1、Chemistry in Aqueous Solutionchapter 15&16 Types of Solutions (Chapter 12.1)Acids,bases&Salts (Chapter 15&16)Precipitates&Solubility (Chapter 16)Solution is a homogenous mixture of solute(in a smaller amount)and solvent(in a larger amount).Suspension is a heterogenous mixture of solute and solvent.A
2、 colloid is an intermediate state between the solution and suspension,heterogenous in nature and kinetically stable.Solution Colloid SuspensionParticle size(nm)1000 nmApparence transparent transparent/cloudy settle down 1.Solutions,suspensions and colloidsH+OH-H2O(1)Arrhenius Ionization Theory(1884,
3、Swedish)(2)Brnsted Proton theory(1932,Danish)NH3 +H2O NH4+OH-acidconjugate basebaseconjugate acidpair Ipair IIAcid-base reaction is a process of proton transfer between two pairs of conjugate acid-base,without concept of salt.2.Definition of acids,bases and saltsAcid +base =salt +water (3)Lewis Elec
4、tron-Pair theory(1932,American)A Lewis acid:it can accept a pair of electrons.H+H O H+OH-acidbaseN HHHH+N HHHH+A Lewis base:It can donate a pair of electronsN HHHF BFF+F BFFN HHHAl2O33.pH:A measure of acidity pH=-log H+H2O(l)H+(aq)+OH-(aq)Kc=H+OH-H2OKcH2O=Kw=H+OH-ion-product constantKw=1.0 x 10-14 a
5、t 25 oC H+=OH-H+OH-H+OH-neutralacidicbasic pH=7pH 7At 250CpH+pOH=pKwOHH+OHHOHHHOH-+autoionization of waterWhat is the concentration of OH-ions in 1.3 M HCl solution?Kw=H+OH-=1.0 x 10-14H+=1.3 MOH-=KwH+1 x 10-141.3=7.7 x 10-15 M15.2pH indicator:a weak acid or base of different colors.HIn(aq)H+(aq)+In
6、-(aq)10HInIn-HIn predominates 10HInIn-In-predominatespH meter:an electrochemical apparatus that can directly measure the pH value of solution.Titration curve of a strong acid with a strong base.Equivalence point 4.Ionization constant of weak acid or base HA(aq)+H2O(l)H3O+(aq)+A-(aq)Ka=H3O+A-HAacid i
7、onization constantKaacidstrength A-(aq)+H2O(l)HA(aq)+OH-(aq)Kb=HAOH-A-base ionization constantKbbasestrengthConjugate acid-base pKa +pKb=pKwWhat is the pH of a 0.30 M HCOOH solution at 250C?HCOOH(aq)H+(aq)+HCOO-(aq)Ka=H+HCOO-HCOOH=1.7 x 10-4Initial(M)equilibrium(M)0.300.000.30-x0.00 xxx20.30-x=1.7 x
8、 10-4x20.30=1.7 x 10-40.30 x 0.30Since Ka Ka2 Ka3(2)First approximation:1+2a+H -H KCH+HS-S2-Ka2(3)Exact solution as follows.12112aa2-+2+aaaS H H CK KKK KH2S H+HS-HS-H+S2-12aHHSKH S22aHSKHS112111212+4+3+2aaaaw+awaaaawH -H +(-)H -(+2)H -=0KK KK CKK KK K CK K KPOHPOPOHPOHPOHPOH-34-24-42434343)(5.Acid S
9、trength and Molecular StructureH XH+X-The stronger the bondThe weaker the acid15.9Intermolecular force and hydrogen bonding?ZOHZO-+H+d-d+15.9H O Cl OOHClO3 HBrO3H O Br OOOxoacidsHClO4 HClO3 HClO2 HClO 6.buffer solutionWhat is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK?HCOOH(aq)H+(
10、aq)+HCOO-(aq)0.300.000.30-x0.52x0.52+xInitial(M)Equilibrium(M)Common ion effect0.30 x 0.300.52+x 0.52Ka=H+A-HA-log H+=-log Ka+logA-HApH=pKa+logA-HApH=3.77+log0.520.30=4.01=9.20Calculate the pH of the solution:(1)0.30 M NH3/0.36 M NH4Cl buffer(80 mL).(2)after(1)is mixed with 20.0 mL 0.050 M NaOH.NH4+
11、(aq)H+(aq)+NH3(aq)pH=pKa+logNH3NH4+pH=9.25+log0.300.36=9.17NH4+(aq)+OH-(aq)H2O(l)+NH3(aq)start(M)end(M)0.290.0100.240.2800.25pH=9.25+log0.250.2816.3A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base.pH=pKa+logconjugate baseacidA buffer
