Test 4 ReviewTest 4 Review

Advanced ChemistryAdvanced Chemistry

EquilibriumEquilibrium mmA + A + nnB B ssP + P + rrQQ

KKeqeq = = [P][P]ss[Q][Q]rr

[A][A]mm[B][B]nn

KKspsp = [A = [A++][B][B--] for dissolving of a solid ] for dissolving of a solid AB AB A A++ + B + B--

KKaa = = [H[H++][A][A--] ] for the dissociation of an acid for the dissociation of an acid [HA][HA]

KKaa is small for a weak acid, large for a strong is small for a weak acid, large for a strong acid.acid.

Never include solids or pure liquids in a KNever include solids or pure liquids in a Keq.eq.

Solubility Product ConstantSolubility Product ConstantAB(s) AB(s) A A++(aq) + B(aq) + B--(aq)(aq)

KKspsp = [A = [A++][B][B--]]

At 25At 25°C the solubility product constant °C the solubility product constant for strontium sulfate, SrSOfor strontium sulfate, SrSO44, is 7.6 x , is 7.6 x 1010-7-7. What is the concentration of Sr. What is the concentration of Sr2+2+ at at 2525°C?°C?

[Sr[Sr2+2+][SO][SO442-2-] = 7.6 x 10] = 7.6 x 10-7-7 [Sr [Sr2+2+]=[SO]=[SO44

2-2-] ]

[Sr[Sr2+2+] = 8.7 x 10] = 8.7 x 10-4-4MM

The solubility product constant for The solubility product constant for strontium fluoride is 7.9 x 10strontium fluoride is 7.9 x 10-10-10. What . What is the molar solubility of SrFis the molar solubility of SrF22 at at 2525°C?°C?

SrFSrF2 2 SrSr+2 +2 + 2F+ 2F--

[Sr[Sr2+2+][F][F--]]22=7.9 x 10=7.9 x 10-10-10

[Sr[Sr2+2+]= x [F]= x [F--]= 2x]= 2x

(x)(2x)(x)(2x)22=7.9 x 10=7.9 x 10-10-10

x = 5.8 x 10x = 5.8 x 10-4-4MM

Strong bases:Strong bases:

hydroxides of groups 1 & 2 hydroxides of groups 1 & 2 (except Be)(except Be)

Strong acids:Strong acids:

HCl, HBr, HIHCl, HBr, HI

HClOHClO44, H, H22SOSO44, HNO, HNO33

Ionization constant, KIonization constant, Ka,a,

for a weak acid for a weak acid

HA HA H H++ + A + A--

KKaa = = [H[H++][A][A--]]

[HA][HA]

KKb =b =

KKaa ∙K ∙Kbb= K= Kww = 1.00 x 10 = 1.00 x 10-14-14

][

]OH-][[

23

33

NHCH

NHCH

What is the [HWhat is the [H++] in 0.100M formic acid?] in 0.100M formic acid?

KKaa for formic acid is 1.77 x 10 for formic acid is 1.77 x 10-4-4

HCOOH HCOOH H H++ + COOH + COOH--

Since this is a weak acid, [HCOOH] Since this is a weak acid, [HCOOH] ͌ ͌

0.100M0.100M

KKaa = = [H[H++][COOH][COOH--]] == 1.77 x 10 1.77 x 10-4-4

[HCOOH][HCOOH]

Let x = [HLet x = [H++] = [COOH] = [COOH--]]

xx22 == 1.77 x 10 1.77 x 10-4-4

0.1000.100X = 4.21 x 10-

3M

Percent ionizationPercent ionization [amount ionized][amount ionized]

[original acid][original acid]

What is the percent ionization of [HWhat is the percent ionization of [H++] ] from the previous problem?from the previous problem?

[H[H++] ] = 4.21 x 10-3M, [HCOOH][HCOOH] = 0.100M0.100M

4.21 x 10-3 = 4.21%

0.100

hydrolysishydrolysis

The reaction of a salt The reaction of a salt with water to form an with water to form an acidic or basic acidic or basic solution.solution.

