Chapter 7 Complete Notes-merged PDF

Chapter 7: Gases Chapter Outline Section 7.1 Gas Pressure Section 7.2 Boyle’s Law Section 7.3 Charles’s Law Section 7.4 The Combined Gas Law Section 7.5 Avogadro’s Law Section 7.6 The Ideal Gas Law Section 7.7 Dalton’s Law of Partial Pressures Section 7.8 Molar Mass and Density in Gas Law Calcula...

Chapter 7: Gases Chapter Outline Section 7.1 Gas Pressure Section 7.2 Boyle’s Law Section 7.3 Charles’s Law Section 7.4 The Combined Gas Law Section 7.5 Avogadro’s Law Section 7.6 The Ideal Gas Law Section 7.7 Dalton’s Law of Partial Pressures Section 7.8 Molar Mass and Density in Gas Law Calculations Section 7.9 Gases in Chemical Reactions Section 7.10 Kinetic Molecular Theory of Gases Section 7.11 Movement of Gas Particles Section 7.12 Behavior of Real Gases 7.1 Gas Pressure When a gas particle strike a surface, it exerts a force against the surface. Gas pressure is the sum of all the forces exerted by gas particles impacting the surface, divided by the area of the surface. Units of pressure: ❑ atmpospheres atm ❑ millimeters of mercury, mmHg ❑ Bar ❑ Pascal, Pa Conversions: ❑ Torr 1 torr = 1 mmHg 1 atm = 760 torr = 760 mmHg 1 atm = 1.01325 × 105 Pa = 101.325 kPa 1 bar = 100 kPa = 1000 mbar 7.2 Boyle’s Law Robert Boyle (1627–1691) discovered that the volume of a given sample of a gas at a constant temperature is inversely proportional to its pressure. In other words, as volume decreases, the gas pressure increases and vice versa. 1 Boyle’s law : V  P 7.2 Boyle’s Law – Calculations If you change that sample of gas from one pressure, P1, to another, P2, at constant temperature, the volume will change from V1 to V2 : 1 1 = PV PV 2 2 Example - A 3.50 L sample of hydrogen gas has a pressure of 0.750 atm. What is its volume if its pressure is increased to 1.50 atm at constant temperature? Rearrange Boyle’s law, P1V1 = P2V2, to solve forV2. V2 = ( 0.750 atm )(3.50 L ) 1.50 atm PV V2 = 1 1 P2 V 2 = 1.75 L 7.3 Charles’ Law In 1787, J. A. C. Charles (1746–1823) discovered that when the pressure is constant, the volume of a sample of gas varies directly with the absolute or Kelvin temperature. In other words, as volume increases, the temperature also increases and vice versa Charles’s law: V ∝ T 7.3 Charles’ Law – Calculations If you change that sample of gas from one temperature, T1, to another, T2, at constant pressure, the volume will change from V1 to V2 : V1 V2 = T1 T2 Example - A 678 mL sample of helium gas, initially at 0°C, is heated at constant pressure. If the final volume of the gas is 0.896 L, what is its final temperature in °C? T1 = 0℃ + 273.15 = 273 K T2 = ( 273 K )( 0.896 L ) = 361 K 0.678 L T1V 2 T2 = Temperatures must be in units of V1 Kelvin when applying gas laws! 361 K − 273.15 = 88C 7.4 The Combined Gas Law Boyle’s law and Charles’s law can be merged into one law, called the combined gas law. PV PV 1 1 = 2 2 T1 T2 Example - Calculate the volume of a sample of hydrogen gas that originally occupies 908 mL at 717 torr and 20°C after its temperature and pressure are changed to 72°C and 1.07 atm, respectively. Initial Conditions Final Conditions T1 = 20℃ + 273.15 = 293 K T2 = 72℃ + 273.15 = 345 K V1 = 908 mL V2 = ?  1 atm  P2 = 1.07 atm P1 = 717 torr   = 0.943 atm  760 torr  Solution PV PV T2 1 1 = 2 2 T2 T1 T2 Rearrange to solve for V2 Multiply both sides by PV 1 1T2 T2 and divide by P2 V2 = P2 P2 P2T1 7.5 Avogadro’s Law Avogadro’s law states that when the pressure and temperature are held constant, the volume of a sample of gas and the number of moles in the sample (n) are directly proportional. In other words, as volume increases, the number of moles increases and vice versa n V V1 V2 = n1 n2 7.5 Avogadro’s Law Example - A sample containing 22.0 g of gaseous carbon dioxide occupies a volume of 15.3 L. What will the new volume be if an additional 15.0 g CO2(g) are added, with no change in temperature or pressure? Initial Conditions Final Conditions 1 𝑚𝑜𝑙 𝐶𝑂2 1 𝑚𝑜𝑙 𝐶𝑂2 n1 = 22.0 𝑔 𝐶𝑂2 = 0.500 𝑚𝑜𝑙 𝐶𝑂2 n2 = 37.0 𝑔 𝐶𝑂2 44.0085 𝑔 𝐶𝑂2 44.0085 𝑔 𝐶𝑂2 = 0.841 𝑚𝑜𝑙 𝐶𝑂2 V1 = 15.3 L V2 = ? Solution V1 V2 Rearrange to solve for V2 V 1 n2 = n2 Multiply both sides by n2 V2 = n1 n2 n1 V2 = ( 15.3 L )( 0.841 mol ) = 25.7 L 0.500 mol 7.6 The Ideal Gas Law Boyle’s, Charles’s, and Avogadro’s laws can be combined into one gas law equation, the ideal gas law. PV = nRT The new constant, R, is known as the ideal gas constant. L  atm R = 0.08206 mol  K 7.6 The Ideal Gas Law Example - Calculate the volume of 42.6 g of oxygen gas at 35°C and 792 torr. Step 1 – Convert the data to units that are compatible of R.  1 mol O2  n = 42.6 g O2   = 1.33 mol O2  32.00 g O2  T = 35C + 273.15 = 308 K  1 atm  P = 792 torr   = 1.04 atm  760 torr  7.6 The Ideal Gas Law Step 2 - Rearrange the ideal gas law, PV = nRT, to solve for V. nRT V = P  L  atm  (1.33 mol )  0.08206 mol  K  (308 K ) V =   = 32.3 L 1.04 atm 7.7 Dalton’s Law of Partial Pressures Dalton’s law of partial pressures states that the total pressure of the mixture is equal to the sum of the individual pressures of the components of a gaseous mixture. Ptotal = P1 + P2 + + Pn The individual pressure, Pi, of each component of a mixture is referred to as a partial pressure. 7.7 Dalton’s Law of Partial Pressures The ideal gas law applies to the total pressure of a mixture of gases, Ptotal, and to each of the components of the mixture separately, Pi. ni RT ntotalRT ni RT Pi V ni Ptotal = and Pi = = = = Xi V V Ptotal ntotalRT ntotal V Xi is the mole fraction, which is the number of moles of a component divided by the total moles in the mixture. 7.7 Dalton’s Law of Partial Pressures Example 1 - A 10.5 L sample of gas at 292 K contains O2 at 0.622 atm and N2 at 0.517 atm. a. Calculate the total number of moles of gas in the sample. Ptotal = PO + PN PtotalV 2 2 ntotal = RT Ptotal = 0.622 atm + 0.517 atm = 1.139 atm ntotal = ( 1.139 atm )(10.5 L ) = 0.499 mol  L  atm   0.08206  ( 292 K )  mol  K 7.7 Dalton’s Law of Partial Pressures Example 1 - A 10.5 L sample of gas at 292 K contains O2 at 0.622 atm and N2 at 0.517 atm. b. Calculate the number of moles of O2 present. PO V nO = ( 0.622 atm )(10.5 L ) = 0.273 mol nO = 2  L  atm   ( 292 K ) 2 2 RT  0.08206  mol  K 7.7 Dalton’s Law of Partial Pressures The vapor pressure is the partial pressure of a vapor (gas) above it liquid The vapor pressure corresponds to the pressure of the vapor the results from the evaporation of the liquid Ptotal = PO2 + PH2O 7.7 Dalton’s Law of Partial Pressures Example 2 - Oxygen gas is collected over water in an apparatus such as that shown in the figure below at a barometric pressure of 759 torr at 25°C. What is the partial pressure of oxygen gas? The vapor pressure of water at 25oC is 23.756 torr. PO = Ptotal − PH O 2 2 PO = 759 torr − 23.756 torr = 735 torr 2 7.8 Molar Mass and Density in Gas Law Calculations Notes: If the identity of the gas is known, use molar mass to convert mass to moles and then determine the P, V, or T of the sample. If the identity of the gas is unknown, use the ideal gas law; the P, V, and T of the sample; and the mass to determine molar mass. The density of a gas, which is given in g/L at a specific temperature and pressure, provides the mass, P, V, and T of a gas sample. If the volume and mass or moles of a gas sample are known, its density can be calculated. 