CHEM 119 Fall 2026 - Exam 2 Review #2 (Chapters 4 -8)
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Overview
Alicia Altemose reviews CHEM 119 Exam 2 topics with worked problems on the photoelectric effect, acids and bases, net ionic equations, solution stoichiometry, titrations, and electron configurations. The examples connect core rules to calculations, including a barium binding energy of 4.30 × 10⁻¹⁹ J, a 127 mL HCl titration, and electron-configuration checks for cobalt and manganese.
Key takeaways
- Photoelectric emission depends on a threshold: shorter wavelength means higher photon energy, and once the metal's binding energy is exceeded, the surplus becomes electron kinetic energy.
- For a photoelectron, calculate KE as photon energy minus binding energy, then use KE = ½mv²; the example yields a speed of about 6.26 × 10⁵ m/s.
- Only aqueous soluble ionic compounds and strong aqueous acids or bases split into ions in a total ionic equation; weak H₃PO₄ and liquid H₂O remain intact.
- Solution stoichiometry proceeds through moles: multiply molarity by volume in liters, apply the balanced-equation mole ratio, then divide by total mixed volume when finding final molarity.
- At acid-base equivalence, match H⁺ and OH⁻ amounts rather than assuming equal acid and base molecule counts; Ba(OH)₂ requires two moles of HCl per mole of base.
- An excited-state configuration must preserve the atom's electron count and obey subshell capacity; cobalt's [Ar] 4s¹ 3d⁸ is excited relative to [Ar] 4s² 3d⁷.
Chapters
- Photons at 450 nm eject electrons while 650 nm photons do not, placing the metal's threshold wavelength between those values.
- Because wavelength and photon energy are inversely related, 400 nm light has enough energy to eject electrons and produces greater electron kinetic energy than 450 nm light.
- Use c = λν and λ = c/ν; a frequency of 8.57 × 10¹⁴ Hz corresponds to approximately 350 nm.
- Convert nanometers to meters before using E = hc/λ, and track units so the result is in joules.
- The 450 nm and 650 nm photons have energies of about 4.42 × 10⁻¹⁹ J and 3.06 × 10⁻¹⁹ J, respectively.
- Barium's binding energy, 4.30 × 10⁻¹⁹ J, lies between those values; silver requires more energy, so 450 nm light is insufficient to eject its electrons.
- For a fixed metal, write photon energy as hc/λ = E_bind + KE, then use the same binding energy in both experiments.
- With λ₁ = 2λ₂ and kinetic energies 1.6 × 10⁻¹⁹ J and 6.4 × 10⁻¹⁹ J, algebra gives hc/λ₂ = 9.6 × 10⁻¹⁹ J and E_bind = 3.2 × 10⁻¹⁹ J.
- The seven strong acids are HCl, HBr, HI, HNO₃, H₂SO₄, HClO₃, and HClO₄; HF is weak, not strong.
- Strong hydroxide bases include LiOH, NaOH, KOH, RbOH, CsOH, Sr(OH)₂, and Ba(OH)₂; the solubility chart lists these ionic compounds.
- Split only aqueous soluble ionic compounds and aqueous strong acids or bases into ions in a total ionic equation.
- Keep weak acids such as H₃PO₄ and liquids such as H₂O intact; use the solubility chart to check ionic compounds and polyatomic-ion formulas.
- For H₃PO₄ + 3KOH → K₃PO₄ + 3H₂O, keep weak H₃PO₄ and liquid water intact while splitting aqueous KOH and K₃PO₄.
- K⁺ appears unchanged on both sides and is the spectator ion; OH⁻ and PO₄³⁻ are not spectators.
- A metal's threshold frequency represents the minimum photon frequency needed for electron ejection, so E_bind = hν_threshold.
- Using the example threshold gives a binding energy of 7.23 × 10⁻¹⁹ J; the frequency units cancel seconds in Planck's constant.
- Subtract the 7.23 × 10⁻¹⁹ J binding energy from a 9.01 × 10⁻¹⁹ J photon to obtain electron kinetic energy of 1.78 × 10⁻¹⁹ J.
