Chapter 11. Dissolution of sugar in water. Chapter Outline

Transcription

1 Chapter Slutins Disslutin f sugar in water Chapter Outline 11.2 Cmpsitin f slutins Nature f disslved species Reactin stichimetry in slutins : acid-base titratins Reactin stichimetry t in slutins : redx titratins Phase equilibrium in slutins : nnvlatile slutes Phase equilibrium in slutins : vlatile slutes Cllidal suspensins

2 11.1 Cmpsitin f Slutins 11.3 slutin: hmgeneus systems with tw r mre substances : a liquid slutin If a slute is disslved in water an aqueus slutin : a slid slutin -- ally slutin : slute + slvent (main cmpnent) mass percentage (weight percentage) : percentage by mass f a given substance in slutin : useful measures - mle fractin, mlarity, mlality Useful measures f cmpsitin 11.4 Mle fractin: # mles f a substance divided by ttal # mles X n = 1 1 X1 + 2 = n1 + n X 2 Mlarity (M): # mles f a slute per Liter f slutin cmmn way f expressing cncentratin. slightly varies n T change mlarity ( M) = mles L Mlality (m): # mles f a slute per kg f slvent independent f T change mles mlality ( m) = kg slvent 1

3 Sample slutins 11.5 What is the mlarity f 100 ml slutin cntaining 15.6g f NaOH? 1 ml NaOH ( 15.6g NaOH ) = ml NaOH 40.0g NaOH ml NaOH M = = 3.90 M NaOH 1 L 100 ml 1000 ml What is the mlarity f a slutin that is 4.0 ppb Cr 6+? 4.0μg 4.0 ppb = L sln μg 1 g Cr 1ml Cr Cr 6 6+ = M L sln. 10 μ g 52.0 gcr 8 6+ example 11.6 A M aqueus slutin f sulfuric acid has a density f g/cm 3. Calculate the mlality, the percentage by mass, and the mle fractin f sulfuric acid in this slitin. m=15.95m, % f sulfuric acid = 61.00%, X SA =0.2232

4 Preparatin f slutins 11.7 In case f diluting slutins : # f mles f slute frm initial slutin = # f mles f slute in final slutin civi c fvf = civi cf = V f 11.2 Nature f Disslved Species 11.8 aqueus slutin f mlecular species : plar mlecules l readily disslved d by water ex) sucrse (C 12 H 22 O 12 ), fructse (fruit sugar, C 6 H 12 O 6 ), ribse (C 5 H 10 O 5 ) : Hydratin prcess Water mlecules surrund the mlecules r ins - nte the psitins f the psitive and negative charges H-bnding interactins with water mlecules and OH grup.

5 Aqueus slutin f inic species 11.9 Electrlytes : inic species disslved in water Strng Electrlytes : Gd cnductrs f electricity in slutin. : Strng acids, strng bases, sluble inic cmpunds. : Cmpletely inized in slutin. Weak Electrlytes : Pr cnductrs f electricity in slutin. : Partially inized. CH 3 COOH, NH 3 Nn-electrlytes : D nt cnduct electricity in slutin. : D nt frm ins in slutin. C 6 H 12 O 6, Aqueus slutin f inic species slubility : max mass that can be disslved in 1 L at 25 C. disslutin rxn K 2 SO 4 (s) 2 K 2+ (aq) + SO 2-4 (aq) by in-diple frces nte the rientatin f ppsite charges K 2 SO 4 1ml 3 mles f ins

