化学原理Chemistry课件-post+8

1、Chemistry in Aqueous Solutionchapter 15 & 16Types 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 a

2、nd solvent.A 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

3、Ionization Theory (1884, 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 salt

4、sAcid + base = salt + water (3) Lewis Electron-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

5、+OH-H2OKcH2O = Kw = H+OH-ion-product constantKw = 1.0 x 10-14 at 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

6、x 10-15 M15.2pH indicator: a weak acid or base of different colors.HIn (aq) H+ (aq) + In- (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 po

7、int 4. Ionization constant of weak acid or base HA (aq) + H2O (l) H3O+ (aq) + A- (aq)Ka =H3O+A-HAacid ionization 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

8、 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 10-4x20.30= 1.7 x 10-40.30 x 0.30Since Ka Ka2 Ka3(2) First approximation:H+ HS-S2- Ka2(3) Exact solution as follows.H2S H+ + HS-HS- H+ + S2-5. Acid Strength and Molecula

9、r 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+ (aq) + HCOO-

10、 (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

11、mL 0.050 M NaOH.NH4+ (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 eithe

12、r acid or base.pH = pKa + logconjugate baseacidA buffer 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 facto

13、r determining the solution pH.pKa = 3.15 pKa1 = 6.38pKa2 = 10.32 7. Titration 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 titrat

14、ed with a strong acid NH3H2O (0.10 M, 25.00 mL) HCl ( 0.10 M )pOH = pKb + logNH4+NH38. acid-base 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) + H2

15、O (l) CH3COOH (aq) + OH- (aq)NaCl(s) Na+ (aq) + Cl- (aq)Acidic Solutions:NH4Cl (s) NH4+ (aq) + Cl- (aq)H2O NH4+ (aq) NH3 (aq) + 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

16、(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. Precipitates & Solubility PbI2(s) Pb2+ (aq) + 2I- (aq)Ksp = Pb2+I-2Ksp is the solubility product constantMgF2 (s) Mg2+ (aq) + 2F- (aq)Ksp = Mg2+F-2Ag2C

17、O3 (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/L at 25 oC?AgCl (s) Ag+ (aq) + Cl- (aq)Ksp = Ag+Cl-Initial (M)Change (M)Equilibrium (M)0+s0+sssKsp = s2s = Ksps =

18、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 10-10 For same type of precipitates, the solubility increases with KspAgCl (s) AgBr(s) AgI(s)Color White 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 de

19、creasing 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 = Ksp= 6.5 10-5 (M)Dissolution of an ionic solid in aqueous solution:Q = KspSaturated solutionQ KspSupersa

20、turated solutionPrecipitate will formAgCl (s) Ag+ (aq) + Cl- (aq)= 1.6 x 10-10at 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 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 =