at 82.7 K and 0.68 atmosphere. (a) Use the data above to draw a phase diagram for argon. Label the axes and label the regions in which the solid, liquid and gas phases are stable. On the phase diagram, show the position of the normal boiling point. (b) Describe any changes that can be observed in a sample of solid argon when the temperature is increased from 40 K to 160 K at a constant pressure of 0.50 atmosphere. (c) Describe any changes that can be observed in a sample of liquid argon when the pressure is reduced from 10 atmospheres to 1 atmosphere at a constant temperature of 100 K, which is well below the critical temperature. Solids, Liquids, & Solutions page 7 (d) Does the liquid phase of argon have a density - A solution of concentrated HNO3 greater than, equal to, or less than the density of (a) Which metal can be easily identified because it is the solid phase? Explain your answer, using inmuch softer than the other two? Describe a chemiformation given in the introduction to this quescal test that distinguishes this metal from the other tion. two, using only one of the reagents above. Write a Answer: balanced chemical equation for the reaction that occurs. (a) 2 (b) One of the other two metals reacts readily with the HCl solution. Identify the metal and write the bal1.8 anced chemical equation for the reaction that oc1.6 curs when this metal is added to the HCl solution. Use the table of standard reduction potentials (at1.4 Solid Liquid tached) to account for the fact that this metal re1.2 acts with HCl while the other does not. 1 re (atm.) Ö Önormal (c) The one remaining metal reacts with the concentrated HNO3 solution. Write a balanced chemical B.P. 0.8 equation for the reaction that occurs. Ö 0.6 (d) The solution obtained in (c) is diluted and a few Gas 0.4 drops of 1 M HCl is added. Describe what would be observed. Write a balanced chemical equation 0.2 for the reaction that occurs. 0 Answer: 81 82 83 84 85 86 87 88 89 (a) Sodium is softest of the three. Temperature (K) Na added to water → gas and base (b) The argon sublimes. Na + H2O → H2 + NaOH (c) The argon vaporizes. (b) Mg reacts with HCl. Mg + 2 H+ → Mg2+ + H2 (d) The liquid phase is less dense than the solid phase. Reduction potentials, E°: Mg = -2.37v; Ag = + Since the freezing point of argon is higher than the 0.80v. Mg, not Ag, reacts with HCl triple point temperature, the solid-liquid equilibri+ + um line slopes to the right with increasing pres- (c) Ag + 4 H + NO3 → 3 Ag + NO + 2 H2O OR Ag + 2 H+ + NO3- → Ag+ + NO2 + H2O sure. Thus, if a sample of liquid argon is compressed (pressure increased) at constant (d) A white precipitate forms: Ag+ + Cl- → AgCl(s) temperature, the liquid becomes a solid. Because increasing pressure favors the denser phase, solid 1990 A argon must be the denser phase. The solubility of iron(II) hydroxide, Fe(OH)2, is 1.43×10-3 gram per litre at 25°C. 1989 D (a) Write a balanced equation for the solubility equiConsider three unlabeled bottles, each contain small librium. pieces of one of the following metals. (b) Write the expression for the solubility product - Magnesium constant, Ksp, and calculate its value. - Sodium (c) Calculate the pH of a saturated solution of - Silver Fe(OH)2 at 25°C. The following reagents are used for identifying the metals. (d) A 50.0 millilitre sample of 3.00×10-3 molar FeSO4 - Pure water solution is added to 50.0 millilitres of 4.00×10-6 - A solution of 1.0 molar HCl molar NaOH solution. Does a precipitate of Solids, Liquids, & Solutions page 8 Fe(OH)2 form? Explain and show calculations to (b) support your answer. Answer: (c) (a) Fe(OH)2 → Fe2+ + 2 OH- Using the data gathered from the freezing-point depression method, calculate the molar mass of the unknown substance. Calculate the mole fraction of benzene in the solution described above. 