7.8 Molar Mass and Density in Gas Law Calculations Example 1 - A 93 g sample of a pure gaseous substance occupies 29.5 L at 27°C and 1.25 atm. What is the molar mass of the gas? Step 1: Calculate the number of moles of gas PV 1.25 𝑎𝑡𝑚 29.5 𝐿 n= n= = 1.50 𝑚𝑜𝑙 RT 𝐿 ⋅ 𝑎𝑡𝑚 0.08206 300 K 𝑚𝑜𝑙 ⋅ 𝐾 7.8 Molar Mass and Density in Gas Law Calculations Example 1 - A 93 g sample of a pure gaseous substance occupies 29.5 L at 27°C and 1.25 atm. What is the molar mass of the gas? Step 2: Use the mass and moles to determine the molar mass 𝑚 93𝑔 M𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 = = = 62 g/mol 𝑛 1.50 𝑚𝑜𝑙 7.8 Molar Mass and Density in Gas Law Calculations Example 2 - Calculate the pressure at which oxygen gas has a density of 1.44 g/L at 22°C. Assume a 1.00 L solution Step 1: Calculate the number of moles of gas using density and molar mass. 𝑚𝑂2 = 𝑑𝑂2 x 𝑣𝑂2 1.44𝑔 n𝑂2 = = 0.0450 mol 32.00𝑔/𝑚𝑜𝑙 g 𝑚𝑂2 = 1.44 x 1.00L = 1.44 g L 7.8 Molar Mass and Density in Gas Law Calculations Example 2 - Calculate the pressure at which oxygen gas has a density of 1.44 g/L at 22°C. Assume a 1.00 L solution Step 2: Use the ideal gas law to determine the pressure nRT P= V ( 0.0450 mol )  L  atm  P=  0.08206  ( 295 K ) = 1.09 atm 1L  mol  K  7.9 Gases in Chemical Reaction Gay-Lussac’s law of combining volumes: When more than one gas is involved in a chemical reaction, the volume ratio of the gases is equal to the mole ratio in the balanced equation and that all the gases must be at the same temperature and pressure. 1 A(g) + 1 B(g) → 3 C(g) + 1 D(g) 1 mol of A produces 3 mol of C. 1 L of A produces 3 L of C. 7.9 Gases in Chemical Reaction 7.9 Gases in Chemical Reaction How many liters of oxygen gas at 21°C and 1.13 atm can be prepared by the thermal decomposition of 0.950 g KClO3? 2 KClO3 ( s ) → 2 KCl ( s ) + 3 O2 ( g ) heat Step 1: Determine the number of moles of O2 produced from KClO3 1 𝑚𝑜𝑙 𝐾𝐶𝑙𝑂3 3 𝑚𝑜𝑙 𝑂2 0.950 𝑔 𝐾𝐶𝑙𝑂3 𝑥 𝑥 = 0.01163 𝑚𝑜𝑙 𝑂2 122.55 𝑔 𝐾𝐶𝑙𝑂3 2 𝑚𝑜𝑙 𝐾𝐶𝑙𝑂3 7.9 Gases in Chemical Reaction How many liters of oxygen gas at 21°C and 1.13 atm can be prepared by the thermal decomposition of 0.950 g KClO3? 2 KClO3 ( s ) → 2 KCl ( s ) + 3 O2 ( g ) heat Step 2: Determine the volume of O2 using the ideal gas law.  L  atm  V = nRT ( 0.01163 mol ) 0.08206  ( 294 K ) mol  K  V =  = 0.248 L O2 P 1.13 atm 7.10 Kinetic Molecular Theory of Gases The kinetic molecular theory explains the gas laws. There are five postulates: 1. Gases are composed of small molecules that are in constant, random motion. 2. The volume that is taken up by the molecules themselves is insignificant compared with the overall volume occupied by the gas. 3. Forces between the molecules are negligible, except when the molecules collide with one another. 4. Molecular collisions are perfectly elastic; that is, no energy is lost when the molecules collide. 5. The average kinetic energy of the gas molecules is directly proportional to the absolute temperature of the gas. 7.11 Movement of Gas Particles The root mean square speed, vrms, is the velocity of a particles possessing the average KE in a gas sample of known molar mass. vrms is obtained by combining the two equations for KE and substituting molar mass (ℳ) for mNA. 7.11 Movement of Gas Particles Speeds of gas particles at 25o C Vrms increases as ℳ decreases 7.11 Movement of Gas Particles – Temperature and Speed The average speeds of gas particles depend on the temperature. At all temperatures, gas particles with very low masses move faster than gas particles of higher masses. Temperature affects the range of speeds also. ❑ At low temperatures, the speeds of all gas particles are close to the average speed, with a narrow range above and below the average. ❑ As temperature increases, the range broadens as the average speed increases. 7.11 Movement of Gas Particles – Diffusion Diffusion occurs when a gas sample is introduced into a larger volume, and the gas particles spread out to occupy the entire volume, mixing with the other gases present. The rate of diffusion depends on the temperature, the mass of the particle, and the collisions with other particles. The mean free path of a particle is the average distance traveled between collisions, and is related to the pressure of the gas particles. 7.11 Movement of Gas Particles – Effusion Effusion is the escape of gas molecules through a tiny hole into a vacuum without collisions. Graham’s law of effusion, derived from the root mean square speed, states that the ratio of rates of effusion of two gases is equal to the square root of the inverse ratio of the molar masses. 7.11 Movement of Gas Particles – Effusion Example 1 - Calculate the relative effusion rates of helium and oxygen gases. Helium atoms effuse at a rate that is 2.83 times faster than the effusion rate of oxygen molecules. 7.11 Movement of Gas Particles – Effusion Example 2 - A gas known to have a formula of NOx was found to effuse at a rate that is 0.834 times the rate of effusion of oxygen. Determine (a) the molar mass of NOx and (b) the value of x. a. NOx effuses more slowly than O2, so NOx must have a higher molar mass. Rearrange Graham's law to solve for the molar mass of NOx. 7.11 Movement of Gas Particles – Effusion Example 2 - A gas known to have a formula of NOx was found to effuse at a rate that is 0.834 times the rate of effusion of oxygen. Determine (a) the molar mass of NOx and (b) the value of x. b. Use the molar mass of NOx to determine the value of x. 1 ( molar mass of N ) + x ( molar mass of O ) = 46.0 g/mol  g  14.007 g/mol + x  15.999  = 46.0 g/mol  mol  46.0 − 14.007 x= = 1.99  2 15.999 Thus, NOx is NO2. 7.12 Real Gases Ideal gas behavior is based on two assumptions: ❑The volume of gas particles is negligible. ❑There are no interactions, attractive or repulsive, between gas particles. At high pressures and low temperatures, these assumptions are not always valid. The van der Waals equation allows you to make more accurate predictions regarding the pressure of real gases. 7.12 Real Gases Real gas particles have finite volumes. Very high pressures force the gas into very small volumes. 7.12 Real Gases The van der Waals equation is a modification of the ideal gas law, introducing two factors that are specific to each gas. The first factor involves volume. Volume is increased by a small factor related to the number of gas particles, n, and a constant specific to each gas, b. nRT V real gas = + nb nRT P V − nb = P 7.12 Real Gases Pressure is decreased by a small factor related to the number of molecules; the volume; and a constant, a, specific to each gas. 2 nRT n Preal gas = −a   V V  Together, the combined P and V equations form the van der Waals equation.  