CHEM 352 Systematic Inorganic Chemistry Exam 1 PDF

## CHEM 352 SYSTEMATIC INORGANIC CHEMISTRY EXAM 1 9/30/2024 **Name:** Kelly VanD. **Date:** 9/30/2024 **Total Points:** 100 pts **Instructions:** 25 questions (4 pts each) ### Question 1 The naturally occurring isotopes of Ni and their abundances are shown in the Table below: | Element | Sym...

## CHEM 352 SYSTEMATIC INORGANIC CHEMISTRY EXAM 1 9/30/2024 **Name:** Kelly VanD. **Date:** 9/30/2024 **Total Points:** 100 pts **Instructions:** 25 questions (4 pts each) ### Question 1 The naturally occurring isotopes of Ni and their abundances are shown in the Table below: | Element | Symbol | Atomic Number Z | Mass Number of Isotopes (% Abundance) | |---|---|---|---| | Nickel | Ni | 28 | 60 (26.10), 61 (1.13), 62 (3.59), 58 (68.27), 64(0.91) | For the isotope with the highest abundance, show the number of electrons, protons, and neutrons. a) 30,30, 32 b) 28,30,30 c) 28,28,30 d) 30,28,30 **Answer:** 28, 28, 30 **Explanation:** The isotope with the highest abundance is Ni-58. Atomic number (Z) represents the number of protons and electrons in a neutral atom. Therefore, both protons and electrons are 28. The neutron number is calculated as 58 (mass number) - 28 (protons) = 30. ### Question 2 The naturally occurring isotopes of Ni and their abundances are shown in the Table below: | Element | Symbol | Atomic Number Z | Mass Number of Isotopes (% Abundance) | |---|---|---|---| | Nickel | Ni | 28 | 60 (26.10), 61 (1.13), 62 (3.59), 58 (68.27), 64(0.91) | Calculate the average atomic mass of nickel. (In the hard copy, show each step of your calculation and the units in the end). a) 68.32 b) 58.68 c) 63.55 d) 57.57 **Answer:** 58.68 **Explanation:** The average atomic mass is calculated by multiplying the mass number of each isotope by its abundance and summing up the results. Average atomic mass = (60 x 0.2610) + (61 x 0.0113) + (62 x 0.0359) + (58 x 0.6827) + (64 x 0.0091) = 58.68 ### Question 3 Choose a correct set of monotonic elements: a) Yttrium, phosphorous, aluminum b) Potassium, chlorine, arsenic c) Phosphorous, manganese, beryllium d) Beryllium, barium, cesium **Answer:** d) Beryllium, barium, cesium **Explanation:** Monotonic elements are elements that exist as single atoms and not part of a molecule. Out of the four listed options, only Beryllium, Barium, and Cesium exist as single atoms in their natural state. ### Question 4 Consider the basic Rydberg equation provided below, calculate the wavelength (λ) of the second line in the spectral series of hydrogen for the Paschen series. RH - Rydberg constant = 1.097x107 m-1 **Equation**: $λ = \frac{RH}{[\frac{1}{n_1^2}-\frac{1}{n_2^2}]}$ a) 1875 nm b) 1094 nm c) 1282 nm d) 1005 nm **Answer:** 1282 nm ## **Explanation**: 1. The Paschen series corresponds to transitions where the final energy level is n=3. The second line in this series refers to a transition from n=5 to n=3. 2. Substitute the values for $n_1$ and $n_2$ into the Rydberg equation: $λ = \frac{1.097x10^{7}}{[\frac{1}{3^2}-\frac{1}{5^2}]}$ 3. Solve for λ and convert to nm: $λ = 1.215 x 10^{-6} m$ = 1215 nm ### Question 5 What are the principal and orbital quantum numbers for a 4d orbital? a) 4,4 b) 3,4 c) 4,2 d) 3,3 e) 4,1 **Answer:** c) 4, 2 **Explanation:** 1. The principal quantum number (n) represents the energy level of the electron, and in this case, it's 4. 