1.43×10−3 g 1 mol −5 mol =1.59 ×10 × L Fe(OH)2 (d) The vapor pressure of pure benzene at 35°C is L 89.9 g (b) 150. millimeters of Hg. Calculate the vapor pressure of benzene over the solution described above = 1.59×10-5 M = [Fe2+] at 35°C. = 3.18×10-5 M = [OH-] Answer: Ksp = [Fe2+][OH-]2 = (1.59×10-5)(3.18×10-5)2 (a) moles / 100 g C H Cl 49.02/12.01 2.743/1.008 48.23/35.45 = 1.61×10-14 = 4.081 = 2.722 = 1.360 1.0 ∞10−14 1.0 ∞10−14 −10 M [H+ ] = = = 3 . 14 ∞ 1 0 mol ratio 3 2 1 [ OH − ] 3.18 ∞10− 5 empirical formula: C H Cl 3 2 (c) (b) ∆Tf = (Kf)(m) pH = -log[H+] = 9.50 OR pOH = -log[OH-] = -log(3.18×10-8) = 4.50 pH = 14 - pOH = 9.50 (d) 50.0 mL of 3.00×10-3 M Fe2+ diluted of 100.0 mL = 1.50×10-3 M Fe2+ 50.0 mL of 4.00×10-6 M OH- diluted of 100.0 mL = 2.00×10-6 M OHQ = [Fe2+][OH-]2 = (1.50×10-3)(2.00×10-6)2 = 6.00×10-15 Precipitate will NOT form since Q < Ksp 4.38°= ( 5.12° molal) × 3.150g molmass 0.02500kg = 147 g mol (c) mol fraction = mol benzene / total mol 25.00 11 = 0.938 = 3.150 78.25. 00 + 147 78.11 (d) vapor pressure = mol fraction × Po = (0.938)(150 mm) = 141 mm Hg 1994 D (Required) Discuss the following phenomena in terms of the chemical and physical properties of the substances involved and general principles of chemical and physical change. 1993 A Elemental analysis of an unknown pure substance indicated that the percent composition by mass is as follows. Element Percent by Mass Carbon 49.02% Hydrogen 2.743% Chlorine 48.23% A solution that is prepared by dissolving 3.150 grams (a) As the system shown above approaches equilibrium, what change occurs to the volume of water in of the substance in 25.00 grams of benzene, C6H6, has beaker A? What happens to the concentration of a freezing point of 1.12°C. (The normal freezing point the sugar solution in beaker B? Explain why these of benzene is 5.50°C and the molal freezing-point dechanges occur. pression constant, Kf, for benzene is 5.12°C/molal.) (a) Determine the empirical formula of the unknown substance. Solids, Liquids, & Solutions page 9 (ii) The TTE will be the bottom layer because its density is greater than the water. (iii) The non-polar iodine will dissolve better in the non-polar TTE and form a pinkish-purple tint. (b) A bell jar connected to a vacuum pump is shown above. As the air pressure under the bell jar decreases, what behavior of water in the beaker will be observed? Explain why this occurs. (c) [see Redox section] 1994 D For each of the following, use appropriate chemical principles to explain the observation. (a) Sodium chloride may be spread on an icy sidewalk in order to melt the ice; equimolar amounts of calcium chloride are even more effective. (b) At room temperature, NH3 is a gas and H2O is a liquid, even though NH3 has a molar mass of 17 grams and H2O has a molar mass of 18 grams. (c) C (graphite) is used as a lubricant, whereas C (diamond) is used as an abrasive. (d) Pouring vinegar onto the white residue inside a kettle used for boiling water results in a fizzing/bubbling phenomenon. Answer: (a) NaCl(s) → Na+(aq) + Cl-(aq) (d) A water solution of I2 is shaken with an equal volume of a nonpolar solvent such as TTE (triCaCl2(s) → Ca+(aq) + 2 Cl-(aq) chlorotrifluoroethane). Describe the appearance of The freezing point of an aqueous solution is lower this system after shaking. (A diagram may be than the freezing point of water. A higher molality helpful.) Account for this observation. of a solution lowers the freezing point more and Answer: an equimolar amount of the two solids gives a (a) volume of water decreases while the concentration larger molal solution from the calcium chloride as of sugar