n  2 P + a    (V − nb ) = nRT  V   7.12 Real Gases  atm  L2   L   atm  L2   L  Gas a  2  b   Gas A  2  b    mol   mol   mol   mol  CH4 2.273 0.0431 CO 1.452 0.0395 CO2 3.610 0.0429 H2S 4.481 0.0434 Cl2 6.260 0.0542 NO 1.351 0.0387 NH3 4.170 0.0371 N2O 3.799 0.0444 H2O 5.465 0.0305 NO2 5.29 0.0443 Xe 4.137 0.0516 SO2 6.770 0.0568 CCl4 19.75 0.1281 HF 9.431 0.0739 O2 1.363 0.0319 HCl 3.648 0.0406 N2 1.351 0.0387 HBr 4.437 0.0442 Kr 5.121 0.0106 HI 6.221 0.0530 Ar 1.336 0.0320 7.12 Real Gases Example - Calculate the pressure of 1.00 mol of argon gas at 2.30 L and 273 K using (a) the ideal gas equation and (b) the van der Waals equation a. Use the ideal gas law, PV = nRT.  0.08206 L  atm  1.00 mol   ( 273 K ) nRT  mol  K  P= = = 9.74 atm V 2.30 L 7.12 Real Gases b. Rearrange the van der Waals equation for P. Find the constants for Ar:  n  2 P + a    (V − nb ) = nRT a = 1.336 atm. L2 /mol2 b =0.0320 L/mol  V    0.08206 L  atm  2 1.00 mol   ( 273 K ) n 2 nRT  mol  K  atm  L  1.00 mol  2 P= −a   P= − 1.336 =  (V − nb ) V   L  2 2.30 L − (1.00 mol )  0.0320 mol  2.30 L    mol  P = 9.63 atm Chapter 19: Electrochemistry Chapter Outline Section 19.1 Redox Reactions Section 19.2 Balancing Redox Equations Section 19.4 Voltaic Cells Section 19.5 Cell Potential Section 19.6 Free Energy and Cell Potential Section 19.7 The Nernst Equation and Concentration Cells Section 19.8 Voltaic Cell Applications: Batteries, Fuel Cells, and Corrosion Section 19.9 Electrolytic Cells and Applications of Electrolysis You are not responsible for knowing section 19.3 Redox Titrations 19.1 Redox Reactions Recall that a redox reaction is a chemical reaction that involves the transfer of electrons from one reactant to another The loss of electrons is oxidation (LEO) and the gain of electrons is reduction (GER) Oxidation corresponds to an increase in oxidation number whereas reduction is a decrease in oxidation number. Refer to chapter 4 for rules for assigning oxidation numbers. 0 -2 0 +3 4 Fe(s) + 3 O2(g) 2 Fe2O3(s) 19.2 Balancing Redox Reactions A redox reaction can be separated into two half-reactions: 2 Al + 3 Cl2 → 2 AlCl3 Oxidation half-reaction: Al → Al3+ + 3 e− Reduction half-reaction: Cl2 + 2 e− → 2 Cl− Balancing a redox reaction involves balancing the mass and charge in each half-reaction. pH is important to consider when balancing redox reactions Acidic solution Basic solution H2O can be added in both acidic and basic solutions to balance hydrogen and Balance for hydrogen by Add OH- ions to balance oxygen atoms adding H+ H+ ions 19.2 Balancing Redox Reactions Example - Balance this reaction below in acidic solution. Cr2O72−(aq) + Sn2+(aq) → Cr3+(aq) + SnO2(s) (+6)(-2) (+2) (+3) (+4)(-2) Step 1: Assign oxidation numbers to see which atoms are oxidized and which are reduced. Separate the redox reaction into the two half reactions. Reduction: Cr2O72−(aq) → Cr3+(aq) Oxidation: Sn2+(aq) → SnO2(s) Step 2: Add water molecules to balance oxygen in both reactions Cr2O72−(aq) → Cr3+(aq) + 7 H2O(l) 2 H2O(l) + Sn2+(aq) → SnO2(s) Step 3: Add H+ to balance hydrogen 14 H+(aq) + Cr2O72−(aq) → 2 Cr3+(aq) + 7 H2O(l) 2 H2O(l) + Sn2+(aq) → SnO2(s) + 4 H+(aq) Step 4: Add electrons to balance the overall charge 6 e− + 14 H+(aq) + Cr2O72−(aq) → 2 Cr3+(aq) + 7 H2O(l) 2 H2O(l) + Sn2+(aq) → SnO2(s) + 4 H+(aq) + 2 e− Step 5: Multiply one or both half-reactions by a small integer to balance the electrons. [2 H2O(l) + Sn2+(aq) → SnO2(s) + 4 H+(aq) + 2 e−] × 3 6 H2O(l) + 3 Sn2+(aq) → 3 SnO2(s) + 12 H+(aq) + 6 e− Step 6: Add the half-reactions, cancelling any species that appear on both sides 2 1 6 e− + 14 H+(aq) +Cr2O72− (aq) → 2 Cr3+(aq)+ 7 H2O(l) 6 H2O(l) + 3 Sn2+(aq) → 3 SnO2(s) + 12 H+(aq ) + 6 e− 2 H+(aq) +Cr2O72− (aq) + 3 Sn2+(aq) → 2 Cr3+(aq)+H2O(l)+ 3 SnO2(s) Step 7: Verify that all atoms and charges are balanced Reactant side: Product side: H: 2 H: 2 Cr: 2 Cr: 2 O: 7 O: 7 Sn: 3 Sn: 3 Charge: +6 Charge: +6 19.4 Voltaic Cells The Daniell cell is a common type of voltaic cell and consist of a redox reaction between copper and zinc: Reduction: Cu2+(aq) + 2 e− → Cu(s) (cathode) There are two main types of electrochemical cells: Oxidation: Zn(s) → Zn2+(aq) + 2 e− (anode) 1. Voltaic Cells - consists of spontaneous chemical reactions to generate electricity Electrons flow from the anode to the cathode 2. Electrolytic Cells – use electricity to drive a nonspontaneous chemical reaction A salt bridge is used to maintain charge balance 19.4 Voltaic Cells – Standard Cell Notation Anode half-reaction: Zn(s) Zn2+(aq) + 2 e− Cathode half-reaction: Cu2+(aq) + 2 e− Cu(s) Overall cell reaction: Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) Salt bridge Anode half-cell Cathode half-cell Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) Electron flow Phase boundary Phase boundary 19.5 Cell Potential The electromotive force associated with the flow of SHE electrons is measured as the cell potential (voltage) Cell potentials for half-reactions are measured relative to a reference cell known as the standard hydrogen electrode, SHE, set to 0.00 V with respect to the half- reaction. Standard reduction potentials, Eored of various half-cell reactions are measure with the SHE are the other half-cell Standard conditions: 298 K, 1.00 atm, 1.00 mol/L  o Half-Reaction E red (V ) F2(g) + 2 e−→ 2 F−(aq) 2.87 Co3+(aq) + e−→ Co2+(aq) 1.92 Table 19.1 MnO4−(aq) + 8 H+(aq) + 5 e−→ Mn2+(aq) + 4 H2O(l) 1.51 Selected Standard Pb4+(aq) + 2 e−→ Pb2+(aq) 1.46 Reduction Potentials Cl2(g) + 2 e−→2 Cl−(aq) 1.36 (25℃) Cr2O72−(aq) + 14 H+(aq) + 6 e−→ 2 Cr3+(aq) + 7 H2O(l) 1.23 Ag+(aq) + e− → Ag(s) 0.80 Fe3+(aq) + e−→ Fe2+(aq) 0.77 Cu+(aq) + e− → Cu(s) 0.52 Cu2+(aq) + 2 e− → Cu(s) 0.34 A more positive Cu2+(aq) + e−→ Cu+(aq) 0.15 value Eo red means that the species is Sn4+(aq) + 2 e−→ Sn2+(aq) 0.15 more likely to be 2 H+(aq) + 2 e−→H2(g) 0.00 (SHE) reduced Pb2+(aq) + 2 e− → Pb(s) −0.13 Sn2+(aq) + 2 e− → Sn(s) −0.14 Ni2+(aq) + 2 e− → Ni(s) −0.26 Cr3+(aq) + e−→ Cr2+(aq) −0.41 Fe2+(aq) + 2 e− → Fe(s) −0.44 Zn2+(aq) + 2 e− → Zn(s) −0.76 2 H2O(l) + 2 e−→H2(g) + 2 OH−(aq) −0.83 Al3+(aq) + 3 e− → Al(s) −1.66 Mg2+(aq) + 2 e−→ Mg(s) −2.38 Na+(aq) + e− → Na(s) −2.71 19.5 Cell Potentials Eo cathode: standard reduction potential for the cathode half- reaction (reduction) The standard cell potential, Eocell is calculated using the formula: Eo anode: standard reduction potential for the anode half- reaction (oxidation) E cell = E cathode − E anode Example – Calculate the cell potential for the voltaic cells comprised of the sets of half-cells. A) Sn4+/Sn2+ and Cr3+/Cr2+ B) Al/Al3+ and Pb/Pb2+ A) Sn4+/Sn2+ and Cr3+/Cr2+ Look up the half-reactions in table 19.1 Sn4+(aq) + 2 e− → Sn2+(aq) E = −0.15 V Cr3+(aq) + e− → Cr2+(aq) E = −0.41 V Since the half-reaction with Sn has a more positive value of Eo , this reaction occurs at the cathode E cell = E cathode − E anode E cell = 0.15 V − ( −0.41 V)= 0.56 V B) Al/Al+3 and Pb/Pb2+ Look up the half-reactions in table 19.1 Al3+(aq) + 3 e− → Al(s) E = −1.66 V Pb2+(aq) + 2e− → Pb(s) E = −0.13 V Since the half-reaction with Pb has a more positive value of Eo , this reaction occurs at the cathode E cell = E cathode − E anode E cell = −0.13 V − ( −1.66 V) = 1.53 V 19.6 Free Energy and Cell Potential The cell potential (Ecell ) is related to the Gibbs free energy (∆G) through the formula: G = −nFEcell In this equation, n is the number of moles of electrons transferred, and F is the Faraday constant, F, which is the charge, in coulombs, C, of a mole of electrons. F = 96,485 C/mol e− Recall that a process is spontaneous when ∆G < 0. Under standard conditions, When Ecell > 0, ∆G < 0. and the reaction is spontaneous. G = −nFEcell 19.7 The Nernst Equation and the Concentration of Cells Recall the relationship between ∆G and the reaction quotient, Q: G = G + RT ln Q Combining this equation with G = −nFEcell yields the Nernst equation: RT 𝐸𝑐𝑒𝑙𝑙 = 𝐸𝑜𝑐𝑒𝑙𝑙 − ln Q nF The Nernst equation can be used to calculate the cell potential when solute concentrations are some value other than 1.00 M. 19.7 The Nernst Equation