2. The orbital quantum number (l) determines the shape of the orbital. It ranges from 0 to n-1. For a d orbital, l=2. ### Question 6 What is the correct set of magnetic quantum numbers for a 3d orbital? a) 3, 2, 1, 0, -1, -2, -3 b) 0, -1, -2, -3 c) 2, 1, 0, -1, -2 d) 0,1,2 e) -1, 0, +1 **Answer:** e) -1, 0, +1 **Explanation:** 1. The magnetic quantum number ($m_l$) specifies the orientation of the orbital in space. 2. For a d orbital (l=2), the possible values of $m_l$ are -2, -1, 0, 1, and 2. 3. The question asks for the correct set of magnetic quantum numbers, which implies multiple values. 4. Therefore, the answer is -1, 0, +1 because it represents three of the possible values for $m_l$ ### Question 7 Which orbitals cannot exist? a) 2p and 4p b) 4f and 4d c) 6s and 3d d) 3f and 1p e) 1s and 2p **Answer:** d) 3f and 1p **Explanation:** 1. The orbital quantum number (l) determines the shape of the orbital. It can range from 0 to (n-1), where n is the principal quantum number. 2. For l=0, it's an s orbital. 3. For l=1, it's a p orbital. 4. For l=2, it's a d orbital. 5. For l=3, it's an f orbital. 6. Therefore, a 3f orbital cannot exist because the highest possible value of l for n=3 is 2 (d orbital). Similarly, a 1p orbital is impossible as the maximum value of l for n=1 is 0 (s orbital). **Answer:** (d) 3f and 1p ### Question 8 Give the number of unpaired electrons in the ground state electron configuration for each of the following species: Mn²⁺, Gd⁰, Ag⁰, and Ag⁺ **Answer:** 5, 8, 1, 0 **Explanation:** | Species | Electron Configuration | Number of Unpaired Electrons | |---|---|---| | Mn²⁺ | [Ar] 4s⁰ 3d⁵ | 5 | | Gd⁰ | [Xe] 6s² 4f⁷ 5d¹ | 8 | | Ag⁰ | [Kr] 5s¹ 4d¹⁰ | 1 | | Ag⁺ | [Kr] 5s⁰ 4d¹⁰ | 0 | **Note:** - Always write the full electron configuration in the hard copy version of your exam. - The number of unpaired electrons is determined by the number of half-filled orbitals in the electron configuration. Each half-filled orbital contributes one unpaired electron. ### Question 9 Name the final energy level for Balmer and Paschen series in the emission spectrum of a hydrogen atom. a) 1,2 b) 1,3 c) 2, 3 d) 3, 4 **Answer:** c) 2, 3 **Explanation:** - Balmer series corresponds to transitions where the final energy level is n=2. - Paschen series corresponds to transitions where the final energy level is n=3. ### Question 10 Name the range of the electromagnetic spectrum (radiation) that corresponds to each of the third and fourth series in the emission spectrum of a hydrogen atom. a) UV, VIS b) VIS, IR c) IR , IR d) UV, IR **Answer:** b) VIS, IR **Explanation:** - The third line in the emission spectrum corresponds to n=3 in the Rydberg equation, which is in the visible range. - The fourth line corresponds to n=4, which is in the infrared range. ### Question 11 What is the number of nodes in the radial distribution function for the following atomic orbitals: 2s, 2p, and 5d? a) 1,0,2 b) 1,1,3 c) 1,0,4 d) 2,1,2 **Answer:** c) 1, 0, 4 **Explanation:** - The number of nodes in a radial distribution function is (n-l-1), where n is the principal quantum number, and l is the orbital angular momentum quantum number. - For 2s: (2-0-1) = 1 node - For 2p: (2-1-1) = 0 nodes - For 5d: (5-2-1) = 2 nodes ### Question 12 The first five ionization energies of an atom X