solution decreases. illustrated by the above equations. Pure water has a higher vapor pressure than does the 10% sugar solution and when equilibrium is (b) Water is more polar than ammonia creating stronger attractions (IMF) between molecules and reached the water will evaporate and the solution making it a liquid. will increase in volume. (b) The water will boil when the pressure in the bell (c) Diamond, the hardest naturally occurring substance, has each carbon atom surrounded by a tetjar reaches the vapor pressure of the water. Boilrahedral arrangement of other carbon atoms (see ing occurs when the vapor pressure of the liquid is drawing). The network solid structure is stabilized in equilibrium with the pressure above the liquid. by covalent bonds, formed by the overlap of sp3 (d) (i) Water and TTE will form separate layers behybridized carbon atomic orbitals. A diamond has cause the polar water is not miscible with the nonuniform very strong bonds in all directions in the polar TTE. crystal. water iodine in TTE Solids, Liquids, & Solutions Graphite has a different kind of bonding based on layers of carbon atoms arranged in fused sixmember rings (see drawing). Each carbon atom in a layer is surrounded by three other carbons in a trigonal planar arrangement with 120° bond angles. The slipperiness is caused by noting that graphite has very strong bonds within layers but little bonding between the layers which allows the layers to slide past one another quite readily. (d) Calcium and magnesium carbonates are left behind from the evaporation of hard water. These carbonates decompose and release carbon dioxide gas when reacted with the acetic acid in the vinegar. CaCO3(s) + MgCO3(s) + CH3COOH(aq) → Ca2+(aq) + Mg2+(aq) + H2O(l) + CO2(g) + CH3COO-(aq) 1995 D page 10 The phase diagram for a pure substance is shown above. Use this diagram and your knowledge about changes of phase to answer the following questions. (a) What does point V represent? What characteristics are specific to the system only at point V?. (b) What does each point on the curve between V and W represent? (c) Describe the changes that the system undergoes as the temperature slowly increases from X to Y to Z at 1.0 atmosphere. (d) In a solid-liquid mixture of this substance, will the solid float or sink? Explain. Answer: (a) Triple point. All three states of the substance coexist (equilibrium); the solid and the liquid have identical vapor pressures. (b) Curve VW represents the equilibrium between the liquid and its vapor. Along this line the liquid will be boiling. The points represent the vapor pressure of the liquid as a function of temperature. (c) At point X the substance is a solid, as its temperature increases (at constant pressure), at point Y the solid is in equilibrium with its vapor and will sublime. From point Y to Z it exist only as a vapor. (d) Sink. A positive slope of the solid-liquid line indicates that the solid is denser than its liquid and, therefore, will sink. 1995 D (repeated in the thermo section) Lead iodide is a dense, golden yellow, slightly soluble solid. At 25°C, lead iodide dissolves in water forming a system represented by the following equation. Solids, Liquids, & Solutions PbI2(s) ↔ Pb + 2 I ∆H = +46.5 kilojoules (a) How does the entropy of the system PbI2(s) + H2O(l) change as PbI2(s) dissolves in water at 25°C? Explain (b) If the temperature of the system were lowered from 25°C to 15°C, what would be the effect on the value of Ksp? Explain. (c) If additional solid PbI2 were added to the system at equilibrium, what would be the effect on the concentration of I- in the solution? Explain. 