Example:Example: FeClFeCl3 3 FeFe3+3+(aq) (aq) + Cl+ Cl--(aq)(aq)

FeFe3+ 3+ + 3OH+ 3OH-- Fe(OH)Fe(OH)33

HH22O O HH++ + OH + OH- -

shifts right, creating more shifts right, creating more HH++

cationcation anionanion solutiosolutionn

StronStrong g basebase

StronStrong acidg acid

NeutrNeutralal

StronStrong g basebase

Weak Weak acidacid

BasicBasic

Weak Weak basebase

StronStrong acidg acid

AcidicAcidic

Weak Weak basebase

Weak Weak acidacid

NeutrNeutralal

[H[H++] ∙ [OH] ∙ [OH--] = 10] = 10-14-14

pH + pOH = 14.0pH + pOH = 14.0

pH = -log[HpH = -log[H33OO++]]

[H[H33OO++] = antilog(-pH)] = antilog(-pH)

Find the pH of a solution with [HFind the pH of a solution with [H33OO++] of ] of 9.85 x 109.85 x 10-8-8M.M.pH = -log (9.85 x 10pH = -log (9.85 x 10-8-8)) = 7.01= 7.01

What is the [HWhat is the [H33OO++] in a solution with pH ] in a solution with pH 7.01?7.01?

[H[H33OO++] = antilog(-7.01)= 9.85 x 10] = antilog(-7.01)= 9.85 x 10-8-8MM

What is the pH of the following What is the pH of the following solutions?solutions?0.1M HCl0.1M HCl0.1M NaOH0.1M NaOH0.1M H0.1M H22COCO33

[H+] = 10-1, pH = 1

[OH-]=10-1, pOH=1, pH=13

Ka = 4.4 x 10-7 = x2

0.1

X = 2.1x10-4 pH = 3.68

TitrationTitration

Standard solution- one whose Standard solution- one whose concentration is knownconcentration is known

Endpoint- the point at which Endpoint- the point at which equivalent amounts of reactants are equivalent amounts of reactants are present.present. M∙V = molesM∙V = moles MMaaVVaa=M=MbbVVb b

Titration curvesTitration curves

Calculating pH of a bufferCalculating pH of a buffer

What is the pH of a buffer that is 0.12M What is the pH of a buffer that is 0.12M lactic acid (HClactic acid (HC33HH55OO33) and 0.10M sodium ) and 0.10M sodium lactate? lactate?

KKaa = 1.4 x 10 = 1.4 x 10-4-4

HCHC33HH55OO33 HH++ CC33HH55OO33--

InitialInitial 0.12M0.12M 00 0.10M0.10M

ChangeChange

equilibriuequilibriumm

+x+x +x+x-x-x

.12-x.12-x xx .10+x.10+x

Henderson-HasselbalchHenderson-Hasselbalch

pH = pKpH = pKaa + log + log [base][base]

[acid][acid]

= -log(1.4x10= -log(1.4x10-4-4) + log ) + log .10.10

.12.12

= 3.85 + (-.079)= 3.85 + (-.079)

= 3.77= 3.77

Conjugate base of the acid

Common Ion (Buffers)Common Ion (Buffers)

Calculate the pH of a 0.100M solution of Calculate the pH of a 0.100M solution of formic acid and 0.020M sodium formate.formic acid and 0.020M sodium formate.

KKaa= 1.77 x 10= 1.77 x 10-4-4

HCOOH HCOOH H H++ + COOH + COOH--

0.100 x 0.0200.100 x 0.020

KKaa = = [H[H++][COOH][COOH--]] == . .020x020x

[HCOOH] 0.100[HCOOH] 0.100X = 8.85 x 10-

4MpH = 3.05

Redox reactions involve a change in Redox reactions involve a change in oxidation numberoxidation number

OxidationOxidation Loss of electronsLoss of electrons

ReductionReduction Gain electronsGain electrons

3Cu3Cu2+2+ + 2Fe + 2Fe 3Cu + 2Fe 3Cu + 2Fe3+3+

copper gains 2 electronscopper gains 2 electronsiron loses 3 electronsiron loses 3 electrons

Reducing agent- is oxidized (iron)Reducing agent- is oxidized (iron)Oxidizing agent- is reduced (copper)Oxidizing agent- is reduced (copper)

(reduced)(reduced)

(oxidized)(oxidized)

Steps for balancing redox Steps for balancing redox reactions using the half-reactions using the half-

reaction methodreaction method1.1. Write Write ionicionic equation for half reactions equation for half reactions