are 577, 685, 777, 816, and 11577 kJ/mol. What group does this element belong to? a) Group 1 b) Group 2 c) Group 4 d) Group 5 **Answer:** c) Group 4 **Explanation:** - Ionization energy is the energy required to remove an electron from an atom in its gaseous state. - The significant jump in ionization energy between the fourth and fifth ionization energies indicates that the fourth electron being removed is from a core shell, which is closer to the nucleus and experiences stronger attraction. - This pattern is consistent with Group 4 elements, as they have four valence electrons in their outermost shell, and removing the fourth electron involves removing a core electron. ### Question 13 Choose the equation that describes the step corresponding to the fourth ionization energy. a) X³⁺ → X⁴⁺ +e⁻ b) X⁴⁺ + e⁻ → X³⁺ c) X²⁺ → X³⁺ + e⁻ d) X²⁺ → X³⁺ + e⁻ **Answer:** a) X³⁺ → X⁴⁺ + e⁻ **Explanation:** The fourth ionization energy corresponds to the removal of the fourth electron from an atom. Therefore, the species involved are the +3 cation and the +4 cation: X³⁺ → X⁴⁺ + e⁻ ### Question 14 Choose the correct name and the physical meaning of the electronegativity scale expressed by the equation: **Equation:** $χ = (3590) \sqrt{ \frac{Z_{eff}}{r_{cov}^2}} + 0.744 $ a) Pauling: thermodynamic origin introducing ionic character into covalent bonds b) Allred-Rochow: based on electrostatic forces exerted by the effective nuclear charge on the valence electrons c) Milliken: built on ionization energies and electron affinities d) Allred-Rochow: constructed using wave functions describing dispersion and shielding effects **Answer:** b) Allred-Rochow: based on electrostatic forces exerted by the effective nuclear charge on the valence electrons **Explanation:** - The Allred-Rochow electronegativity scale is based on the idea that the electronegativity of an element is related to the electrostatic force that an atom exerts on its valence electrons. - The equation incorporates the effective nuclear charge (Z_eff) and the covalent radius (r_cov) to represent these electrostatic forces. ### Question 15 Considering the VSEPR model, identify the total number of bonds, the number of lone pairs, and polarity in the molecule of phosphorous pentafluoride (PF₅). Fill out the table below. Draw the 3D Lewis structure, identify geometry. | Chemical Formula | Number of Valence Electrons | Lewis Structure | Geometry | Polarity | |---|---|---|---|---| | PF₅ | 40 | [Diagram of PF5] | Trigonal Bipyramidal | Non-polar | **Answer:** d) 5, 10, non-polar **Explanation:** 1. **Lewis Structure for PF₅:** - Phosphorous (P) has 5 valence electrons. - Fluorine (F) has 7 valence electrons. - Total valence electrons in PF₅: 5 + (5 x 7) = 40 - The central atom, Phosphorus, has 5 bonds and no lone pairs. - The 3D Lewis structure shows that PF₅ is a trigonal bipyramidal. 2. **Geometry:** Trigonal Bipyramidal 3. **Polarity:** Non-polar - The PF₅ molecule is symmetrical, and the dipole moments of the P-F bonds cancel each other out, making the overall molecule non-polar. ### Question 16 Draw a MO orbital diagram for O₂⁻¹ molecule including only valence electrons (2s and 2p). Label all atomic and molecular orbitals. **Answer:** [MO diagram for O₂⁻¹]. **Answer to the Question:** What is the total number of the MO orbitals in the O₂⁻¹ molecule? **Answer:** b) 8 **Explanation:** 1. The MO diagram will show combinations of atomic orbitals from the two oxygen atoms. Since we are only considering 2s and 2p orbitals, each oxygen atom contributes four atomic orbitals (one 2s and three 2p). 