12、 solution is a solution containing a weak acid or a weak base and the corresponding salt!Which of the following are buffer systems?(a)KF/HF,(b)KBr/HBr,(c)Na2CO3/NaHCO3Answer:(a)buffer solution (b)no buffer (c)buffer solutionMajor factor determining the solution pH.pKa=3.15 pKa1=6.38pKa2=10.32 7.Titr
13、ation of acid and baseA strong acid titrated with a strong base.Equivalence point A weak acid titrated with a strong base CH3COOH(25.00 mL,0.10 M)NaOH(0.10 M)pKa 4.75pH=pKa+logCH3COO-CH3COOHpKb 4.75 A weak base titrated with a strong acid NH3H2O(0.10 M,25.00 mL)HCl(0.10 M)pOH=pKb+logNH4+NH38.acid-ba
14、se properties of salts Neutral Solutions:a salt produced by strong acid and strong base.Basic Solutions:a salt formed from strong base and weak acid.CH3COONa(s)Na+(aq)+CH3COO-(aq)H2OCH3COO-(aq)+H2O(l)CH3COOH(aq)+OH-(aq)NaCl(s)Na+(aq)+Cl-(aq)Acidic Solutions:NH4Cl(s)NH4+(aq)+Cl-(aq)H2O NH4+(aq)NH3(aq
15、)+H+(aq)a salt derived from strong acid and weak base.Al(H2O)6(aq)Al(OH)(H2O)5(aq)+H+(aq)3+2+The products are complex,greatly depending on OH/Al ratio,pH,T.etc.Al(H2O)63+Al2(OH)2(H2O)44+AlO4Al12(OH)24(H2O)127+Al(OH)4-very acidic oligomers,polymers,polyoxocations very basicPb2+(aq)+2 I-(aq)PbI2(s)9.P
16、recipitates&Solubility PbI2(s)Pb2+(aq)+2I-(aq)Ksp=Pb2+I-2Ksp is the solubility product constantMgF2(s)Mg2+(aq)+2F-(aq)Ksp=Mg2+F-2Ag2CO3(s)2Ag+(aq)+CO32-(aq)Ksp=Ag+2CO32-Ca3(PO4)2(s)3Ca2+(aq)+2PO43-(aq)Ksp=Ca2+3PO33-225 oC Ksp=1.4 10-8 Ksp and SolubilityWhat is the solubility of silver chloride in g/
17、L at 25 oC?AgCl(s)Ag+(aq)+Cl-(aq)Ksp=Ag+Cl-Initial(M)Change(M)Equilibrium(M)0+s0+sssKsp=s2s=Ksps=1.3 x 10-5Ag+=1.3 x 10-5 MCl-=1.3 x 10-5 M1.3 x 10-5 mol AgCl1 L soln143.35 g AgCl1 mol AgClxKsp=1.6 x 10-10p For same type of precipitates,the solubility increases with KspAgCl(s)AgBr(s)AgI(s)Color Whit
18、e pale yellow yellowKsp 1.6 10-10 7.7 10-13 8.3 10-17S(M)1.3 10-5 8.8 10-6 9.3 10-8 decreasingp Among different type of precipitates,the solubility must be calculated from one to another.Ag2CrO4(s)Ag+(aq)+CrO42-(aq)Initial(M)0 0Equilibrium(M)2s sKsp=1.1 10-12s(2s)2=Ksp34spKs=6.5 10-5 (M)Dissolution
19、of an ionic solid in aqueous solution:Q=KspSaturated solutionQ KspSupersaturated solutionPrecipitate will formAgCl(s)Ag+(aq)+Cl-(aq)=1.6 x 10-10+-AgClQCC+-Ag Cl spKat 25 oCPrecipitate and ionsIf 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 M CaCl2,will a precipitate form?The ions present in
20、solution are Na+,OH-,Ca2+,Cl-.Only possible precipitate is Ca(OH)2(solubility rules).Is Q Ksp for Ca(OH)2?Ca2+0=0.100 MOH-0=4.0 x 10-4 MKsp=Ca2+OH-2=8.0 x 10-6Q=Ca2+0OH-02=0.10 x(4.0 x 10-4)2=1.6 x 10-8Q KspTherefore,No precipitate will formDissolution and Transformation of Precipitates1.Addition of
21、 acid or baseAl3+(aq)OH-OH-H+H+Al(OH)4-(aq)Al(OH)3(s)ZnS(s)+H+(aq)Zn2+(aq)+H2S(aq)2.Addition of complexing reagentCu2+(aq)NH3NH3H+H+Cu(NH3)42+(aq)Cu(OH)2(s)Ag+(aq)Cl-NH3Ag(NH3)2+(aq)AgCl(s)Br-AgBr(s)Ag(S2O3)23-S2O32-AgI(s)I-4.Addition of oxidant(or reductant)CuS(s)+H+(aq)+NO3-(aq)Cu2+(aq)+S(s)+NO(g)+H2O(l)3 8 2 3 3 2 43.Transfer into less soluble precipitatesCaSO4(s)+CO32-(aq)CaCO3(s)+SO42-(aq)Ksp 7.1 10-5 8.7 10-95.Decomposition into other species CuS(s)+O2(g)CuO(s)+SO2(g)D D15.6815.9116.9716.10116.11016.115Exercises