and the Concentration of Cells Example - Calculate the potential of a zinc/silver cell in which the zinc electrode is immersed in 0.100 M zinc nitrate and the silver electrode is in 0.500 M silver nitrate. Step 1: Write the half-reactions, and determine Eo ∘ ∘ ∘ E𝑐𝑒𝑙𝑙 = E𝑐𝑎𝑡ℎ𝑜𝑑𝑒 − E𝑎𝑛𝑜𝑑𝑒 = 0.80 𝑉 − (−0.76 𝑉) = 1.56 𝑉 Step 1: Write the half-reactions, and determine n Since there are two moles of electrons that are transferred in the Zn2+(aq) + 2 e− → Zn(s) E = −0.76V reaction, n = 2 + (Ag+(aq) + e− → Ag(s) E = +0.76V) x 2 Zn + 2 Ag+ → Zn2+ + 2 Ag Step 3: Calculate Q using the concentration of Zn+2 1 1 −1 Q= = = 10.0 M [Zn2+ ] 0.100 M Step 4: Calculate Q using the concentration of Zn+2  J  RT  8.3145 (298 K)  E =E −  lnQ E = −0.76 V −  K  mol  ln 10 nF  −  96 , 485 C    (2 mol e ) −    1 mol e  E = −0.79 V 19.8 Voltaic Cell Applications: Batteries and Fuel Cells Lead storage car battery Alkaline Dry Cell Hydrogen Fuel Cells Anode: Anode: Anode: H2(g) → 2 H+(aq) + 2 e− Pb(s) + SO4 2−(aq) → PbSO4(s) + 2 e− Zn(s) + 2 OH−(aq) → Zn (OH)2(s) + 2 e− Cathode: Cathode: Cathode: O2(g) + 4 H+(aq) + 4 e− MnO2(s) + H2O(l) + e− MnO2(s) + H2O(l) + e− → 2 H2O(l) → MnO(OH)(s) + 2 OH−(aq) → MnO(OH)(s) + 2 OH−(aq)  19.8 Voltaic Cell Applications: Corrosion The oxidation of metal structures is Corrosion can be prevented by adding a more known as corrosion active metal to the iron surface. The more active metal becomes the anode via cathode protection Anode (oxidation): Fe(s) → Fe2+(aq) + 2 e− Zinc is more active than iron (recall activity series) and is used as a sacrificial anode Cathode (reduction): O2(g) + 4 H3O+(aq) + 4 e−→ 6 H2O(l) 2 Fe2+(aq) + ½ O2(g) + 6 H2O+(l) → Fe2O3(s) + 4 H3O+(aq) Oxidation: Zn(s) → Zn2+(aq) + 2 e− Reduction: O2(g) + 4 H3O+(aq) + 4 e−→ 6 H2O(l) 19.9 Electrolytic Cells and Applications of Electrolysis Electrolytic cells apply electrical currents to cause a nonspontaneous reaction to occur. 19.9 Electrolytic Cells and Applications of Electrolysis Electrolysis calculations are stoichiometric calculations based on the number of moles of electrons passing through the electrolytic cell. The number of electrons available is determined by the current, which is measured in amperes, A, defined as the amount of charge in coulombs per second, C/s. 1C 1 A= and 1 C = 1 A  s s 1 mol of electrons has a total charge of 96,485 C, the Faraday constant. 1 mol e− = 96,485 C The total charge (q) is calculated by multiplying current (I) by time (t). q=It 19.9 Electrolytic Cells and Applications of Electrolysis Example - Calculate the mass of copper metal that will be deposited by passing a 10.0 A current through a solution of copper(II) sulfate for 3.00 h. Step 1: Calculate the total charge, q t = 3.00 h x 3600 s/1 h = 1.08 x 104 s q=It q = 10.0 A x 1.08 x 104 s = 1.08 x 105 A.s = 1.08 x 105 C Step 2: Convert charge to moles of electrons using Faraday’s constant 1.08 x 105 C x = 1.12 mol e- Step 3: Convert moles of electrons to moles of copper Cu2+(aq) + 2 e− → Cu(s) 1.12 mol e- x = 0.556 mol Cu Step 4: Convert moles of copper to mass of copper 0.556 mol Cu x = 35.6 g Cu Chapter 18 Chemical Thermodynamics CHY 102 Chapter Outline Section 18.1 Entropy and Spontaneity Section 18.2 Entropy Changes Section 18.3 Entropy and Temperature Section 18.4 Gibbs Free Energy Section 18.5 Free-Energy Changes and Temperature Section 18.6 Gibbs Free Energy and Equilibrium  18.1: Entropy and Spontaneity Entropy (S) is a measure of the degree of disorder or randomness in a system Entropy of a system increases whenever there is: A phase change from a more condensed to less condensed phase (solid à liquid, liquid à gas) Ø Solids are highly organized, liquids less so, gas particles move randomly An increase in temperature within a given phase Ø More kinetic energy increases the motion of the particles An increase in the number of gas particles or dissolution of a solid in an aqueous solution Ø Increase in number of particles provides more ways for the system’s energy to be distributed  18.1: Entropy and Spontaneity Example - Determine the sign of ΔS for the following processes: a. 2 KClO3(s) → 2 KCl(s) + 3 O2(g) ΔS = + The number of particles increases, and a gas is formed b. CO2(g) → CO2(s) ΔS = − A gas becomes a solid c. ΔS = + A solid becomes a gas d. condensation appearing on bathroom mirror after a shower ΔS = − A gas becomes a liquid  18.1: Entropy and Spontaneity Entropy increases when there are more energetically equivalent ways to arrange the components of that system. Entropy was defined mathematically by Boltzman in the 1870s k = Boltzman constant = 1.38 x 1023 J/K S = k ln W W = the number of energetically equivalent arrangements (microstates) possible for the system What are microstates? Consider a box with a divider down the middle containing two particles on the left side A on top B on top A and B both A and B both B on bottom A on bottom on top on bottom Since there are four possible arrangements for this system, W=4  18.1: Entropy and Spontaneity If, in the last example, the divider were to be removed: 16 microstates 1 macrostate There are now 16 possible arrangements, therefore W=16. For any system, we can calculate W as: X = # of locations n W=X n = # of particles In the above example, there are two particles and four possible locations in which the particles can be, so W = 42 = 16 Equivalent microstates form a macrostate, and the most probable macrostate is the one containing the highest number of microstates. Each circle represents a microstate. Which macrostate is most probable? B A A. A and B together A B A B B. A and B in separate locations A B B A A C. A in top left, B in bottom area B D. All of the above are equally probable. A B B B A A Each circle represents a microstate. Which macrostate is most probable? B A A. A and B together 3 A B A B B. A and B in separate 6 locations A B B A A C. A in top left, B in bottom 1 area B D. All of the above are equally probable. A B B B A A  18.1: Entropy and Spontaneity In general, the tendency towards disorder is a statistical probability; a system will spontaneously move to a more spread-out arrangement Ø A more disordered system is more probable than an ordered one The criteria for spontaneity are not solely based on the entropy change for the system. The entropy change of the surroundings must also be considered, and how the system and surroundings together affect the total entropy of the universe Sfewer moles < Smore moles Ø Processes that decrease the entropy of a system can still be spontaneous The Second Law of Thermodynamics states that 2NH3 à N2 + H2 spontaneous processes always result in an overall increase in entropy, S, of the universe  18.2 Entropy Changes The third law of thermodynamics states that the entropy of a pure, perfectly ordered, crystalline substance at absolute zero is zero: Scrystal (0 K) = 0 At T = 0K, there is no thermal energy and the particles in a perfectly ordered crystal would exist in a single microstate: Scrystal = k ln W = k ln 1 = 0 Standard molar entropies, S°, are calculated based on the third law of thermodynamics, and these can be used to calculate standard changes in entropy, ΔS°: DS rxn  = å m[S  (products)] - å n[S (reactants)] Where m and n are the coefficients of the products and reactants from the balanced chemical equation, and standard thermodynamic conditions are 