6 Precipitatin rxn Precipitatin rxn 11.12

7 Aqueus slutin f inic species Mlecular equatin (cmpnents written as cmpunds): AgNO 3 (aq) + NaCl(aq) AgCl(s) + NaNO 3 (aq) Cmplete Inic Equatin (electrlytes tes written as ins): Ag + (aq) + NO 3 (aq) + Na + (aq) + Cl (aq) AgCl(s) + Na + (aq) + NO 3 (aq) Net inic equatin (ins participating in the reactin Ag + (aq) + Cl (aq) AgCl(s) (Na + and NO 3 are spectatr ins.) 11.3 Rxn Stichimetry in Slutins :Acid-base titratins Rxn in slutin : cnservatin laws in balanced chemical eq. mass mles vlumes f slutin In case f preparing elemental brmine frm its salts in sl. 2Br - (aq) + Cl 2 (aq) 2Cl - (aq) +Br 2 (aq) What vlume f a M slutin f Cl 2 is needed t react cmpletely l with Br -? There is ml f M sl f NaBr. # mles f Br- = M x L = 3.00x10-3 ml # mles f Cl 2 required = ml Br Vlume f Cl 2 required = - 1ml Cl2 3 = ml Cl ml Br ml Cl 2 = L M [Cl - ] = ml / ( )L = M

8 Titratin Titratin :Cntrlled additin f a slutin f knwn cncentratin t react with a slutin f unknwn cncentratin. Equivalence pint The pint when all reactants are cmpletely cnsumed. End pint Clse t the equivalence pint marked by an indicatr. Backgrund n Acid-Base Rxn Svante Arrhenius: Acid-Base cncept acid: a substance when disslved in water prduces hydrnium in HCl(g) + H 2 O(l) H 3 O + (aq) + Cl - (aq) base: a substance when disslved in water prduces hydrxide in NaOH(s) Na + (aq) + OH - (aq) autdissciatin i ti f water 2H O(l) H 3 O (aq) OH (aq) Neutralizatin reactin: when acid is mixed with base HCl + NaOH H 2 O + NaCl H 3 O + (aq) + OH - (aq) 2H 2 O(l) net equatin reverse f water inizatin reactin

9 Acid-Base Rxn H 2 SO 4 HCl HNO 3 Cmmn acids CH 3 COOH CH 3 COOH(aq) + OH-(aq) CH 3 COO - (aq) + H 2 O(l)

10 Sample Prblem What vlume f M HCl slutin is need t neutralize ml f a M KOH slutin? The balanced mlecular equatin is: HCl(aq) + KOH(aq) KCl(aq) + H 2 O(l) mlkoh 1ml OH 1 L ( 25.0 ml) = 1 L 1 ml KOH 1000 ml ml OH at neutralizatin, ml OH ( ml H + ) 56.3 ml HCl - = ml H - + 1ml HCl 1 L 1000 ml + 1 ml H = ml HCL 1 L 11.4 Rxn Stichimetry in Slutins Oxidatin-Reductin Titratins Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) xidatin reactin: lss f electrns : increase f xidatin number Zn Zn 2+ (aq) + 2e - reductin reactin: gain f electrns : decrease f xidatin number Cu 2+ (aq) + 2e - Cu

11 Redx reactin xidatin f Mg 2Mg(s)+O(g) 2 (g) MgO Assign xidatin numbers t all atms in the fllwing: CO 2, K 2 Cr 2 O 7, PCl 5 CO 2 xygen is -2 2 = -4, making C +4. K 2 Cr 2 O 7 xygen -2 7 = -14. K (grup 1A) is +1 2 =+2 2Cr must equal +12 = each Cr = +6 PCl 5 Cl (grup VIIA) = -1 5 = -5 P= Balancing Redx rxn Eq in Aqueus Slutins reactin between slid cpper sulfide and aq nitric acid slutin CuS(s) + NO (aq) Cu (aq) SO 4 (aq) + NO(g) 1. write tw unbalanced half reactins xidatin: CuS(s) Cu 2+ (aq)+so 2-4 (aq) reductin: NO - 3 (aq) NO(g) 2. Insert cefficnets fr all atms (except H, O) 3. Fr O : add H 2 O, CuS + 4H +SO 2 O Cu NO 3- NO + 2H 2 O 4. Adjust H number by adding H 3 O + (if basic sl, add OH - ) CuS + 12H 2 O Cu 2+ +SO H 3 O + NO 3- +4H 3 O + NO + 6H 2 O 5. Charge balance by adjusting electrn numbers CuS + 12H 2 O Cu 2+ + SO H 3 O + + 8e - NO H 3 O + + 3e - NO + 6H 2 O 6. Cmplete equatin by adjusting electrn numbers and add tw equatins Overall: 3CuS + 8NO H 3 O + 3Cu SO NO+12H 2 O

