Chem that idk

• Finding the pH for an acid when given the Ka and initial concentration of the acid: Ka = ((H)(A))/HA

  1. write an acid dissociation equation of HA -- H+ + A-
  2. write the Ka equation based on the reaction
  3. make an ICE table for the concentration of H gained or lost
  4. Plug in the x's (H+ and A-) and the initial concentration (HA) to equal Ka
  5. solve for x and -log it for the pH
• pKb (-log of Kb, the base dissociation constant) measures the strength of a base in solution. so a high pKb means a strong base. pOH also measures a solution's basicity but is from the negative log of OH concentration. pKb = pOH at half-equivalence point of a titration bc half of the weak base has reacted with the strong acid and turned to the conjugate acid, meaning B = BH+ meaning Kb = OH and pKb = pOH
• diluting an aqueous system at equilibrium causes shift to the side with more aqueous species because solution volume increases and moles stay the same. this decreases pressure so moles increases to bring pressure back up
• Stirring a reaction causes an increase in reaction rate because more movement = more collisions. this only works for heterogenous mixtures

  where all parts of a mixture are NOT identical
• if identical volume and temp, determine greatest pressure by greatest moles (find by multiplying given grams by molar mass)
• if different gases have same mass, the one with the most moles is the one with the smallest molar mass bc the moles are "completed" more often
• at the exact point of color change, the concentration of the protonated form with be equal to/than the deprotonated form
• lattice energy is based on coulomb's law. greater lattice energy means greater melting point.
• noble gases can bond with highly electronegative atoms
• if the acid is a strong acid, that means it dissociates completely, so concentration of the acid will equal the concentration of H+, so the pH of a strong acid can be found using the negative log of [HA] not just [H+]
• adding a gas at constant pressure to sealed container alr at equilibrium increases overall moles, which decreases partial pressures of other substances. Use the Q law of mass expression to determine if Q's pressure will initially decrease (products partial presure decreases) or increase (reactants' partial pressure decreases) and the shift it takes to re-establish equilibrium
• due to the common ion effect, solids are least soluble in solutions they share an ion with
• in a saturated solution, the concentration of each component ion will always remain constant. if water evaporates in a saturated solution, precipitate will increase so ions can decrease and maintain concentration ratio to the decreasing water
• larger battery = larger voltage = larger current = more ions being plated out of solution into a form of solid atop the surface of the other solid.
• the conjugate bases of strong acids are ineffective because strong acid dissociate completely, so strong acids can't form buffers
• henderson hasselbach ratios

  pH = pKa + log (A / HA). To decrease pH by 1 from pKa, the log of (A/HA) must be -1, so (A/HA) must = 1/10. to increase pH by 1 more than pKa, log (A/HA) = 1, so A/HA = 10
• diluting a buffer does not change the pH significantly because acid and conjugate base concentration change together, not changing the ratio between them and therefore in henderson hasselbach not changing the log (A/HA) part of pH = pKa + log (A/HA) much.
• Retention factor

  when using paper chromatography to seperate a solution's components, you can find how far the ink travels along the paper by retention. Rf = distance traveled by solute / distance traveled by solvent. the stronger the attraction between the polar water solvent and the solute, the larger the Rf value will be.
• Electroplating

  Electrolytic cells are used for electroplating, the process used to coat a metal object with a thin layer of another metal, done by passing an electric current through a solution with dissolved metal ions and the object to be plated, causing the metal ions to deposit onto the surface of the object.

  Solving electrolysis problems

  Given current & time

  Calculate charge: I = q / t

  I = current (amperes, A)

  q = charge (coulombs, C)

  t = time (seconds, s)
• higher elevation (low atmospheric pressure), boiling point ↓
• -∆H, +∆S, low T = negative ∆G, is favored
• -∆H, +∆S, high T = negative ∆G, is favored
• +∆H, -∆S, low T = positive ∆G, not favored
• +∆H, -∆S, high T = positive ∆G, not favored
• +∆H, +∆S, low T = positive ∆G, not favored
• +∆H, +∆S, high T = negative ∆G, is favored
• -∆H, -∆S, low T = negative ∆G, is favored
• -∆H, -∆S, high T = positive ∆G, not favored
• finding ∆G when you know Keq

  ∆G = -(8.31 J/mol K )(Kelvin)(ln Keq)
• gas deviations from ideal behavior happens when IMFs significant enough to affect their behavior. greater IMFs = greater attraction = less movement / pressure. so gases with greater IMFs have less pressure and more likely to deviate from ideal behavior
• when compounds have same volume, moles, temperature, the pressure difference goes to the IMFs - strongest IMFs have strongest attractions which lowers pressure
• during the equivalence point of a strong acid and strong base titration, the dominant species at equiv point is water and salt. in a titration of a weak acid and strong base, the dominant species at equiv point is the conjugate base of the weak acid
• the voltage of a cell is dependent to how close it is to equilibrium. the closer to equilibrium, the lower the voltage becomes.
• delta G = 0 at phase change. if you know the delta H and delta S of something at fusion or vaporization, you can find the melting or boiling point of that substance by setting delta G to 0.
• the more negative a reduction potential, the more likely it is to be oxidized bc u flip the sign and get a high ox potential. the larger a reduction potential, the more likely it is to occur
• if a metal loses electrons that means lower reduction potential than the solution, and it got oxidized and the solution got reduced
• the salt bridge maintains the electrical neutrality in the cell. at the cathode, the solution becomes less positively charged, so positive cations from the salt bridge flow into the half-cell. at the anode, where oxidation happens, solution becomes more positively charged so negative anions from the salt bridge flow into the half-cell. if a salt bridge was removed, the solutions in the half-cells would be electrically imbalanced and the voltage of the cell would drop to 0.
• current - the flow of electrons from one place to another.
• galvanic/voltaic cell - where a favored redox reaction is used to generate flow of current
• when A- = HA, pH = Pka and same for bases. When choosing an acid for a buffer solution, it is best to pick an acid with a pKa that is close to the desired pH to have almost equal amounts of conjugate base and acid, making the buffer flexible in neutralizing both H+ and OH-