Selective Review Part Dos Mr. Gates AP Chemistry

Unit 6: Buffers & Titrations Ch. 15

What is a buffer? Buffered Solution: Resists change despite the addition of an acid or a base. Made from a weak acid or base and its salt.

What causes buffers to work? Common Ion Effect: a shift in the equilibrium position that occurs because of the addition of an ion already involved in the equilibrium reaction. Example: NH4Cl(s) +H2O(l) → NH4+(aq) + Cl-(aq) NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)

pH of a Buffer Problem What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, and 0.50M sodium acetate, NaC2H3O2? Ka = 1.8 x 10-5 ? Step #1: Write the dissociation equation: HC2H3O2  C2H3O2- + H+

pH of a Buffer Problem Step #2: ICE it! HC2H3O2  C2H3O2- + H+ I C E 0.50 0.50 - x +x +x 0.50 - x 0.50+x x

pH of a Buffer Problem Step #3: Set up the law of mass action HC2H3O2  C2H3O2- + H+ E 0.50 - x 0.50+x x

This illustrates an important point: pH of a Buffer Problem Step #4: Calculate x and pH: x = 1.8 x 10-5 pH = 4.74 This illustrates an important point: Buffer pH at equal concentrations of salt and weak acid (or base) equals the pKa (or pKb) of that substance, in other words, the –log of the Ka or Kb for that acid or base!

Henderson-Hasselbalch Equation This is an exceptionally powerful tool, and it’s use will be emphasized in our problem solving.

Buffer Recap A buffer contains a large concentration of both a conjugate acid an the conjugate base. This is usually made with either a base or an acid and a salt of the conjugate species. Weak acids and their conjugate bases make for good buffers, strong acids and bases do not! When the pH > pKa, the base form has a higher concentration, but when pH < pKa, the acid form has a higher concentration.

Titrations Adding a solution of a known concentration (titrant) to a solution of an unknown concentration (analyte or titrate) by a buret until the unknown solution has been consumed. This is a common way to find the concentration of an acid or base Reactions of acids and bases are called neutralization reactions, and these reactions generally have K>1, and thus can be considered to go to completion.

Neutralization Reactions cont. When a strong base is added to a solution of a weak acid, a neutralization reaction occurs: conjugate acid + OH- → conjugate base + H2O When a strong acid is added to a solution of a weak base, a neutralization reaction occurs: conjugate base + H3O+ → conjugate acid + H2O

Titration Curves (pH curves) Graph displaying the pH changes with respect to the addition of the titrant. As base is added to either a strong or weak acid solution, the hydronium concentration does not change much. The change in pH is less than ~1.5 for the region where 10-90 percent of the base needed to reach the equivalence point has been added.

Titration Curves (pH curves) A titration technique exists for neutralization reactions. At the equivalence point, the moles of titrant and the moles of titrate are present in stoichiometric proportions. In the vicinity of the equivalence point, the pH rapidly changes. This can be used to determine the concentration of the titrate.

Equivalence Points With both strong acids and bases titrated together the equivalence point will be at pH = 7. With a weak acid titrated with a strong base the equivalence point will be at a pH > 7. With a weak base titrated with a strong acid the equivalence point will be at a pH < 7.

Solubility as an Equilibrium Position Solubility can represent the equilibrium position. CuBr (s)  Cu+ (aq) + Br- (aq) Ksp=[Cu+]m [Br-]n m and n are the coefficients of the respective ions. [Cu+] and [Br-] represent the “solubility” of the solid. Ksp is called the solubility-product constant This is an equilibrium constant and is not in itself the solubility! The solubility equilibrium position is not determined by pure liquids or solids.

CaF2(s)  Ca2+(aq) + 2F-(aq) Common Ions With the presence of common ions, the solubility of a solid decreases. What is the solubility of CaF2 (Ksp=4.0x10-11) in a 0.025M NaF solution? CaF2(s)  Ca2+(aq) + 2F-(aq) Ksp =4.0x 10-11= [Ca2+][F-]2

Common Ions/Practice cont. ICE box Ca2+ F- I 0.025M C +x +2x E x 0.025M+2x

Common Ions/Practice cont. Ksp =4.0x 10-11= [x][0.025+2x]2 We assume that x << 0.025M Ksp =4.0x 10-11= [x][0.025]2 Solve for x, x = 6.4 x 10-8M Thus the solubility of the CaF2 in 0.025M NaF solution is 6.4 x 10-8mol/L

Equilibrium Position and Q The dissolution of a substance in a solvent is a reversible reaction, and so has an associated equilibrium constant. For dissolution of a salt, the reaction quotient, Q, is referred to as the solubility product, and the equilibrium constant for this reaction is denoted as Ksp, the solubility –product constant. If Q > Ksp then precipitation will occur. If Q < Ksp then no precipitation will occur.

Solubility Rules All sodium, potassium, ammonium, and nitrate salts are soluble in water.

Unit 7: Thermochemistry Chapter 6

Enthalpy Changes The enthalpy change of a reaction gives the amount of energy released (for negative values) or absorbed (for positive values) by a chemical reaction at constant pressure. Hess’ Law: in going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or in many.

Characteristics of Enthalpy Changes If a reaction is reversed, the sign of ΔH is also reversed. The magnitude of ΔH is directly proportional to the quantities of reactants and products in a reaction. If the coefficients in a balanced reaction are multiplied by an integer, the value of ΔH is multiplied by the same integer.