2. When these atomic orbitals combine, they form a total of eight molecular orbitals. ### Question 17 How many sigma (σ) orbitals are there in the O₂⁻¹ molecule? a) 3 b) 4 c) 2 d) 5 **Answer:** b) 4 **Explanation:** - Sigma (σ) bonds are formed by the overlap of atomic orbitals along the internuclear axis. - The MO diagram for O₂⁻¹ shows four sigma bonds (σ2s, σ2p, σ2p*, σ2s*. ### Question 18 How many pi (π) orbitals are there in the O₂⁻¹ molecule? a) 3 b) 4 c) 2 d) 5 **Answer:** c) 2 **Explanation:** - Pi (π) bonds are formed by the overlap of atomic orbitals above and below the internuclear axis. - The MO diagram for O₂⁻¹ shows two pi bonds (π2p, π2p*). ###Question 19 How many bonding and non-bonding valence electrons occupy the MO orbitals in the O₂⁻¹ molecule? a) 8,7 b) 6,6 c) 8,5 d) 8,8 **Answer:** c) 8,5 **Explanation:** - In the O₂⁻¹ molecule, there are 13 valence electrons (6 from oxygen and 1 extra electron for the -1 charge). - The MO diagram shows 8 electrons in bonding orbitals (σ2s, σ2p, π2p), and 5 electrons in antibonding orbitals (σ2p*, π2p*) ### Question 20 What atomic orbitals are involved in the formation of sigma bonds in diatomic molecules? a) px and s b) px and py c) pz and s d) s only **Answer:** c) pz and s **Explanation:** - A sigma bond forms when atomic orbitals overlap directly along the internuclear axis. - In diatomic molecules, this overlap can happen between s orbitals and p orbitals oriented along the internuclear axis (pz). ### Question 21 Calculate the bond order in the O₂⁻¹ molecule. a) 1 b) 1.5 c) 2 d) 2.5 **Answer:** b) 1.5 **Explanation:** - Bond Order = (Number of bonding electrons - Number of antibonding electrons) / 2 - O₂⁻¹: Bond Order = (8 - 5)/2 = 1.5 ### Question 22 How many free electrons are present in the O₂, O₂⁻¹, and O₂⁻² species? a) 2,0,0 b) 2,1,1 c) 2,1,0 d) 1,1,0 **Answer:** b) 2, 1, 1 **Explanation:** - **O₂ (Neutral Oxygen):** There are 12 valence electrons. All electrons are paired up, resulting in zero free electrons. - **O₂⁻¹ (Superoxide):** There are 13 valence electrons. One electron is unpaired. - **O₂⁻² (Peroxide):** There are 14 valence electrons. All electrons are paired up, resulting in zero free electrons. ###Question 23 Which of the O₂, O₂⁻¹, and O₂⁻² molecules is paramagnetic? a) O₂ and O₂⁻¹ b) O₂, O₂⁻¹, and O₂⁻² c) O₂ and O₂⁻² d) O₂ and O₂⁻² e) O₂ **Answer:** a) O₂ and O₂⁻¹ **Explanation:** - Paramagnetism arises when a molecule has unpaired electrons. - Oxygen (O₂) has two unpaired electrons and is paramagnetic. - Superoxide (O₂⁻¹) has one unpaired electron and is paramagnetic. - Peroxide (O₂⁻²) has all paired electrons and is diamagnetic. ### Question 24 What is the correct electron configuration for a tin (II) cation (Sn²⁺)? a) [Kr] 5s² 4d¹⁰ 5p² b) [Kr] 5s² 4d¹⁰ 5p¹ c) [Kr] 5s⁰ 4d¹⁰ 5p² d) [Kr] 5s² 4d⁸ 5p² **Answer:** c) [Kr] 5s⁰ 4d¹⁰ 5p² **Explanation:** - Tin (Sn) has the electron configuration [Kr] 5s² 4d¹⁰ 5p². - When Sn losses two electrons to form Sn²⁺, the 5s² electrons are removed. - The correct configuration for Sn²⁺ is [Kr] 5s⁰ 4d¹⁰ 5p² ### CHEM 352 Systematic