12 Sample prblem Ex1) Cr 2 O NO(g) Cr 3+ (aq) +NO 3- (aq) in acidic slutin Cr 2 O NO(g)+6H + 2Cr 3+ (aq) +2NO 3- (aq)+3h 2 O(l) disprprtinatin in case when a single substance is bth xidized and reduced ex2) MnO - 4- (aq)+mn 2+ (aq) MnO 2 (s) In acidic slutin 2MnO 4- (aq)+3mn 2+ (aq) +2H 2 O(l) 5MnO 2 (s)+4h + (aq) If in basic slutin 2MnO 4- (aq)+3mn 2+ (aq) +4OH - 5MnO 2 (s)+2h 2 O(l)

13 Redx Titratin MnO 4- (aq) + 5Fe 2+ (aq) + 8H 3 O + (aq) Mn 2+ (aq) + 5Fe 3+ (aq) + 12H 2 O(l) deep purple clrless MnO 4- (aq) Fe 2+ (aq) 2 5ml Fe + Amunt f Fe 2+ reacting = mle f MnO 4- x - 1ml MnO 4 Indirect titratin Indirect determinatin f [Ca 2+ (aq)]: multistep reactins precipitatin acid/base redx Ca 2+ (aq) + C 2 O 2-4 (aq) CaC 2 O 4 (s) (adding ammnium xlate) CaC 2 O 4 (s) + 2H 3 O + (aq) Ca 2+ (aq) + H 2 C 2 O 4 (aq) + 2H 2 O (precipitates it t are washed and redisslved d in acid) precipitatin acid/base 2MnO 4- (aq) + 5H 2 C 2 O 4 (aq) + 6H 3 O + (aq) 2Mn 2+ (aq) + 10CO 2 (g) + 14H 2 O(l) redx xalic acid is titrated with permanganate

14 11.5 Phase Equilibrium in Slutins : Nnvlatile slutes vapr pressure f nnvlatile slute abve the slutin is negligible : sucrse in water slvent vapr pressure nn zer, changes with cmpsitin f sl. if mle fractin f slvent (X 1 1) =1, P 1 P 1 (vapr p f pure slvent) if X 1 appraches 0 (pure slute), P 1 0 Rault s law : P 1 =X 1 P 1 Clligative prperties clligative prperties (in Latin wrd, clligare t cllect tgether) : depend n the cllective effect f the number f disslved particles rather than chemical nature. lwering f vapr pressure f slutin elevatin f biling pint depressin f freezing pint smtic pressure

15 Vapr-pressure lwering Change in vapr pressure f the slvent is prprtinal t the mle f slute ΔP 1 = P 1 -P 1 = X 1 P 1 -P 1 = -X 2 P 1, (X 1 = 1 - X 2 ) negative sign vapr-pressure lwering ex) What will be the vapr pressure f a slutin made by disslving 6.25g f glucse, C 6 H 12 O 6, in 50.0g f water at 25 C? Hw much was the vapr pressure f the pure water lwered? The vapr pressure f water at 25 C is 23.8 trr 1 ml ml glucse = 6.25 g = ml glucse 180. g 1 ml ml water = 50.0 g = 2.78 ml water 18.0g 2.78 χwater = = P = χ P = trr = 23.5 trr sln water water ( )( ) vapr pressure lwering = 23.8 trr trr = 0.2 trr Biling-pint elevatin because disslved slute reduces vapr pressure, temperature f sl must be increased t make the slutin bil ΔT b - ΔP 1 X 2 m ; (dilute slutin) ΔT b = K b m K b : cnstant fr bp elevatin m: mlality(ml/kg)) fr inic species ΔT b = ik b m i = sum f the cefficients f the ins (i = 1 fr mlecular cmpunds)