Standards Standard Enthalpy of Formation: change of enthalpy that accompanies the formation of one mole of a compound from its elements with all of its elements in their standard states. The standard state of a substance is the state at which the substance will exist under standard pressure and temperature. Ex. Oxygen would be O2(g); water would be H2O (l) 0.5N2(g) +1.5H2(g)  NH3(g) ΔH◦f = -46 kJ/mol

Laws of Thermodynamics 1st Law of Thermodynamics: the energy of the universe is constant. 2nd Law of Thermodynamics: the entropy of the universe is increasing. For spontaneous reactions

Increasing Entropy Entropy increases when: Matter is dispersed (solid to liquid, etc.) Particles become free to move Number of individual particles increases during a chemical reaction With respect to gases… an increase in volume. Gas particles can move in a larger space Energy is dispersed Temperature increases

Favorable Reactions “Thermodynamically Favored” means the products are favored at equilibrium. In regards to chemistry, this phrase has the same meaning as “spontaneous.” In some reactions, it is necessary to consider both enthalpy and entropy to determine if a reaction will be thermodynamically favorable. The freezing of water and the dissolution of sodium nitrate in water provide good examples of such situations.

Gibbs Free Energy Free energy is a function that describes spontaneity/thermodynamic favor. Free energy is especially helpful with temperature dependence. Symbolized by “G” G = H-TSsys Thus… ΔG = ΔH – TΔS …at constant temperature and pressure.

So how does this help? If ΔG<0, the products are favored at equilibrium and the forward process is said to be “thermodynamically favored.” If ΔG>0, the reactants are favored at equilibrium and the reverse reaction is said to be “thermodynamically favored.” If ΔG=0, the reaction is at equilibrium and there is no driving force in either direction.

Rate of Favoritism Do thermodynamically favored reactions occur quickly or slowly? Yes “Thermodynamically favored” doesn’t imply anything about the rate and so some thermodynamically favored reactions will occur quickly while other will go slowly. Many processes that are thermodynamically favored do not occur to any measurable extent, or they are at extremely slow rates.

Calculating ΔG◦ Method #1: ΔG◦ = ΔH◦ – TΔS◦ We can calculate the ΔH for the reaction, the temperature in Kelvins and calculate the ΔS of the system. With these pieces of information, we can solve for ΔG. To calculate ΔH and ΔS, we must look at the standard enthalpy of formation as well as the entropy of each of the products and reactants. These values are usually found in a table.

Calculating ΔG◦ Method #2… an adaptation of Hess’ Law: Cdiamond(s) + O2(g)  CO2(g) G0 = -397 kJ Cgraphite(s) + O2(g)  CO2(g) G0 = -394 kJ Cdiamond(s) + O2(g)  CO2(g) G0 = -397 kJ CO2(g)  Cgraphite(s) + O2(g) G0 = +394 kJ Cdiamond(s)  Cgraphite(s) G0 = -3 kJ

Calculating ΔG◦ Method 3… Using standard free energy of formation (ΔGf◦ ): ΔGf◦ of a pure element in its standard state is zero.

∆𝐺 ° =−𝑅𝑇 ln 𝐾 Thus… 𝐾= 𝑒 − ∆𝐺 ° 𝑅𝑇 The thermal energy (RT) at room temperature is 2.4kJ/mol. Processes where ∆𝑮 ° < 0, favor products and K > 1. Processes where ∆𝑮 ° > 0, favor reactants and K < 1.

Electrochemistry Chapter 4.9 and Chapter 17

Oxidation Numbers Pure Element: oxidation # is zero Monatomic ion: oxidation number is the charge Neutral compound: sum of oxidation #’s is zero Polyatomic ion: sum of #’s is ion’s charge

Oxidation-Reduction Oxidation Involves the Loss of electrons… OIL Oxidation is the loss of e- thus causing the oxidation number to increase (become more positive) Reduction is the gain of e- thus causing the oxidation number to decrease (become more negative) Oxidation Involves the Loss of electrons… OIL Reduction Involves the Gain of electrons… RIG Reduction reduces the oxidation number.

Example Find the oxidation states for each species below: CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) Reactants: H: +1 C: -4 O: 0 Products: C: +4 O:-2

Balancing Redox Reactions Write separate equations (half-reactions) for oxidation and reduction For each half reaction Balance elements involved in e- transfer If needed add water and hydronium for acidic solutions and hydroxide and water for basic solutions Balance number e- lost and gained To balance e- multiply each half-reaction by whole numbers

Types of Cells Voltaic (galvanic) cells: Electrolytic cells: A spontaneous reaction generates electrical energy. Electrolytic cells: Absorb free energy from an electrical source to drive a nonspontaneous reaction.

Electrodes Oxidation occurs at the anode and reduction occurs at the cathode for all electrochemical cells. Oxidation at the Anode (vowels) Reduction at the Cathode (consonants)

Cell Potential, E◦cell E◦cell = cell potential under standard conditions Elements in their standard states (298K) Solutions are 1M Gases are at 1atm E◦cell = E◦cathode - E◦anode

Cell Potential and Free Energy E◦cell > 0 ΔG◦ < 0 Spontaneous E◦cell < 0 ΔG◦ > 0 Not Spontaneous E◦cell = 0 ΔG◦ = 0 At Equilibrium

Cell Potential and Equilibrium Under standard conditions and at equilibrium: ΔG◦ = 0, K = Q, Temp=298K

Concentration Cell The cell will run until the solutions on both sides are in equilibrium.