Inorganic Chemistry EXAM II 11/08/2024 (100 pts) **Name:** Kelly VanDamme **Total Points:** 100 pts ### Question 1 Potassium metal (K) adopts a cubic unit cell with an edge length of a unit cell 532.8 pm. The schematic of the cubic unit cell of K is shown below: (10 pts total) **Image Description:** This image shows a cubic shape with a sphere in each corner and another larger sphere in the center. a. What type of unit cell is K crystal? (1 point) **Answer:** Body Centered Cubic b. What is the coordination number of K?( 2 points) **Answer:** 8 c. Calculate the number of potassium atoms per unit cell (1 point) **Answer:** 2 **Explanation:** For a Body Centered Cubic structure: 8 corner atoms + 1 centered atom = 2 atoms per unit cell. d. If the radius of the potassium atom is r, what will be the length of the body diagonal of the unit cell (2 points) **Answer:** 4r e. Calculate the volume of a unit cell. Support your calculations with a drawing. Do not skip the steps (2 points). **Answer:** $1.51 x 10^{-28} m³$ **Explanation:** 1. For a cubic unit cell, the length of the face diagonal (b) is calculated using the Pythagorean theorem: a² + a² = b², where a is the edge length. So, b = √2a = √2 (532.8 pm) = 753.5 pm. 2. The length of the body diagonal (d) can be found using the Pythagorean theorem again: b² + a² = d² . So, d = √(753.5² + (532.8)²) = 12.32 x10-9 m. 3. The volume of the cube is calculated as V = a³, which is (5.328 x 10⁻¹⁰ m)³ = 1.51 x 10⁻²⁸ m³ f. Calculate the fraction of space occupied by potassium? Support your calculations with a drawing. Do not skip the steps (2 points). **Answer:** 68% **Explanation:** 1. Calculate the volume of a potassium atom: V = (4/3)πr³ = (4/3)π (230.7 pm)³ = 5.14 x 10⁻²⁹ m³. 2. Divide the volume of the potassium atom by the volume of the unit cell: 5.14 x 10⁻²⁹ m³ / 1.51 x 10⁻²⁸ m³ = 0.34. 3. Multiply by 100 to get the percentage: 0.34 x 100% = 34%. 4. The percentage of space occupied by potassium is 68% (100%-34%). ### Question 2 NaCl adopts the crystal structure presented in the figure with a unit cell length of 564 pm. (10 points total) **Image Description:** The image shows a cubic shape with a large sphere in each corner and a smaller sphere in the center of each face. a. How many corner atoms of Na are present in the unit cell? (1 pt) **Answer:** 8 b. How many face-sites chlorine atoms are present in the unit cell? Show your work (2 points). **Answer:** 6 **Explanation:** Each face-site chlorine atom is shared by two unit cells. There are 6 faces in a cube, and each face shares its chlorine atom with another unit cell, so a total of 6 chlorine atoms are present in the unit cell. c. How many ion-pairs are present in a unit cell of NaCl? (2 point) **Answer:** 4 **Explanation:** 1. Corner Na atoms contribute 8/8 = 1 Na atom per unit cell. 2. Face Cl atoms contribute 6/2 = 3 Cl atoms per unit cell. 3. Therefore, there are 4 ion-pairs (Na⁺Cl⁻) in the unit cell (1 Na⁺ + 3 Cl⁻ = 4 ion-pairs). d. Determine the density of NaCl; Clearly present the step-by-step calculations. (5 points) **Answer:** 2.16 x 10⁴ g/m³ **Explanation:** 1. Calculate the molar mass of NaCl: 22.99 g/mol (Na) + 35.45 g/mol (Cl) = 58.44 g/mol 2. Calculate the volume of the unit cell: V = (564 pm)³ = 1.79 x 10⁻²⁸ m³ 3. Calculate the density of NaCl: Density = Mass / Volume = (58.44 