298 K, 1 atm gas pressure, and 1 M solute concentration. Standard Entropies at 298 K of Selected Substances Compound J Compound J S S (state) mol × K (state) mol × K CO2(g) 213.8 HCl(g) 186.9 CaO(s) 38.1 Hg(l) 75.9 CaCO3(s) 91.7 Hg(g) 175.0 H2(g) 130.7 I2(s) 116.1 H2O(l) 70.0 I2(g) 260.7 H2O(g) 188.8 O2(g) 205.2  18.2 Entropy Changes Example - Predict the sign of ΔS and calculate the standard entropy change associated with the reaction of gaseous hydrogen and gaseous oxygen to form water vapor. Step 1: Write out the balanced chemical equation 2 H2(g) + O2(g) → 2 H2O(g) Prediction: The number of moles of gases decreases in the reaction, so ΔS will be negative (−). Step 2: Calculate the standard change in entropy using standard entropy data from the table DS rxn  = å m[S  (products)] - å n[S (reactants)] DS rxn  = 2S  ( H 2 O(g)) - [2S  ( H2(g)) + S  (O2(g))] æ J ö é æ J ö æ J öù DS  = 2 mol ç 188.8 ÷ - ê2 mol ç 130.7 mol × K ÷ + 1 mol ç 205.2 mol × K ÷ ú rxn è mol × K ø ë è ø è øû J DS rxn  = -89.0 K  18.3: Entropy and Temperature An overall change in entropy is equal to the change in entropy of the system plus the change in entropy of the surroundings: ΔS univ = ΔS sys + ΔS surr For a process to be spontaneous, ∆Suniv must be positive The entropy change for the system may be negative, but the process still be spontaneous  18.3: Entropy and Temperature For an isothermal process (i.e., no heat transfer between system and surroundings), and at constant pressure the entropy change of the surroundings is related to the thermal energy change of the system by: -DH sys ΔS surr = T Example – Calculate ΔSsurr and ΔSuniv for the reaction of gaseous hydrogen and gaseous oxygen to form 2 mol of water vapor at 298 K Step 1: Write the balanced chemical equation 2 H2(g) + O2(g) → 2 H2O(g) Step 2: Calculate DH rxn  Ø Look up DH f values in Appendix A.1 of textbook: H2O(g) = -241.8 kJ/mol, H2 and O2 are in their elemental state = 0 kJ/mol DH rxn  = å m[ DH f (products)] - å n[ DH f (reactants)] DH rxn  = 2 ( DH f (H2O(g)) - [2 ( DH f (H2(g)) + DH f(O2 )(g))] æ kJ ö é æ kJ ö æ kJ ö ù DH  rxn = 2 mol ç -241.8 ÷ - ê2 mol ç 0 ÷ + 1 mol ç 0 mol ÷ ú è mol ø ë è mol ø è øû DH rxn  = -483.6 kJ Step 3: calculate ΔSsurr -DH sys ΔS surr = T -( -483.6 kJ) 1000 J DS surr  = ´ 298 K 1 kJ J DS surr  = 1.62 ´ 103 K Step 4: Calculate ΔSsys and ΔSuniv Ø We calculated ΔSsys for this reaction in the previous example; ΔSsys = -89.0 J/K ΔS univ = ΔS sys + ΔS surr J J ΔS univ = -89.0 + 1620 K K J ΔS univ = 1530 K  18.4 Gibbs Free Energy Gibbs free energy, G, is a state function that an be used for predicting if a process is spontaneous at a given temperature: G = H - TS For a chemical reaction occurring under standard conditions but nonstandard temperature, the change in Gibbs free-energy is: DG  = DH  -T DS  ∆G° = - à Reaction is spontaneous in the forward direction ∆G° = + à Reaction is spontaneous in the reverse direction ∆G° = 0 à Reaction is at equilibrium Example - Determine ΔG° for the reaction of hydrogen and oxygen to form water given ΔS° = -89J/K and ΔH° = -483.6 kJ. 2 H2(g) + O2(g) → 2 H2O(g) æ Jö æ 1 kJ ö DG  = DH  -T DS  DG  = -483.6 kJ - 298 K ç -89.0 ÷ ç 1000 J ÷ DG  = -457.1 kJ è Kø è ø  18.4 Gibbs Free Energy The standard free energy of formation, DG f , is the free-energy change associated with the o formation of a substance from its component elements, in their standard states. DG rxn  = å m[ DG f (products)] - å n[ DG f (reactants)] Example - Calculate DG rxn  for the combustion of propane. Compound ∆Gf° (kJ/mol) C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) C3H8(g) -23.4 CO2(g) -394.4 Solution: Use standard free energy of formation values to calculate ∆Grxn° H2O(l) -237.1 DG rxn  =[3DG f (CO2(g)) + 4DG f(H2O(l))] - [ DG f(C3H8(g)) + 5DG f(O2(g))] é æ kJ ö æ kJ ö ù é æ kJ ö æ kJ ö ù DG  rxn = ê3 mol ç -394.4 ÷ + 4 mol ç -237.1 ÷ - ú ê1 mol ç -23.4 ÷ + 5 mol ç 0 mol ÷ ú ë è mol ø è mol ø û ë è mol ø è øû DG rxn  =( -2131.6 kJ) - ( -23.4 kJ)= -2108.2 kJ   Determine DG rxn for the following reaction: 2 A(g) + B2(g) → 2 AB(g) A. −400 kJ B. −250 kJ C. −200 kJ D. −100 kJ E. −50 kJ Compound(state) 𝚫𝑮∘𝐟 (kJ/mol) A(g) 0 B2(g) −100 AB(g) −150   Determine DG rxn for the following reaction: 2 A(g) + B2(g) → 2 AB(g) DG rxn  = å m[ DG f (products)] - å n[ DG f (reactants)] A. −400 kJ B. −250 kJ 𝑘𝐽 𝑘𝐽 𝑘𝐽 Δ𝐺°!"# = 2 𝑚𝑜𝑙 −150 − 2 𝑚𝑜𝑙 0 + 1 𝑚𝑜𝑙 −100 𝑚𝑜𝑙 𝑚𝑜𝑙 𝑚𝑜𝑙 C. −200 kJ D. −100 kJ 𝚫𝑮°𝒓𝒙𝒏 = −𝟐𝟎𝟎 𝒌𝑱 E. −50 kJ Compound(state) 𝚫𝑮∘𝐟 (kJ/mol) A(g) 0 B2(g) −100 AB(g) −150  18.5 Free-Energy Changes and Temperature Standard values are measured at 298K. ∆H and ∆S values only vary slightly with temperature, so standard values can be used for them at temperatures other than 298K One way ∆G can be calculated at temperatures that are not 298K is by calculating ∆H° and ∆S°, then using ΔG = ΔH° – TΔS° We can also determine the temperature at which a reaction becomes spontaneous, i.e., when ∆G=0: ΔG = ΔH – TΔS = 0 DH T= DS T will provide the minimum temperature at which the forward reaction is spontaneous.  18.5 Free-Energy Changes and Temperature Example - Determine the minimum temperature for the spontaneous vaporization of liquid mercury using data from the table below Hg(l) → Hg(g) Compound ∆S° (J/mol・K) Solution Hg(g) 175.0 J J J Hg(l) 75.9 DS = 175.0  - 75.9 DS  = 99.1 mol × K mol × K mol × K Compound ∆H° (kJ/mol) Hg(g) 61.4 𝑘𝐽 𝑘𝐽 𝑘𝐽 Δ𝐻°!"# = 1 𝑚𝑜𝑙 61.4 − 1 𝑚𝑜𝑙 0 = 61.4 Hg(l) 0 (elemental state) 𝑚𝑜𝑙 𝑚𝑜𝑙 𝑚𝑜𝑙 kJ 61.4 DH mol T= = = 620 K DS J æ 1 kJ ö 99.1 ç K è 1000 J ÷ø ∴ The vaporization of mercury becomes spontaneous at temperatures above 620 K.  18.6 Gibbs Free Energy and Equilibrium For other nonstandard conditions, the change in free energy for a reaction can be calculated as: ΔG = ΔG° + RT ln Q Reaction Quotient Gas constant Temperature (QC for aqueous 8.314 J mol-1 K-1 (K) QP for gas) Recall that Q occurs under nonequilibrium conditions. The result of the above equation tells us the direction in which the reaction will proceed: Ø ΔG = negative à reaction proceeds spontaneously Ø ΔG = positive à reaction does not proceed spontaneously Ø ΔG = 0 à reaction is at equilibrium At equilibrium, Q = K and ΔG = 0, so the above equation becomes: ΔG° = -RT ln K K < 1 à The equilibrium position lies closer to the lower- energy reactants When Q = 1, ΔG > 0 à The reverse process is spontaneous K = Q & ΔG = 0 à No net change in forward or reverse direction K > 1 à The equilibrium position lies closer to the lower- energy products When Q = 1, ΔG < 0 à The forward process is spontaneous  18.6 Gibbs Free Energy and Equilibrium Example – a) Calculate ΔG for the Haber process at 365 K for a mixture of 1.5 atm N2, 4.5 atm H2, and 0.75 atm NH3, b) In which direction is the reaction currently proceeding? N2(g) + 3 H2(g) ⇌ 2 NH3(g) a) Step 1: Start by calculating ΔG° from DG fvalues found in Appendix A.1. æ kJ ö DG  = 2 mol ç -16.4 ÷ - [0 + 0]= -32.8 kJ è mol ø Step 2: Calculate Q (PNH )2 Qp = 3 (PN )(PH )3 2 2 (0.75)2 -3 Qp = = 4.1 ´ 10 (1.5)(4.5)3 Step 3: Plug your calculated values for ΔG° and 𝑄 into the ∆G equation ΔG = ΔG° + RT ln Q kJ æ J 1 kJ ö DG = -32.8 + ç 8.3145 ´ ÷ (365 K) ln(4.1 ´ 10-3 ) mol è mol × K 1000 J ø DG = -49.5 kJ/mol b) Since ∆G is negative, the system is not at equilibrium and the reaction is occurring spontaneously in the forward direction Given the ∆G° value calculated for