2.2. Balance Balance chemicallychemicallya)a) Balance non- O and H atomsBalance non- O and H atoms

b)b) Add HAdd H22O to balance O’sO to balance O’s

c)c) Add HAdd H++ to balance H’s (in a basic solution add OH to balance H’s (in a basic solution add OH--))

3.3. Balance Balance electricallyelectrically- add e- add e--’s to the more + side’s to the more + side

4.4. CheckCheck for balanced charges on both sides for balanced charges on both sides

5.5. CombineCombine half reactions and cancel common half reactions and cancel common itemsitems

6.6. Add spectator ions and balanceAdd spectator ions and balance

Activity SeriesActivity Series

lithiumlithium

potassiumpotassiummagnesiummagnesiumaluminumaluminumzinczincironironnickelnickelleadleadHYDROGRENHYDROGRENcoppercoppersilversilverplatinumplatinum

goldgold

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 CellVoltaic CellAnodeAnode

attracts anionsattracts anions where oxidation occurswhere oxidation occurs

CathodeCathode attracts cationsattracts cations where reduction occurswhere reduction occurs

Salt bridge Salt bridge connects the two half cells connects the two half cells contains a strong electrolyte contains a strong electrolyte

Zn Zn+ Cu2+ Cu in shorthand

Two half cells connected by a salt Two half cells connected by a salt bridgebridge

Reduction half reactions

F2 is the strongest oxidizing agent

Li is the strongest reducing agent

Reduction Potentials (E)Reduction Potentials (E) If E is positive, the reaction is If E is positive, the reaction is

spontaneous.spontaneous. If E is negative, the reverse reaction is If E is negative, the reverse reaction is

spontaneous.spontaneous. EEoo is the standard electrode potential is the standard electrode potential

all ions are 1M and gases are 1 atmall ions are 1M and gases are 1 atm The net EThe net Eo o is the sum of the Eis the sum of the Eoo of the of the

half reactionshalf reactions The stronger oxidizing agent reduces.The stronger oxidizing agent reduces. Reverse the sign of the substance Reverse the sign of the substance

oxidized.oxidized.

What is the voltage produced from the What is the voltage produced from the reaction of Zn metal with Cureaction of Zn metal with Cu2+2+ ions? ions?

Zn(Zn(ss) + Cu) + Cu2+2+((aqaq) ) Zn Zn2+2+((aqaq) + Cu() + Cu(ss))

ZnZn2+2+ + 2e + 2e-- Zn -0.7628 Zn -0.7628

CuCu2+2+ + 2e + 2e-- Cu 0.3402 Cu 0.3402

-(-0.7628) + 0.3402 = 1.103 volts-(-0.7628) + 0.3402 = 1.103 volts

Will happen spontaneouslyWill happen spontaneously

ZnZn2+2+ + Ni( + Ni(ss) ) Zn( Zn(ss) + Ni) + Ni2+2+

-(-0.23) + (-0.7628) = -.53-(-0.23) + (-0.7628) = -.53

Will not occur spontaneouslyWill not occur spontaneously

Faraday’s LawFaraday’s Law

Coulombs = amperes x secondsCoulombs = amperes x seconds 1 C = 1amp1 C = 1amp·1·1secsec

96,485 coulombs = 1 mole e96,485 coulombs = 1 mole e--

What mass of copper will be What mass of copper will be deposited by a current of 7.89 deposited by a current of 7.89

amps flowing for 1200 amps flowing for 1200 seconds?seconds? CuCu2+2+ + 2e + 2e-- Cu at the cathode Cu at the cathode

7.89A x 1200s x 7.89A x 1200s x 1C1C x x 1 mol e1 mol e-- == .0981 mol .0981 mol

ee--

AA·s 96,485C·s 96,485C

.0981 mol e.0981 mol e-- x x 1 mol Cu1 mol Cu xx 63.5g Cu 63.5g Cu = 3.1g = 3.1g

Cu Cu

2 mol e2 mol e- 1- 1 mole Cu mole Cu

Nernst EquationNernst Equation

EEo Voltage under standard conditions 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 equationAt non-standard conditions, use Nernst equation

E = E° - E = E° - 0.059160.05916 log [ log [products]products]

n [reactants]n [reactants]

n = no. of electrons transferredn = no. of electrons transferredCoefficients in front of reactants or products are used as powers of their concentrations.Coefficients in front of reactants or products are used as powers of their concentrations.