2+ - (d) At equilibrium, ∆G = 0. What is the initial effect on the value of ∆G of adding a small amount of Pb(NO3)2 to the system at equilibrium? Explain. Answer: (a) Entropy increases. At the same temperature, liquids and solids have a much lower entropy than do aqueous ions. Ions in solutions have much greater “degrees of freedom and randomness”. (b) Ksp value decreases. Ksp = [Pb2+][I-]2. As the temperature is decreased, the rate of the forward (endothermic) reaction decreases resulting in a net decrease in ion concentration which produces a smaller Ksp value. (c) No effect. The addition of more solid PbI2 does not change the concentration of the PbI2 which is a constant (at constant temperature), therefore, neither the rate of the forward nor reverse reaction is affected and the concentration of iodide ions remains the same. (d) ∆G increases. Increasing the concentration of Pb2+ ions causes a spontaneous increase in the reverse reaction rate (a “shift left” according to LeChatelier’s Principle). A reverse reaction is spontaneous when the ∆G>0. 1996 B Concentrated sulfuric acid (18.4-molar H2SO4) has a density of 1.84 grams per milliliter. After dilution with water to 5.20-molar, the solution has a density of 1.38 grams per milliliter and can be used as an electrolyte in lead storage batteries for automobiles. (a) Calculate the volume of concentrated acid required to prepare 1.00 liter of 5.20-molar H2SO4. (b) Determine the mass percent of H2SO4 in the original concentrated solution. page 11 (c) Calculate the volume of 5.20-molar H2SO4 that can be completely neutralized with 10.5 grams of sodium bicarbonate, NaHCO3. (d) What is the molality of the 5.20-molar H2SO4? Answer: (a) M1V1 = M2V2 ; (5.20 M)(1.00 L) = (18.4 M)(V2) V2 = 0.283 L (b) 18.4 mol/1 L × 98.1 g H2SO4/1 mol = 1805 g H2SO4 1 L × 1.84 g/1 mL × 1000 mL/1 L = 1840 g solution mass % = 1805 g/1840 g × 100 = 98.1% (c) 2 NaHCO3 + H2SO4 → 2 H2O + Na2SO4 + 2 CO2 10.5 g NaHCO3 × 1 mol NaHCO3/84.0 g NaHCO3 × 1 mol H2SO4/2 mol NaHCO3 × 1 L/5.20 mol H2SO4 = 0.0120 L (d) molality = mol solute/1 kg solvent 5.20 mol/1 L × 98.1 g H2SO4/1 mol = 510 g H2SO4 1 L × 1.38 g/1 mL × 1000 mL/1 L = 1380 g solution 1380 g solution - 510 g H2SO4 = 870 g solvent 5.20 mol H2SO4/0.870 kg solvent = 5.98 m 1998 A (Required) Solve the following problem related to the solubility equilibria of some metal hydroxides in aqueous solution. (a) The solubility of Cu(OH)2(s) is 1.72×10-6 gram per 100. milliliters of solution at 25°C. (i) Write the balanced chemical equation for the dissociation of Cu(OH)2(s) in aqueous solution. (ii) Calculate the solubility (in moles per liter) of Cu(OH)2 at 25°C. (iii) Calculate the value of the solubility-product constant, Ksp, for Cu(OH)2 at 25°C. (b) The value of the solubility-product constant, Ksp, for Zn(OH)2 is 7.7×10-17 at 25°C. (i) Calculate the solubility (in moles per liter) of Zn(OH)2 at 25°C in a solution with a pH of 9.35. Solids, Liquids, & Solutions page 12 (ii) At 25°C, 50.0 milliliters of 0.100-molar (c) When 1.570 grams of the compound is vaporized at 300°C and 1.00 atmosphere, the gas occupies a Zn(NO3)2 is mixed with 50.0 milliliters of volume of 577 milliliters. What is the molar mass 0.300-molar NaOH. Calculate the molar conof the compound based on this result? centration of Zn2+(aq) in the resulting solution once equilibrium has been established. As- (d) Briefly describe what occurs in solution that acsume that volumes are additive. counts for the difference between the results obAnswer tained in parts (b) and (c). 