the Haber process in the last example, what is the equilibrium constant for the Haber process at 298 K. A. 5.6 x 105 B. 1.01 C. 4.74 x 108 D. 1.02 x 103 Given the ∆G° value calculated for the Haber process in the last example, what is the equilibrium constant for the Haber process at 298 K. ΔG  - ΔG° = -RT ln K K=e RT A. 5.6 x 105 B. 1.01 DG  -( -32.8 kJ) 1000 J Calculate exponent: - = ´ = 13.24 RT æ J ö 1 kJ C. 4.74 x 108 ç 8.3145 mol × K ÷ (298 K) è ø D. 1.02 x 103 Now solve for 𝐾: K = e 13.24 K = 5.6 × 105 Chapter 17: Aqueous Equilibrium Chapter Outline Section 17.1 Introduction to Buffer Solutions Section 17.2 The Henderson–Hasselbach Equation Section 17.3 Titrations of Strong Acids and Strong Bases Section 17.4 Titrations of Weak Acids and Weak Bases Section 17.5 Indicators in Acid–Base Titrations Section 17.6 The Solubility Product Constant, Ksp Section 17.7 The Common-Ion Effect and the Effect of pH on Solubility Section 17.8 Precipitation: Q Versus Ksp You are not responsible for knowing sections 17.9 and 17.10 17.1 Introduction to Buffer Solutions A buffer solution consists of a weak acid and its conjugate base: Buffer solutions resist a change in pH, even when a large amount of a strong acid or strong base is added. HA(aq) + H2O(l) ⇌ H3O+(aq) + A−(aq) Conjugate base Weak acid 17.1 Introduction to Buffer Solutions HA(aq) + H2O(l) ⇌ H3O+(aq) + A−(aq) How does a buffer solution work? A buffer has significant quantities of both HA and its conjugate base, A−, allowing it to shift in either direction when either acid or base is added. 17.1 Introduction to Buffer Solutions Consider the equilbirium reaction below: HNO2(aq) + H2O(l) ⇌ NO2−(aq) + H3O+(aq) If NaNO2 is added to the reaction, then what will happen to the equilbirum? According to Le Chatelier’s principle, the reaction will shift to the left producing more reactants if additional NO2- is added to the equilibrium. NO2- is a common ion in this reaction and suppresses the ionization of the weak acid, HNO2. This is known as the common-ion effect 17.1 Introduction to Buffer Solutions Example - Calculate the pH of (a) an aqueous solution containing 0.30 M HNO2 and (b) a solution containing 0.30 M HNO2 and 0.20 M NaNO2. The Ka of HNO2 is 5.6 × 10−4. HNO2(aq) + H2O(l) ⇌ NO2−(aq) + H3O+(aq) (a) 0.30 M HNO2 1. Start by constructing an ICE table and use the equilibrium expression to solve for x: HNO2(aq) + H2O ⇌ H3O+(aq) + NO2−(aq) Initial (M) 0.30 ~0 0 Change (M) −x +x +x Equilibrium (M) 0.30 − x x x [H3O+ ][NO2− ] ( x )2 ( x )2 Ka = =  = 5.6  10−4 [HNO2 ] (0.30 − x ) 0.30 x = (5.6  10−4 )(0.30) = [H3O+ ] = 0.013 M 2. Verify that x < 5% of the initial concentration of HNO2 to validate the assumption. 0.013  100% = 4.3% 0.30 3. Calculate pH using pH = −log[H3O+]. pH = −log[H3O+] = −log(0.013) = 1.89 (b) 0.30 M HNO2 and 0.20 M NaNO2 1. Start by constructing an ICE table and use the equilibrium expression to solve for y: HNO2(aq) + H2O ⇌ H3O+(aq) + NO2−(aq) Initial (M) 0.30 ~0 0.20 Change (M) −y +y +y Equilibrium (M) 0.30 − y y 0.20 + y [H3O+ ][NO2− ] ( y )(0.20 + y ) ( y )(0.20) 5.6  10−4 (0.30) Ka = =  = 5.6  10−4 y= = [H3O+ ] = 0.00084 M [HNO2 ] (0.30 − y ) 0.30 0.20 2. Verify that y ≪ 0.20 M and that y ≪ 0.30 M. 3. Calculate pH using pH = −log[H3O+]. 0.00084  100% = 0.42% 0.20 pH = −log[H3O+] = −log(0.00084) = 3.08 0.00084  100% = 0.28% 0.30 17.2 The Henderson-Hasselbach Equation Consider the equilibrium reaction for the ionization of the general weak acid, HA; [H3O+ ][A − ] HA(aq) + H2O(l) ⇌ H3O+(aq) + A−(aq) Ka = [HA] The pH of a buffer solution formed from A- and HA can be calculated using the Henderson- Hasselbach equation: pKa = -log Ka [A − ] pH = pK a + log [HA] [HA]: concentration of weak acid [A-]: concentration of conjugate base 17.2 The Henderson-Hasselbach Equation The Henderson-Hasselbach equation can be used to assess the value of the pH relative to the pKa of the weak acid: [A − ] pH = pK a + log [HA] When [A−] > [HA], the ratio of [𝐴− ] , and pH > pKa. >𝟏 [𝐻𝐴] [𝑨− ] When [HA] > [A−], the ratio of < 𝟏 , and the pH < pKa. [𝑯𝑨] [𝐴− ] When [HA] = [A−], the ratio of = 1 , and the pH = pKa. [𝐻𝐴] 17.2 The Henderson-Hasselbach Equation Example 1 - Calculate the relative concentrations of acetate ion and acetic acid that must be used to make a buffer solution with a pH of 4.00. The Ka of acetic acid is 1.8 × 10−5. Rearrange the Henderson-Hasselbach equation to isolate for [A-]/[HA] [A − ] [A − ] log = − 0.74 pH = pK a + log [HA] [HA] [A − ] [A − ] log = pH − pK a = 4.00 − log(1.8  10−5 ) = 10−0.74 = 0.18 [HA] [HA] 17.2 The Henderson-Hasselbach Equation Example 2 - Calculate the pH of the following a. A buffer containing 0.150 mol of acetic acid (HC2H3O2) and 0.195 mol of sodium acetate (NaC2H3O2) in 1.00 L. HC2H3O2 (aq) + H2O(l) ⇌ C2 H3O2-(aq) + H3 O+ (aq) b. After 0.0300 mol of solid NaOH is added to the buffer from part (a) with no change in volume. a. A buffer containing 0.150 mol of acetic acid (HC2H3O2) and 0.195 mol of sodium acetate (NaC2H3O2) in 1.00 L. [A − ] pH = pK a + log [HA] −5 0.195 pH = −log(1.8  10 ) + log = 4.85 0.150 b. After 0.0300 mol of solid NaOH is added to the buffer from part (a) with no change in volume. 1. Use an ICE table to determine the equilibrium concentrations after neutralization. When NaOH is added to the buffer, the weak acid reacts to neutralize it. HC2H3O2(aq) + NaOH(aq) → NaC2H3O2 (aq) + H2O (l) Initial (M) 0.150 0.0300 0.195 Change (M) -0.0300 -0.0300 +0.0300 Equilibrium (M) 0.120 0 0.225 2. Use the Henderson-Hasselbach equation to calculate the pH NaOH is the limiting reagent and is completely [𝐴− ] 𝑝𝐻 = 𝑝K 𝑎 + 𝑙𝑜𝑔 consumed in the reaction [𝐻𝐴] 0.225 𝑝𝐻 = −𝑙𝑜𝑔 1.8 × 10−5 + 𝑙𝑜𝑔 = 𝟓. 𝟎𝟐 0.120 17.2 The Henderson-Hasselbach Equation The effectiveness of a buffer solution is based on two factors: buffer capacity and effective pH range. Buffer capacity refers to the ability of the buffer solution to resist a change in pH when an acid or a base is added. Buffer capacity is dependent on the concentration of HA and A−. Buffers are most effective at pH values within 1 pH unit of the p𝐾a of the buffer’s conjugate acid. The buffer′s pH range is p𝐾a ± 1. 17.3 Titrations of Strong Acids and Strong Bases Recall that a titration is an experimental method for adding a solution of a known concentration (the titrant) to a solution of unknown concentration. We will examine two types of titrations: 1. Strong acid and strong base 2. Weak acid and weak base 17.3 Titrations of Strong Acids and Strong Bases In the titration of a strong base with a strong acid, a titration curve shows how the pH changes with volume of strong base or strong acid added: Titration of strong acid to strong base Titration of strong base to strong acid 17.3 Titrations of Strong Acids and Strong Bases Region A: the initial pH of the solution Region B: segment between initial pH and the equivalence point. Region C: the equivalence point, acid has neutralized the base. For a strong acid- base titration, the pH at the equivalence point is equal to 7. Region D: segment after the equivalence point. 