- Substitute into KE = ½mv² with electron mass 9.11 × 10⁻³¹ kg; the resulting speed is about 6.26 × 10⁵ m/s.
- The exam covers electron configurations for s- and p-block ions, including groups 1, 2, and 13 through 18, but not transition-metal ions.
- For example, P³⁻ has the same 18-electron configuration as argon, written as [Ne] 3s² 3p⁶.
- Mixing 0.0300 L of 0.100 M CaCl₂ with 0.0150 L of 0.100 M AgNO₃ forms a precipitate and removes some chloride ions from solution.
- CaCl₂ initially supplies twice its own molarity in chloride ions because each formula unit contains two Cl⁻ ions.
- Convert each solution to moles using molarity × volume: 0.00300 mol CaCl₂ and 0.00150 mol AgNO₃.
- The balanced reaction requires 1 mol CaCl₂ per 2 mol AgNO₃, so AgNO₃ is limiting and consumes 0.00075 mol CaCl₂; subtract this from the initial amount to find the excess.
- Convert the remaining CaCl₂ to chloride using the 2:1 Cl⁻-to-CaCl₂ ratio, giving 0.0045 mol Cl⁻ in solution.
- Divide by the combined solution volume, 0.0450 L, to obtain 0.100 M Cl⁻; using CaCl₂ molarity without the ion ratio would give the wrong quantity.
- At the acid-base equivalence point, moles of H⁺ equal moles of OH⁻, even when moles of the intact acid and base are not equal.
- For HCl reacting with Ba(OH)₂, two HCl molecules are required per Ba(OH)₂ because each base unit contains two OH⁻ ions.
- A 30.0 mL sample of 0.475 M Ba(OH)₂ contains 0.01425 mol Ba(OH)₂, requiring 0.0285 mol HCl at the 2:1 ratio.
- Divide by the HCl concentration of 0.225 M to get 0.127 L, or 127 mL, of acid to reach equivalence.
- Slide 24 is the last slide with Exam 2 material; practice includes iodine configurations, identifying groups, and configurations of ions.
- Magnetism slides were moved to after Exam 2, while solution stoichiometry question 11 does not require a limiting-reactant analysis.
- Neutral cobalt has 27 electrons, so a configuration written with an argon core must contain nine additional electrons.
- Reject configurations that place three electrons in an s subshell or have the wrong total; [Ar] 4s² 3d⁷ is ground state, while [Ar] 4s¹ 3d⁸ is an excited state.
- Ground-state manganese has configuration [Ar] 4s² 3d⁵; Hund's rule places one electron in each of the five 3d orbitals before pairing.
- For n = 3 and mₗ = 2, only the 3d orbital with that magnetic quantum number qualifies; in ground-state Mn it contains one electron, so the answer is one.
Transcript excerpt
0:01 Okay. So, our practice quiz here, the questions change a little bit, but not too much. Um so, the same question that I find for the photoelectric effect might be a little different than yours, but it should still be the same number. Um So, it Oh, looks like it's the last one. Um so, for this one, this one you don't have me to do anything for the um kinetic energy of the electron. Um so, we can look at a different one
0:49 for that, but this one um is good to understand the concept as well. So, a device uses the photoelectric effect to detect incident light. In a test experiment, photons with a wavelength of 450 nm are found to eject electrons from the metal inside the device, while photons with a wavelength of 650 nm do not eject electrons. And the um relationship between wavelength and energy, remember that is an inverse relationship. So, 450 nm is short enough to eject electrons. Meaning it has enough energy.
1:43 Or high enough energy, I should say. So, wavelength is short when energy is high. So, 450 is short enough, Which means it has enough energy. 650 is not. So, 650 nanometers not enough energy. So, that concept is good to know. Um that the energy the binding energy of the metal must be somewhere um if we converted it to wavelength, must be somewhere between 450 and 650. Um So, 650 is not enough, but 450 is. So, based on this light with a
2:33 wavelength of 400 400 is even shorter. So, if we're talking even shorter even shorter means it should definitely be enough. Um So, it's based on that concept of short enough. Even shorter, great. So, this one uh will certainly eject electrons. It's a high enough energy. Uh it's even
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Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Alicia Altemose.