2+ (a)i Cu(OH)2 → Cu + 2 OH Answer − 6 24.96 65.60 9.44 1.72 ∞10 g 1 mol mol = 5.462 = 1.56 = 9.37 ∞ =1.76 ∞10 −7 16.00 12.011 1.0079 97.5 g 0.100 L L (a) ii 5.462 9.37 1.56 = 3.50 = 6.00 = 1.00 iii Ksp = [Cu2+][OH-]2 1.56 1.56 1.56 = [1.76×10-7][3.53×10-8]2 = 2.20×10-20 (C3.5H6O1)2 = C7H12O2 (b)i Zn(OH)2 → Zn2+ + 2 OHg solvent mol solute molar mass Ksp = [Zn2+][OH-]2 kg solvent kg solvent (b) ∆T = kâm =kâ = kâ pH 9.35 = pOH 4.65; [OH-] = 10-pOH (g solute)( k∂ ) [OH-] = 10-4.65 = 2.24×10-5 M FP [Zn2+] = solubility of Zn(OH)2 in mol/L K sp 7.7 ∞10 −1 7 [Zn 2 + ] = = =1.5 ∞10 −7 M [OH – ]2 ( 2.24 ∞10 −5 )2 ii 50 mL [Zn2+] = 0.100M × 100 mL = 0.0500 M 50 mL [OH-] = 0.300M × 100 mL = 0.150 M X = conc. loss to get to equilibrium Ksp = 7.7×10-17 = (0.0500 - X)(0.150 - 2X)2 [Zn2+] = 0.0500 - X = 3.1×10-14M init init (kg solvent)( ∆TFP ) molar mass = kg _K ) (1.570 g)(40.0 mol = (0.01608 kg)(15.2_C) = 257 g/mol C7H12O2 = 128 g/mol ; 257/128 = 2 ∴, (C7H12O2)2 = C14H24O4 grams (c) PV = molar mass RT grams ∞ R ∞ T P ∞V molar mass = ⊇ L • atm _ ∞ 573K (1.570 g) ∞ℑ0.08205 ⊄ mol • K ↓ 1998 B 1.00 atm ∞ 0.577 L An unknown compound contains only the three ele= 128 g/mol ments C, H, and O. A pure sample of the compound is (d) since the apparent molar mass from freezing point analyzed and found to be 65.60 percent C and 9.44 change is twice that determined by the vapor, the percent H by mass. molecules of the compound must associate in (a) Determine the empirical formula of the comcamphor to form dimers, OR the compound must pound. dissociate in the gas phase. (b) A solution of 1.570 grams of the compound in 16.08 grams of camphor is observed to freeze at a 1998 D temperature 15.2 Celsius degrees below the nor- Answer each of the following using appropriate chemmal freezing point of pure camphor. Determine ical principles. the molar mass and apparent molecular formula of (a) Why does it take longer to cook an egg in boiling the compound. (The molal freezing-point depreswater at high altitude than it does at sea level? sion constant, Kâ, for camphor is 40.0 kg•K•mol-1.) Answer Solids, Liquids, & Solutions (a) At a high altitude the pressure is lower and the boiling point of water is lower. Since the temperature of the boiling water is less, it takes longer to cook the egg. page 13 20 15 2000 D Required The molar mass of an unknown solid, which is nonvol10 atile and a nonelectrolyte, is to be determined by the freezing-point depression method. The pure solvent used in the experiment freezes at 10°C and has a 5 known molal freezing-point depression constant, Kƒ. Assume that the following materials are also available. • test tubes • stirrer • pipet 0 0 Time • stopwatch • graph paper (b) Information from these graphs may be used to de• thermometer • balance • beaker termine the molar mass of the unknown solid. • ice • hot-water bath (i) Describe the measurements that must be (a) Using the two sets of axes provided below, sketch made to determine the molar mass of the uncooling curves for (i) the pure solvent and for (ii) known solid by this method. the solution as each is cooled from 20°C to 0.0°C (ii) Show the setup(s) for the calculation(s) that Pure Solvent must be performed to determine the molar mass of the unknown solid from the experi20 mental data. (iii) Explain how the difference(s) between the two graphs in part (a) can be used to obtain 15 information needed to calculate the molar mass of the unknown solid. (c) Suppose that during the experiment a significant 10 but unknown amount of solvent evaporates from the test tube. What effect would this have on (he calculated value of the molar mass of the solid 5 (i.e., too large, too small, or no effect)? Justify your answer. 