Exam 3 study guide Speed is the key to do well on this exam. The exam isn’t any longer or shorter than the other exams, but it covers a much broader range of problem types. For example, I won’t ask you to work through a full set of titration calculations, but you should be able to determine very quickly if a mixture of an acid and base is before, after or at the equivalence point (and solve accordingly). Know this.. Memorize strong acids, and be able to recognize them instantly. - HCl, HBr, HI, HNO3, H2SO4, HClO4Be able to convert between pH, pOH, [H+], and [OH-] - pH = -log[H+]- [H+] = 10^-pH- pOH = -log[OH-]- [OH-] = 10^-pOH- pH + pOH = 14.00- Kw = [H+][OH-]Be able to convert between pKa and pKb, as well as Ka and Kb (using pKw and Kw)- Ka * Kb = Kw- pKa = -log(Ka) & pKb = -log(Kb)- pKa * pKb = pKwGeneral concepts and calculations Acids and conjugate bases, and their salts - Conjugate base of an acid is the species that remains when one proton has been removed from the acid. - A salt derived from a strong acid (such as HCl) and a weak base (such as NH3) creates an acidic solution when dissolved in waterBases and conjugate acids, and their salts - Conjugate acid is the species that results from the addition of a proton to a base.- Salta derived from a strong base and weak acid produced a basic solutionAmphoteric chemicals - Be(OH)2, Al(OH)3, Sn(OH)2, Pb(OH)2, Cr(OH)3, Cu(OH)2, Zn(OH)2, Cd(OH)2- Can react with both acids and bases- All amphoteric hydroxides are insolublePercent ionization - percent ionization = (ionized acid concentration at equilibrium/initial concentration of acid) * 100%Calculate solubility, molar solubility, using Ksp and concentrations (and vice versa) solubility of compoundmolar solubility of compoundconcentrations of cations and anionsKsp of compoundKsp of compountconcentrations of cations and anionsmolar solubility of compoundsolubility of compound- Example 16.8 and 16.9Titrations between: Strong acid-strong base - Equivalence point where equimolar amounts of acid and base have been reacted.- pH = 7.00Weak acid-strong base - At equivalence point pH is greater than 7 as a result of the excess OH- ions formed from the hydrolysis of the weak acid’s conjugate base- ex: CH3COOH + NaOH CH3COONa + H2O simplified to:CH3COOH + NaOH CH3COO- + H2OCH3COO- goes through hydrolysis:CH3COO- + H2O CH3COOH + OH-Strong acid-weak base - pH at equivalence point is less than 7 due to hydrolysis of weak base’s conjugate acid- ex: HCl + NH3 NH4Cl simplified to:H+ + NH3 NH4+NH4+ goes through hydrolysis:NH4+ + H2O NH3 + H+Le Chatelier’s principle, and how the equilibrium is affected in different solutions: Acids and salts of their conjugate bases/bases and salts of their conjugate acids - Creates a buffer solution, which resists changes in pH- Also presence of a common ion suppresses the ionization of the weak acid or base and pushes equilibrium to the elftJust salts of acids and bases - Salt that produces neutral solution:o Salts containing an alkali metal ion or alkaline earth metal ion and the conjugatebase of a strong acid do not undergo hydrolysis to a great extent and the solutions are neutral (Ex. NaNO3)- Salt that produces basic solution:o Salt derived from a strong base and a weak acid creates a basic solution (Ex. CH3COONa)- Salt that produces acidic solution:o When a salt derived from a strong acid and a weak base is dissolved in water, the resulting solution is acidico In principle, all metal ions react with water to produce an acidic solutionMixtures of two acids or two bases (with different pKa or pKb values)- Kb > Ka solution is basic- Kb < Ka solution is acidic- Kb = Ka solution is nearly
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