Test 4 Review Advanced Chemistry

Equilibrium mA + nB  sP + rQ mA + nB  sP + rQ K eq = [P] s [Q] r K eq = [P] s [Q] r [A] m [B] n [A] m [B] n K sp = [A + ][B - ] for dissolving of a solid AB  A + + B - K a = [H + ][A - ] for the dissociation of an acid K a = [H + ][A - ] for the dissociation of an acid [HA] [HA] K a is small for a weak acid, large for a strong acid. K a is small for a weak acid, large for a strong acid. Never include solids or pure liquids in a K eq. Never include solids or pure liquids in a K eq.

Solubility Product Constant AB(s)  A + (aq) + B - (aq) K sp = [A + ][B - ] At 25°C the solubility product constant for strontium sulfate, SrSO 4, is 7.6 x What is the concentration of Sr 2+ at 25°C? [Sr 2+ ][SO 4 2- ] = 7.6 x [Sr 2+ ]=[SO 4 2- ] [Sr 2+ ] = 8.7 x M [Sr 2+ ] = 8.7 x M

The solubility product constant for strontium fluoride is 7.9 x What is the molar solubility of SrF 2 at 25°C? SrF 2  Sr F - SrF 2  Sr F - [Sr 2+ ][F - ] 2 =7.9 x [Sr 2+ ]= x [F - ]= 2x (x)(2x) 2 =7.9 x x = 5.8 x M

Strong bases: Strong bases: hydroxides of groups 1 & 2 (except Be) Strong acids: Strong acids: HCl, HBr, HI HClO 4, H 2 SO 4, HNO 3

Ionization constant, K a, for a weak acid HA  H + + A - HA  H + + A - K a = [H + ][A - ] K a = [H + ][A - ] [HA] [HA] K b = K b = K a ∙K b = K w = 1.00 x K a ∙K b = K w = 1.00 x

What is the [H + ] in 0.100M formic acid? K a for formic acid is 1.77 x HCOOH  H + + COOH - Since this is a weak acid, [HCOOH] ͌ 0.100M K a = [H + ][COOH - ] = 1.77 x [HCOOH] [HCOOH] Let x = [H + ] = [COOH - ] x 2 = 1.77 x x 2 = 1.77 x X = 4.21 x M

Percent ionization [amount ionized] [amount ionized] [original acid] [original acid] What is the percent ionization of [H + ] from the previous problem? [H + ] [HCOOH]0.100M [H + ] = 4.21 x M, [HCOOH] = 0.100M 4.21 x = 4.21% 0.100

hydrolysis The reaction of a salt with water to form an acidic or basic solution. The reaction of a salt with water to form an acidic or basic solution. Example: Example: FeCl 3  Fe 3+ (aq) + Cl - (aq) FeCl 3  Fe 3+ (aq) + Cl - (aq) Fe OH -  Fe(OH) 3 H 2 O  H + + OH - H 2 O  H + + OH - shifts right, creating more H + shifts right, creating more H + cationanionsolution Strong base Strong acid Neutral Strong base Weak acid Basic Weak base Strong acid Acidic Weak base Weak acid Neutral

[H + ] ∙ [OH - ] = pH + pOH = 14.0 pH = -log[H 3 O + ] [H 3 O + ] = antilog(-pH) Find the pH of a solution with [H 3 O + ] of 9.85 x M x M. pH = -log (9.85 x ) = 7.01 = 7.01 What is the [H 3 O + ] in a solution with pH 7.01? [H 3 O + ] = antilog(-7.01)= 9.85 x M

What is the pH of the following solutions? 0.1M HCl 0.1M NaOH 0.1M H 2 CO 3 [H + ] = 10 -1, pH = 1 [OH - ]=10 -1, pOH=1, pH=13 K a = 4.4 x = x X = 2.1x10 -4 pH = 3.68

Titration Standard solution- one whose concentration is known Standard solution- one whose concentration is known Endpoint- the point at which equivalent amounts of reactants are present. Endpoint- the point at which equivalent amounts of reactants are present. M∙V = moles M∙V = moles M a V a =M b V b M a V a =M b V b