0 (d) Show the setup for the calculation of the percent0 Time age error in a student’s result if the student obtains Solution a value of 126 g mol-1 for the molar mass of the solid when the actual value is 120. g mol-1. Answer: (a) (i) pure solvent Solids, Liquids, & Solutions page 14 20 20 15 15 10 10 5 5 0 0 Time (ii) solution 20 15 10 0 0 T Time (c) Too small. If solvent evaporates then its mass decreases and the recorded denominator in the equation in (b)(i) is larger than the expt. value and the resulting answer decreases. (d) % error = Ошибка! × 100% 2001 A Required Answer the following questions relating to the solubility of the chlorides of silver and lead. (a) At 10°C, 8.9 × 10-5 g of AgCl(s) will dissolve in 100. mL of water. (i) Write the equation for the dissociation of AgCl(s) in water. 0 0 Time (ii) Calculate the solubility, in mol L–1, of AgCl(s) in water at 10°C. (b) (i) mass of pure solvent; freezing point of pure solvent; mass of unknown solid solute added to (iii) Calculate the value of the solubility-product pure solvent; freezing point of resulting solution constant, Ksp for AgCl(s) at 10°C. (ii) determine the change in freezing point, ∆T (b) At 25°C, the value of Ksp for PbCl2(s) is 1.6 × 10-5 5 ∆T = Kƒ • m, where m = Ошибка! and moles solute = Ошибка! therefore, molar mass = Ошибка! (iii) the change in temperature is the difference in the flat portions of the graph. and the value of Ksp for AgCl(s) is 1.8 × 10-10. (i) If 60.0 mL of 0.0400 M NaCl(aq) is added to 60.0 mL of 0.0300 M Pb(NO3)2(aq), will a precipitate form? Assume that volumes are additive. Show calculations to support your answer. (ii) Calculate the equilibrium value of [Pb2+(aq)] in 1.00 L of saturated PbCl2 solution to which 0.250 mole of NaCl(s) has been added. Assume that no volume change occurs. (iii) If 0.100 M NaCl(aq) is added slowly to a beaker containing both 0.120 M AgNO3(aq) and 0.150 M Pb(NO3)2(aq) at 25°C, which will Solids, Liquids, & Solutions page 15 precipitate first, AgCl(s) or PbCl2(s)? Show (d) Which solution could be used to oxidize the Cl– calculations to support your answer. (aq) ion? Identify the product of the oxidation. Answer: (e) Which solution would be the least effective con+ – ductor of electricity? Explain. (a) (i) AgCl → Ag + Cl Answer: (ii) Ошибка! × Ошибка! = 6.2 × 10-6 M (a) solution 1, Pb(NO3)2. This compound will dissoci(iii) Ksp = [Ag+]•[Cl–] = (6.2 × 10-6)2 = 3.9 × 10-11 ate into three ions with the highest total particle (b) (i) No molality. The greater the molality, the higher the boiling point. Solutions 2, 3, and 5 will produce Ошибка! × Ошибка! = 0.0200 M [Cl–] two ions while solution 4 is molecular. Ошибка! × Ошибка! = 0.0150 M [Pb+] (b) solution 5, KC2H3O2. The salt of a weak acid (in trial Ksp = [Pb2+]•[Cl–]2 = (0.0150)(0.0200)2 = this case, acetic acid) produces a basic solution, 6.0×10-6; trial Ksp < Ksp, no ppt. and, a higher pH. 2+ – 2 2+ – 2 (ii) Ksp = [Pb ]•[Cl ] , [Pb ] = Ksp/[Cl ] (c) solution 1, Pb(NO3)2, and solution 2, NaCl. PbCl2 [Pb2+] = Ошибка!= 2.6×10-4 M (d) solution 3, KMnO4 , ClO3– (iii) AgCl (e) solution 4, C2H5OH. Ethyl alcohol is covalently –] = Ошибка! = Ошибка! = 1.5×10-9 M bonded and does not form ions in water. There[Cl fore, the solution is not a better conductor of elecPbCl2 tricity than water, which is also covalently bondK sp 1.6 × 10−5 ed. 2+ = 0.150 [Cl–] = [Pb ] = 0.0103 M 2003 D Required AgCl will form a ppt when the [Cl–] exceeds 1.5×10-9 M. The formation of PbCl2(s) requires a For each of the following, use appropriate chemical principles to explain the observations. Include chemi[Cl–] that is over 6 million times more concentratcal equations as appropriate. ed. (a) In areas affected by acid rain, statues and structures made of limestone (calcium carbonate) often 2001 D Required show signs of considerable deterioration. (b) When table salt (NaCl) and sugar (C12H22O11) are dissolved in water, it is observed that (i) both solution have higher boiling points than pure water, and (ii) the boiling point