17.4 Titrations of Weak Acids and Weak Bases pH calculations in different regions: Region A: the initial pH is the pH of a weak acid solution HA(aq) + H2O(l) ⇌ A−(aq) + H3O+(aq) [A − ][H3O+ ] x2 Ka =  [HA] [HA] pH = −log[H3O+ ] = −log K a [HA] 17.4 Titrations of Weak Acids and Weak Bases pH calculations in different regions: Region B: buffer region. The pH is calculated using the Henderson-Hasselbach equation HA(aq) + H2O(l) ⇌ A−(aq) + H3O+(aq) [A − ] pH = pK a + log [HA] 17.4 Titrations of Weak Acids and Weak Bases pH calculations in different regions: Region C: at the equivalence point, only the conjugate base A- remains. A−(aq) + H2O(l) ⇌ HA(aq) + OH−(aq) Kw x2 Kb = = − [OH− ] = x = K b[A − ] K a [A ] pH = 14.00 − pOH = 14.00 − ( −log[OH − ]) 17.4 Titrations of Weak Acids and Weak Bases pH calculations in different regions: Region D: after the equivalence point, the pH depends on the concentration of excess OH- added. pH = 14.00 − pOH = 14.00 − ( −log[OH − ]) 17.4 Titrations of Weak Acids and Weak Bases Example - Suppose that 40.0 mL of a 0.500 M formic acid (HCHO2, Ka = 1.8 × 10−4) solution is titrated with a 0.500 M solution of NaOH. Determine the pH at these points. a. prior to the addition of NaOH b. at the equivalence point c. After the equivalence point has been reached and 2.0 mL of excess NaOH has been added to the solution a. prior to the addition of NaOH CHO2−(aq) + H2O(l) ⇌ HCHO2(aq) + OH−(aq) The pH is calculated from the weak acid 𝑝𝐻 = −𝑙𝑜𝑔 K 𝑎 [𝐻𝐶𝐻𝑂2 ] = −𝑙𝑜𝑔 (1.8 × 10−4 )(0.500) = 𝟐. 𝟎𝟐 b. at the equivalence point nOH- = nCHO2- = nHCHO2 = 0.0400 L x 0.500 mol/L =0.0200 mol VOH- = 0.0200 mol / 0.500 mol/L = 0.0400 L Vtotal = VHCHO2 + VOH- = 0.0400 L + 0.0400 L = 0.0800 L b. at the equivalence point CHO2−(aq) + H2O(l) ⇌ HCHO2(aq) + OH−(aq) 𝑚𝑜𝑙 0.0400 𝐿 0.500 𝐿 [𝐶𝐻𝑂2 − ] = = 0.250 𝑚𝑜𝑙/𝐿 0.0800 𝐿 1.0  10−14 −11 Kb = = 5.6  10 1.8  10−4 𝑝𝑂𝐻 = −𝑙𝑜𝑔[𝑂𝐻 − ] = −𝑙𝑜𝑔 5.6 × 10−11 0.250 = −log(3.7 × 10−4 ) = 5.43 pH = 14.00 − 5.43 = 8.57 c. after the equivalence point has been reached and 2.0 mL of excess NaOH has been added to the solution 𝑚𝑜𝑙 0.0020 𝐿 0.500 𝐿 𝑂𝐻 − 𝑒𝑥𝑐𝑒𝑠𝑠 = = 0.0122 𝑚𝑜𝑙/𝐿 0.0820 𝐿 𝑚𝑜𝑙 𝑂𝐻 − 𝑡𝑜𝑡𝑎𝑙 = 0.0122 + 3.3 𝑥 10−6 M = 0.0122 mol/L 𝐿 𝑝𝑂𝐻 = −𝑙𝑜𝑔[0.0122] = 1.91 𝑝𝐻 = 14.00 − 𝑝𝑂𝐻 = 14.00 − 1.91 = 𝟏𝟐. 𝟎𝟗 17.4 Titrations of Weak Acids and Weak Bases Addition of a weak acid to a weak base The pH calculations are similar to the addition of weak base to a weak acid, however excess acid remains after the equivalence point in these titrations. 17.5 Indicators in Acid-Base Titrations An indicator can be used to determine the equivalence point of a titration. Most of the indicators used for acid–base titrations are weak acids, where the protonated (acidic) form is one color, and the unprotonated (basic) form is another color. HA(aq) + H2O(l) ⇌ A−(aq) + H3O+(aq) color 1 color 2 Phenolphthalein is commonly used for strong acid–strong base titrations. It is colorless in acidic solution and is a pink color at pH ≥ 9. 17.5 Indicators in Acid-Base Titrations Phenolphthalein works well for strong acid–strong base titrations and for most weak acid–strong base titrations. However, if the weak acid has a very small Ka, phenolphthalein may not be useful and an indicator with a higher indicator range is used.Weak base–strong acid titrations require indicators that change color at lower pH values. Indicator Name pH Range and Color Change Methyl Orange (red) 3.2–4.4 (yellow) Bromocresol Green (yellow) 3.8–5.4 (blue) Methyl Red (red) 4.8–6.0 (yellow) Bromothymol Blue (yellow) 6.0–7.6 (blue) Phenolphthalein (colorless) 8.2–10.0 (pink) 17.6 The Solubility Product Constant, Ksp In saturated solutions of ionic compounds, the dissolved ions are in equilibrium with the undissolved solid compound, forming a heterogeneous equilibrium. The equilibrium constants for these processes are known as solubility product constants, Ksp. For example, a saturated solution PbI2 is described by the equilibrium reaction: PbI2(s) ⇌ Pb2+(aq) + 2 I−(aq) Ksp = [Pb2+][I−]2 17.6 The Solubility Product Constant, Ksp Some Solubility Product Constants at 25℃ Formula Ksp Formula Ksp AgCl 1.77 × 10−10 Ca(OH)2 5.02 × 10−6 Al(OH)3 4.6 × 10−33 Ca(IO3)2 6.47 × 10−6 BaCO3 2.58 × 10−9 Ca3(PO4)2 2.07 × 10−33 BaF2 1.84 × 10−7 CaSO4 4.93 × 10−5 Ba(NO3)2 4.64 × 10−3 Cr(OH)3 3 × 10−29 Ba3(PO4)2 3.40 × 10−23 Co(OH)2 5.92 × 10−15 BaSO4 1.08 × 10−10 CoS 4.0 × 10−21 BaSO3 5.0 × 10−10 CuBr 6.27 × 10−9 CdF2 6.44 × 10−3 CuCl 1.72 × 10−7 CdS 1 × 10−27 CuI 1.27 × 10−12 CaCO3 3.36 × 10−9 CuS 6.3 × 10−26 17.6 The Solubility Product Constant, Ksp Example 1 - A saturated solution of iron(II) sulfide contains 4.0 × 10−10 M of Fe2+ and 4.0 × 10−10 M of S2−. Calculate Ksp for iron(II) sulfide. FeS(s) ⇌ Fe2+(aq) + S2−(aq) Ksp = [Fe2+][S2−] Ksp = (4.0 × 10−10)(4.0 × 10−10) = 1.6 × 10−19 17.6 The Solubility Product Constant, Ksp Notes of the value of Ksp Ksp is a measure of an ionic compound’s solubility. Lower Ksp values indicate a lower solubility of ionic compounds. Ksp can be used to calculate the molar solubility, which is the number of moles of substance that can dissolve in 1 L of solution. ICE tables can be used to determine the molar solubility, x. For example, consider the solubility of Ca(OH)2 Ksp = [Ca2+][OH−]2 Ca(OH)2(s) ⇌ Ca2+(aq) + 2 OH−(aq) Ksp = [x][2x]2 = 4x3 Initial (M) Excess 0 0 Change (M) −x +x +2x K sp Equilibrium (M) x 2x molar solubility = x = 3 4 17.6 The Solubility Product Constant, Ksp Example 2 – The solubility of Ba(NO3)2 is 27.50 g/L. Calculate the Ksp of Ba(NO3)2 Step 1 - Start by converting the solubility into molarity. 27.50 g  1 mol    = 0.1052 M 1 L  261.338 g  Step 2 - Construct an ICE table with molar solubility as x. Ba(NO3)2(s) ⇌ Ba2+(aq) + 2 NO3−(aq) Initial (M) Excess 0 0 Change (M) −x +x +2x Equilibrium (M) x 2x Step 3 - Write the Ksp expression, and use the molar solubility to determine the value of Ksp. Ksp = [Ba2+][NO3−]2 Ksp = (x)(2x)2 = 4x3 Ksp = 4(0.1052)3 = 4.657 × 10−3 17.7 The Common-Ion Effect and the Effect of pH on Solubility Consider the solubility of AgCl: AgCl(s) ⇌ Ag+(aq) + Cl−(aq) Ksp = 1.33 × 10−5 M. When 0.200 M NaCl is added to the AgCl solution, the additional Cl− shifts the equilibrium back to AgCl(s). Cl- is a common ion Example - Calculate the molar solubility of AgCl in a solution of 0.200 M NaCl, and compare it to the molar solubility of AgCl in pure water (1.33 × 10−5 M). The Ksp of AgCl is 1.77 x 10-10 Initial concentration of AgCl(s) ⇌ Ag+(aq) + Cl−(aq) Cl- from 0.200 M NaCl. Initial (M) Excess 0 0.200 Change (M) −x +x +x Equilibrium (M) x 0.200 + x Ksp = [Ag+][Cl−] = 1.77 × 10−10 M Ksp = [x][0.200 + x]≈ [x][0.200] Ksp = [x][0.200] = 1.77 × 10−10 x = 8.85 × 10−10 M 17.8 Precipitation: Q Versus Ksp 1. If Q < Ksp, the solution is not yet 1. 2. 3. saturated. More of the solid ionic compound can be dissolved. 2. If Q = Ksp, the solution is saturated and at equilibrium. 