Titration curves

Calculating pH of a buffer What is the pH of a buffer that is 0.12M lactic acid (HC 3 H 5 O 3 ) and 0.10M sodium lactate? What is the pH of a buffer that is 0.12M lactic acid (HC 3 H 5 O 3 ) and 0.10M sodium lactate? K a = 1.4 x K a = 1.4 x HC 3 H 5 O 3 H+H+H+H+ C3H5O3-C3H5O3-C3H5O3-C3H5O3- Initial0.12M00.10M Change equilibrium +x+x-x.12-xx.10+x 

Henderson-Hasselbalch pH = pK a + log [base] pH = pK a + log [base] [acid] [acid] = -log(1.4x10 -4 ) + log.10 = -log(1.4x10 -4 ) + log = (-.079) = (-.079) = 3.77 = 3.77 Conjugate base of the acid

Common Ion (Buffers) Calculate the pH of a 0.100M solution of formic acid and 0.020M sodium formate. K a = 1.77 x HCOOH  H + + COOH x x K a = [H + ][COOH - ] =.020x [HCOOH] [HCOOH] X = 8.85 x M pH = 3.05

Redox reactions involve a change in oxidation number Redox reactions involve a change in oxidation number Oxidation Oxidation Loss of electrons Loss of electrons Reduction Reduction Gain electrons Gain electrons 3Cu Fe  3Cu + 2Fe 3+ copper gains 2 electrons iron loses 3 electrons Reducing agent- is oxidized (iron) Oxidizing agent- is reduced (copper) (reduced) (oxidized)

Steps for balancing redox reactions using the half-reaction method 1. Write ionic equation for half reactions 2. Balance chemically a) Balance non- O and H atoms b) Add H 2 O to balance O’s c) Add H + to balance H’s (in a basic solution add OH - ) 3. Balance electrically- add e - ’s to the more + side 4. Check for balanced charges on both sides 5. Combine half reactions and cancel common items 6. Add spectator ions and balance

Activity Series lithiumpotassiummagnesiumaluminumzincironnickelleadHYDROGRENcoppersilverplatinumgold Oxidizes easily Reduces easily Less active More active The metal must be above (more active than) the ion for it to be a spontaneous reaction.

Voltaic Cell Anode attracts anions attracts anions where oxidation occurs where oxidation occursCathode attracts cations attracts cations where reduction occurs where reduction occurs Salt bridge connects the two half cells connects the two half cells contains a strong electrolyte contains a strong electrolyte

Zn Zn + Cu 2+ Cu in shorthand Two half cells connected by a salt bridge

Reduction half reactions F 2 is the strongest oxidizing agent Li is the strongest reducing agent

Reduction Potentials (E) If E is positive, the reaction is spontaneous. If E is positive, the reaction is spontaneous. If E is negative, the reverse reaction is spontaneous. If E is negative, the reverse reaction is spontaneous. E o is the standard electrode potential E o is the standard electrode potential all ions are 1M and gases are 1 atm all ions are 1M and gases are 1 atm The net E o is the sum of the E o of the half reactions The net E o is the sum of the E o of the half reactions The stronger oxidizing agent reduces. The stronger oxidizing agent reduces. Reverse the sign of the substance oxidized. Reverse the sign of the substance oxidized.

What is the voltage produced from the reaction of Zn metal with Cu 2+ ions? Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s) Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s) Zn e -  Zn Cu e -  Cu ( ) = volts Will happen spontaneously Zn 2+ + Ni(s)  Zn(s) + Ni 2+ -(-0.23) + ( ) = -.53 Will not occur spontaneously

Faraday’s Law Coulombs = amperes x seconds Coulombs = amperes x seconds 1 C = 1amp·1sec 1 C = 1amp·1sec 96,485 coulombs = 1 mole e - 96,485 coulombs = 1 mole e -

What mass of copper will be deposited by a current of 7.89 amps flowing for 1200 seconds? Cu e -  Cu at the cathode Cu e -  Cu at the cathode 7.89A x 1200s x 1C x 1 mol e - =.0981 mol e - A·s 96,485C A·s 96,485C.0981 mol e - x 1 mol Cu x 63.5g Cu = 3.1g Cu 2 mol e - 1 mole Cu 2 mol e - 1 mole Cu

Nernst Equation E o Voltage under standard conditions E o Voltage under standard conditions (1M solutions at 25°C and 101.3kPa) (1M solutions at 25°C and 101.3kPa) At non-standard conditions, use Nernst equation E = E° log [products] n [reactants] n [reactants] n = no. of electrons transferred n = no. of electrons transferred Coefficients in front of reactants or products are used as powers of their concentrations.