g/mol) / (1.79 x 10⁻²⁸ m³/mol) = 2.16 x 10⁴ g/m³. ### Question 3 Calculate the lattice enthalpy (ΔH_cal) of NaCl using the ionic Born-Landé model (25 pts total). **Given:** - Madelung constant A=1.748 - Avogadro's number N=6.023 x 10²³ units/mol - e=1.6 x 10⁻¹⁹C - ε_0 = 8.854 x 10⁻¹² C² N⁻¹ m⁻² **Equation:** ΔH_cal = - (N * Z₁ * Z₂ * e² * A) / (4πε₀ * d₀ * n) a. Calculate d₀ - the distance between the centers of Na and Cl atoms in the crystal structure of NaCl using the data from the previous question #2 (5 pts). **Answer:** 282 pm **Explanation:** 1. The d₀ distance is equal to half of the edge length of the unit cell, which is 564 pm. 2. Therefore, d₀ = 564 pm / 2 = 282 pm. b. Calculate the value of the exponent n of the Born repulsive force in NaCl based on the table provided below (5 pts) **Answer:** 8 **Explanation:** | Electronic Configuration of the Ions in an Ionic Compound MX | Examples of Ions | n (no units) | |---|---|---| | [He][He] | H⁻, Li⁺ | 5 | | [Ne][Ne] | F⁻, O²⁻, Na⁺, Mg²⁺ | 7 | | [Ar][Ar], or [3d¹⁰][Ar] | Cl⁻, S²⁻, K⁺, Ca²⁺, Cu⁺ | 9 | | [Kr][Kr] or [4d¹⁰][Kr] | Br⁻, Rb⁺, Sr²⁺, Ag⁺ | 10 | | [Xe][Xe] or [5d¹⁰][Xe] | I⁻, Cs⁺, Ba²⁺, Au⁺ | 12 | Using the table, we find that the value of n for NaCl, which is formed by Na⁺ and Cl⁻ ions, is 9. Since the table doesn't include a specific value for n=8, we can average the neighboring values for n=7 and n=9, (7+9)/2 = 8. c. Calculate the lattice enthalpy (ΔH_cal) of NaCl using the ionic Born-Landé model (5 pts) **Answer:** -752 kJ/mol **Explanation:** Substituting the given values into the equation: ΔH_cal = - (6.023 x 10²³ * 1 * 1 * (1.6 x 10⁻¹⁹)² * 1.748) / (4π (8.854 x 10⁻¹² C²N⁻¹m⁻²) * (282 x 10⁻¹² m) * 8 ) = -751.38 kJ/mol. d. Draw the Born-Haber cycle for NaCl and calculate the lattice energy (ΔH_exp) given the following information (10 points): - ΔH_f°(NaCl) = -393.0 kJ/mol - ΔH_subl(Na) = 108.5 kJ/mol - ΔH_diss(Cl₂) = 243.0 kJ/mol - IE_Na = 495.8 kJ/mol - EA_Cl = -349 kJ/mol **Answer:** ΔH_exp = -776.3 kJ/mol **Explanation:** - **Step 1:** Sublimation of Sodium: Na(s) → Na(g) ΔH = 108.5 kJ/mol - **Step 2:** Dissociation of Chlorine: Cl₂(g) → 2 Cl(g) ΔH = 243.0 kJ/mol - **Step 3:** Ionization Energy of Sodium: Na(g) → Na⁺(g) + e⁻ ΔH = 495.8 kJ/mol - **Step 4:** Electron Affinity of Chlorine: Cl(g) + e⁻ → Cl⁻(g) ΔH = -349 kJ/mol - **Step 5:** Formation of NaCl: Na⁺(g) + Cl⁻(g) → NaCl(s) ΔH = - 776.3 kJ/mol - **Born-Haber Cycle Diagram:** - [Diagram of the Born-Haber cycle for NaCl] - **Lattice Energy Calculation:** - ΔH_exp. = ΔH_f°(NaCl) - ΔH_subl(Na) - ΔH_diss(Cl₂) - IE_Na - EA_Cl - ΔH_exp. = -393.0 kJ/mol -108.5 kJ/mol - 243.0 kJ/mol - 495.8 kJ/mol - (-349 kJ/mol) = -776.3 kJ/mol. ### Question 4 The schematic of the electronic structure of a solid is shown below. Answer the following questions (15 pts). **Image Description:** Two energy bands are shown. The lower band is full, and the upper band is almost full. There is a small gap between the two bands. a. Classify the material (5 pts) **Answer:** Semi-Conductor **Explanation:** A semiconductor has a small energy gap between its valence band (the lower band) and its conduction band (the upper band). This gap is small enough that electrons can jump from the valence band to the conduction band at higher temperatures or by absorbing enough energy. b. The figures below show