3. If Q > Ksp, the solution is past the point of saturation, and the excess ionic compound will precipitate from solution until Q = Ksp 17.8 Precipitation: Q Versus Ksp Example - Equal volumes of a solution of 0.0015 M AgClO4 and a solution of 0.0015 M NaCl are mixed. Will AgCl precipitate? The Ksp of AgCl is 1.77 x 10-10 AgCl(s) ⇌ Ag+(aq) + Cl−(aq) Q = [Ag+][Cl−] Q = (0.0015/2)(0.0015/2) = 5.63 × 10−7 Q > Ksp; therefore, AgCl will precipitate Chapter 16: Acids and Bases CHY102 F24  Chapter Outline Section 16.1 Ionization Section 16.6 Polyprotic Acids Reactions of Acids and Bases Section 16.7 Acid–Base Properties of Salts Section 16.2 Brønsted–Lowry Theory Section 16.8 Relating Acid Strength to Structure Section 16.3 Autoionization of Water Section 16.9 Lewis Acids and Bases Section 16.4 pH Calculations Section 16.5 Weak Acids and Bases  16.1: Ionization Reactions of Acids and Bases Arrhenius definition of acids and bases: Arrhenius acid: increases the concentration of hydrogen ions when dissolved in water HCl(aq) → H+(aq) + Cl−(aq) Arrhenius base: increases the concentration of hydroxide ion when dissolved in water NaOH(aq) → Na+(aq) + OH−(aq) H+ and OH− ions are formed by the ionization or dissociation of electrolytes, where a neutral compound is broken into ions. H+ consists of only a proton and is more likely to exist in water as a hydronium ion H3O+. We use H+ and H3O+ interchangeably  16.1: Ionization Reactions of Acids and Bases Acid Ionization: Acid formulas typically begin with H to indicate ionizable H atoms Strong acids dissociate completely in water (single arrow) HCl(aq) → H+(aq) + Cl−(aq) HCl(aq) + H2O(l) → H3O+(aq) + Cl−(aq) Weak acids do not dissociate completely in water (double arrow) HF(aq) ⇌ H+(aq) + F−(aq) HF(aq) + H2O(l) ⇌ H3O+(aq) + F−(aq)  16.1: Ionization Reactions of Acids and Bases Base Ionization Strong bases dissociate completely in water (single arrow) Produces OH- and a counterion NaOH(aq) → Na+(aq) + OH−(aq) Weak bases do not dissociate completely in water (double arrow) Typically are derivatives of NH3, ionize in water NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH−(aq)  16.1: Ionization Reactions of Acids and Bases List of Strong Acids: Strong Bases: HCl Strong bases are group 1 and heavier group 2 hydroxides HBr LiOH, NaOH, KOH, RbOH, HI CsOH, Ca(OH)2, Ba(OH)2, and HNO3 Sr(OH)2. H2SO4 Ca(OH)2, Ba(OH)2, and Sr(OH)2 HClO4 are dibasic and produce 2 OH− ions per formula unit. HClO3 Always check to see if you are dealing with a strong acid/base  16.1: Ionization Reactions of Acids and Bases Example – Write the ionization reactions for HNO3(aq), HNO2(aq), Ba(OH)2(aq), and CH3NH2(aq). Solution HNO3(aq): HNO2(aq): HNO3 (aq) → H+(aq) + NO3−(aq) HNO2 (aq) ⇌ H+(aq) + NO2−(aq) or or HNO3 (aq) + H2O(l) → H3O+(aq) + NO3−(aq) HNO2 (aq) + H2O(l) ⇌ H3O+(aq) + NO2−(aq) Ba(OH)2(aq): CH3NH2(aq): Ba(OH)2(aq) → Ba2+(aq) + 2 OH−(aq) CH3NH2(aq) + H2O(l) ⇌ CH3NH3+(aq) + OH−(aq)  16.2 Brønsted–Lowry Theory Brønsted-Lowry definitions of acids and bases were later introduced due to some circumstances where Arrhenius definitions of acids and bases do not apply A Brønsted acid is a proton donor. Molecule or ion that can release H+ into solution. A Brønsted base is a proton acceptor. Molecule or ion that can form a chemical bond with H+ Some compounds can both donate and accept protons (e.g., water), and can therefore act as both a Brønsted acid and a Brønsted base. Amphoteric substance  16.2 Brønsted–Lowry Theory Conjugate Acid-Base Pairs When a weak acid donates a proton, the remaining part of the acid is the conjugate base, which can now accept a proton. When a weak base accepts a proton, it becomes the conjugate acid of that base and can now donate a proton.  16.2 Brønsted–Lowry Theory Acid and base strength determines to what extent an acid or a base reacts with water to form hydronium ions or hydroxide ions. Strong acids and bases react entirely with water, whereas weak acids and bases ionize to varying degrees. Thus, conjugate bases and acids also react (as bases or acids) to different extents. The stronger the weak acid or weak base is, the weaker its conjugate base or conjugate acid. The weaker the weak acid or weak base is, the stronger its conjugate base or conjugate acid. Example - The following weak acids are listed in order of acid strength. HClO > HBrO > HIO Identify their conjugate bases, and list them in order of base strength from strongest to weakest. Solution The conjugate bases are ClO−, BrO−, and IO−. Order of base strength: IO− > BrO− > ClO−  16.3: Autoionization of Water There are always some hydronium ions and some hydroxide ions present in every aqueous solution, whether it is acidic, basic, or neutral This is called autoionization of water H2O(l) + H2O(l) ⇌ H3O+(aq) + OH−(aq) The water ionization constant, Kw is an equilibrium constant specific to the autoionization of water: Kw = [H3O+][OH−] At 25℃, for any solution, Kw = 1.0 × 10−14 At 25℃ , in pure water, [H3O+] = [OH−] = 1.0 × 10−7  16.3: Autoionization of Water The acidity or basicity of a solution depends on the relative amounts of hydronium and hydroxide ions: [H3O+] > [OH−] Acidic solution always true regardless of [H3O+] < [OH−] Basic solution temperature [H3O+] = [OH−] Neutral solution Example - Calculate [OH−] of Ketchup if [H3O+] = 1.3 × 10−4 M and classify it as acidic, basic, or neutral Solution - Solve the Kw expression for [OH−] Kw = [H3O+][OH−] = 1.0 × 10−14 Kw 1.0×10−14 OH  = − = −4 = 7.7 ×10 −11 M H3O  1.3×10 + [H3O+] > [OH−], so the solution is acidic.  16.4: pH Calculations The pH scale was developed as a convenient way to express H3O+ concentration. pH = −log[H+] = −log[H3O+] pH values range from 1 to 14 pH < 7 Acidic solution only true at 25°C pH = 7 Neutral solution (assume 25°C if T is not specified) pH > 7 Basic solution Example – Determine the pH of the following solutions: 1.0 × 10−5 M HCl 7.2 × 10−2 M NaOH Solution: First, solve Kw for [H3O+]; then use the pH definition pH = −log[H3O+] = −log(1.0 × 10−5 M) = 5.00 Kw 1.0×10−14 −13 [H3O ] = + = −2 = 1.4 ×10 M → acidic OH  7.2×10 − pH = −log[H3O+] = −log(1.4 × 10−13 M) = 12.86 →basic  16.4: pH Calculations If we know the pH of a solution, we can calculate the hydrogen ion concentration using the relationship: [H3O+] = 10−pH Similar to the pH scale, the pOH scale quantifies the hydroxide ion concentration of a solution: pOH = −log[OH−] [OH−] = 10−pOH And pH + pOH = 14.00  16.4: pH Calculations Example - Calculate the hydronium ion concentration and the pH in a) a 0.1 M HCl solution, and b) a 0.1 M NaOH solution Solution a) HCl is a strong acid, therefore: [H3O+] = [HCl] = 0.1 M pH = −log [H3O+] = −log(0.1 M) = 1.0 b) NaOH is a strong base, therefore: [OH−] = [NaOH] = 0.1 M. + Kw 1.0  10−14 [H3O ] = − = = 1  10−13 M [OH ] 0.1 pH = −log [H3O+] = −log(1 × 10−13 M) = 13.0  16.5: Weak Acids and Bases Calculating [H3O+] in strong acid solutions is a straightforward process using the concentration of the strong acid, because it 100% dissociates Weak acids and weak bases that only partially ionize are in an equilibrium with an equilibrium constant For the general ionization reaction of a weak acid (HA) into its conjugate base (A-) HA(aq) ⇌ H+(aq) + A−(aq) The acid ionization constant, Ka, applies: [H3O+ ][ A − ] Ka = [HA ]  16.5: Weak Acids and Bases For the general ionization reaction of a weak base (B) to form its conjugate acid (BH+): B(aq) + H2O(l) ⇌ BH+(aq) + OH−(aq) The base ionization constant, Kb, applies: [BH+ ][OH− ] Kb = [B] In general, larger Ka and Kb values indicate more complete ionization and can therefore be used to compare strengths of different weak acids or weak bases Larger Ka = stronger acid Larger Kb = stronger base  Acid Formula Ka Acetic acid HC2H3O2 1.8 × 10−5 Strongest acid in table Boric acid HBH2O3 5.8 × 10−10 Formic acid HCHO2 1.7 × 10−4 Strongest base in table Hydrofluoric acid HF 6.8 × 10−4 Base Formula Kb Nitrous acid HNO2 4.0 × 10−4 Ammonia NH3 1.8 × 10−5 Phenol HOC6H5 1.1 × 10−10 Methylamine CH3NH2 5.0 × 10−4 Dimethylamine (CH3)2NH 5.4 × 10−4 Weakest acid in table Trimethylamine (CH3)3N 6.3 × 10−5 Pyridine C5H5N 1.7 × 10?