16 Freezing pint depressin the change in the freezing pint is: ΔT f = ik f m i = sum f the cefficients f the ins (i = 1 fr mlecular cmpunds) K f = freezing pint depressin cnstant t m = mlality Sample Prblem Calculate the biling pint elevatin and the freezing pint depressin f a slutin made by disslving 12.2g f KCl in 45.0g f water. K b = C/m and K f = 1.86 C/m i = 2 fr KCl K + + Cl 1 ml ml KCl = 12.2g = ml 74.6g ml KCl ml mkcl = = = 3.64 m kg water kg Δ T = ik m = C/ m 3.64 m = 3.73 C b b ( )( )( ) ( )( )( ) Δ = i m = m m = T f i K fm C/ m 3.64 m 13.5 C

17 Osmtic pressure the pressure n the slutin side f membrane > atm pressure f surface by π=ρgh (ρ: slutin density, g: gravity) Osmtic pressure π=ρgh (ρ: slutin density, g: gravity) By van t Hff discvered later as π = icrt R: gas cnst. smtic p: π (atm) c: mlarity c =n/v πv = nrt (n: mles, V: vlume) i : number f ins salinatin f cabbage, measurement f mlar mass desalinizatin f sea water reverse smsis

18 11.35 Desalinatin f sea water 11.6 Phase Equilibrium in Slutins : Vlatile Slutes vapr pressure f each species present is prprtinal t its mle fractin P i = X i P i (P i = vapr p f pure substance i) fr an ideal mixture f tw vlatile substances, vapr pressure f cmpnent 1 P 1 = X 1 P 1 vapr pressure f cmpnent 2 P 2 = X 2 P 2 = (1- X 1 )P 2 Henry s law at sufficiently lw X 2 (X 1 1) P 2 = k 2 X 2

19 Fractinal Distillatin a prcess in which the cmpnents are successively evaprated and recndensed. d islatin f ne cmpnent frm a mixture. Hexane (C 6 H 14 ) & heptane (C 7 H 16 ) : ideal slutin ver whle range f mle fractin at 25C, VP f pure hexane P 10 =0.198atm pure heptane P 20 =0.0600atm a slutin with 4.00mle f hexane & 6.00ml f heptane X 1 =0.4, X 2 =0.6 (mle fractins in liq state) The vapr in equilibrium with this ideal slutin P hexane = P 1 = X 1 P 1 = (0.400)(0.198 atm) = atm P heptane = P 2 = X 2P 2 = (0.600)( atm) = atm frm Daltn s law, P ttal = P 1 + P 2 = atm Fractinal Distillatin mle fractin at vapr, X 1, X 2 ' atm ' X 1 = = X 2 = = atm the vapr is enriched in the mre vlatile cmpnent (hexane) if sme f vapr is remved and cndensed t a liquid, the vapr in eq with this new slutin wuld be still richer in the mre vlatile cmpnent, repeating islatin. fractinal distillatin

20 Fractinal Distillatin cmpnent with lwer vapr pressure (2) jas a higher biling pint (T b ) Nnideal slutin negative deviatins frm Rault s law (strng slute-slvent frces) : shws biling pint maximum. maximum-biling azetrpe (H 2 O/HCl) psitive deviatins minimum-biling azetrpe (C 2 H 5 OH/H 2 O) : ethanl and water frm azetrpe with bp C & 4% (w/w) water attractins between (ethanl-ethanl), ethanl) (water-water) water) > (ethanl-water) then slutin bils at a lwer temp than either pure cmpnent.

21 11.7 Cllidal suspensin cllid : mixture f tw r mre susbstances in which ne phase is suspended as a large number f small particles in a secnd phase - particles are in cnstant mtin : Brwnian mtin temperature dependent examples: aersl sprays (liq suspened in gas) smke (slid particles in air) milk (fat drplets and slids in water) maynnaise (water drplets in il) paint (slid pigment in il, r in water) gemstne pal (water in silicn dixide) Cllids Precipitatins frm cllids heating flcculatin aggregatin sedimentatin

22 Hmewrk ,8,10,12,14,16,24,28,32,33,34,39,44,46,54,64