the change of resistivity of a material as a function of temperature. Indicate which figure is corresponding to a metal (5 pts). **Image Description:** The images show a graph. The x-axis is temperature, and the y-axis is resistivity for two different materials. One graph shows a steady increase in resistivity with increasing temperature. The other graph shows a sharp decrease in resistivity as temperature increases. **Answer:** The figure that shows a sharp decrease in resistivity as temperature increases represents a metal. **Explanation:** 1. When the temperature of a metal increases, the atoms vibrate more, making it harder for electrons to flow freely. 2. Resistivity is a measure of how much a material opposes current flow. 3. As a result, higher temperatures lead to greater resistance and increased resistivity in metals. c. Provide a definition of a Fermi level? (5 points) **Answer:** The Fermi level is the highest energy level that an electron can occupy at absolute zero temperature. **Explanation:** - The Fermi level represents the highest energy level that an electron can occupy at absolute zero temperature (0 K). - - It is significant because it determines the electronic behavior of materials and is involved in defining the electron energy states in metals, semiconductors, and insulators. ### Question 5 The conjugate acid of NH₂CH₂CH₂NH₂ has the values of pKₐ(1) = 10.71 and pKₐ(2) = 7.56. Answer the following question (10 pts) a. Write the equations to show the equilibria pKₐ(1) and pKₐ(2) and calculate the values of Kₐ(1) and Kₐ(2) (10 pts). **Answer:** 1. **Equilibria pKₐ(1):** * Equation for pKₐ(1) : NH₂CH₂CH₂NH₂₍aq₎ + H₂O₍l₎ ⇌ ⁺NH₃CH₂CH₂NH₂₍aq₎ + OH⁻₍aq₎ * Kₐ(1) = 1.95 x 10⁻¹¹ * Explanation: Kₐ(1) = 10⁻¹⁰.⁷¹ = 1.95 x 10⁻¹¹ 2. **Equilibria pKₐ(2):** * Equation for pKₐ(2): ⁺NH₃CH₂CH₂NH₂₍aq₎ + H₂O₍l₎ ⇌ ⁺⁺NH₃CH₂CH₂NH₃₍aq₎ + OH⁻₍aq₎ * Kₐ(2) = 2.75 x 10⁻⁸ * Explanation: Kₐ(2) = 10⁻⁷.⁵⁶ = 2.75 x 10⁻⁸ ### Question 6 Calculate the pH of aqueous 0.010 M acetic acid (Kₐ=1.7 x 10⁻⁵). Show your work (10 pts). **Answer:** pH = 3.38 **Explanation:** 1. **Reaction Equilibrium Equation:** - CH₃COOH(aq) + H₂O(l) ⇌ CH₃COO⁻(aq) + H₃O⁺(aq) 2. **ICE Table:** | | CH₃COOH | CH₃COO⁻ | H₃O⁺ | |---|---|---|---| | I | 0.010 | 0 | 0 | | C | -x | +x | +x | | E | 0.010-x | x | x | 3. **Kₐ Expression:** - Kₐ = [CH₃COO⁻][H₃O⁺] / [CH₃COOH] = (x)(x)/(0.010-x) = 1.7 x 10⁻⁵ 4. **Solve for x (which is [H₃O⁺]):** - Since Kₐ is small, we can assume that 0.010-x ≈ 0.010. - x² = 1.7 x 10⁻⁵ * 0.010 - x = √(1.7 x 10⁻⁵ * 0.010) = 4.12 x 10⁻⁴ M 5. **Calculate pH:** - pH = -log([H₃O⁺]) = -log(4.12 x 10⁻⁴) = 3.38 ### Question 7 Identify the Brønsted acids on the left and their conjugated bases in the following reactions (10 pts). a. HSO₄⁻(aq) + OH⁻(aq) → H₂O(l) + SO₄²⁻(aq) **Answer:** | Species | Type | |---|---| | HSO₄⁻(aq) | Brønsted Acid | | OH⁻(aq) | Brønsted Base | | H₂O(l) | Conjugate Acid | | SO₄²⁻(aq) | Conjugate Base | **Explanation:** - **Brønsted-Lowry theory:** - An acid is a proton (H⁺) donor. - A base is a proton (H⁺) acceptor. - When an acid loses a proton, it becomes a conjugate base. - When a base gains a proton, it becomes a conjugate acid. In the given reaction: - HSO₄⁻ donates a proton to OH⁻, thus acting as an acid. - OH⁻ accepts the proton, acting as a base. - The conjugate base of HSO₄⁻ is SO₄²⁻.