of 0.10 M NaCl(aq) is higher than that of 0.10 M C12H22O11(aq). (c) Methane gas does not behave as an ideal gas at Answer the questions below that relate to the five low temperatures and high pressures. aqueous solutions at 25°C shown above. (d) Water droplets form on the outside of a beaker (a) Which solution has the highest boiling point? Excontaining an ice bath. plain. Answer: (b) Which solution has the highest pH? Explain. (a) limestone reacts with acid to produce a soluble (c) Identify a pair of the solutions that would produce substance, a gas, and water which wash away a precipitate when mixed together. Write the forCaCO3(s) + H+(aq) → Ca2+(aq) + CO2(g) + H2O(l) mula of the precipitate. (b) (i) a solution made from a non-volatile solute has a higher boiling point than the pure solvent be- Solids, Liquids, & Solutions page 16 cause the solution has a lower vapor pressure than the water (Raoult’s Law) . the temperature of the solution has be higher to produce enough vapor pressure to equal the atmospheric pressure (i.e., boiling) (ii) the amount of boiling point elevation depends on the number of non-volatile particles in solution. since the salt dissociates into 2 particles for every NaCl that dissolves, it will increase the boiling point more that an equal concentration of sugar (a molecular cpd) that does not dissociate or ionize. (c) at low temperatures and high pressures, the methane molecules are slow and closer together. under these conditions, van der Waal forces become measurable and significant and creates a deviation from ideal behavior. at high pressure the volume of a real molecule is also significant. (d) a water vapor molecules collide with the cool beaker, the molecules lose kinetic energy, slow down, attract others, and condense into a liquid tration of is M. The equilibrium constant expression for the dissolving of Ag3PO4 is shown below: Ksp = [Ag+]3[PO43–] 2004 A Required Answer the questions relating to the solubility’s of two silver compounds, Ag2CrO4 and Ag3PO4 (8.7×10–5 mol L–1)(331.74 g mol–1) = = 0.0287 g L–1 in 100 mL, 100 mL × Ошибка!= 0.0029 g decrease, since the silver nitrate dissolves and produces an increase in [Ag+], this causes a LeChatelier shift to the left and results in a decrease in the products, including chromate ions Ag3PO4(s) √ 3 Ag+(aq) + PO43–(aq) [PO43–] = Ошибка! Ksp = [Ag+]3[PO43–] = (5.3×10–5)3( Ошибка!) = 2.6×10–18 [Ag+] = 5.3×10–5 M. In a saturated solution the molar concentration is independent of the total volume. As the volume evaporates and becomes half, half of the Ag+ will precipitate out. Ag+ 5.3×10–5 (e) Write the balanced equation for the dissolving of Ag3PO4 in water. (f) Calculate the value of Ksp for Ag3PO4 in water. (g) A 1.00 L sample of saturated Ag3PO4 solution is allowed to evaporate at 25˚C to a final volume of 500. mL. What is [Ag+] in the solution? Justify your answer. Answer: (a) Ksp = [Ag+]2[Cr2O72–] (b) [Cr2O72–] = X; [Ag+] = 2X (2X)2(X) = Ksp = 2.6×10–12 X = 8.7×10–5 mol L–1 2X = [Ag+] = 1.7×10–4 mol L–1 (c) amt. Ag2CrO4 = X = 8.7×10–5 mol L–1 Silver chromate dissociates in water according to the (d) following equation: Ag2CrO4(s) ↔ 2 Ag+(aq) + CrO42–(aq) Ksp = 2.6×10–12 at 25˚C (a) Write the equilibrium constant expression for the (e) dissolving of Ag2CrO4. (f) (b) Calculate the concentration in mol L–1, of Ag+ in a saturated solution of Ag2CrO4 at 25˚C. (c) Calculate the maximum mass in grams of (g) Ag2CrO4 that can dissolve in 100. mL of water at 25˚C. (d) A 0.100 mol sample of solid AgNO3 is added to 1.00 L saturated solution of Ag2CrO4. Assuming no volume change, does [CrO42–] increase, decrease, or remain the same? Justify your answer. In a saturated solution of Ag3PO4 at 25˚C, the concen-