5.2.1 Ionization equilibrium of water HO (1)+H2O(1)-H,O(aq)+OH (aq) or H20(①)=H(aq)+OH(aq) K-(HO)c(OH) ce ce or K=c(H;O)c(OH) K9一ion-product constant of water(水的离 子积常数) 25℃：c(H)=c(OH-)=1.0×10-moL1 K9=1.0×10-14 100℃： K=5.43×103T个，K个WNh3
5.2.1 Ionization equilibrium of water H 2O (l) + H 2O(l) H 3 O + (aq) + OH －(aq) or H 2O (l) H + (aq) + OH －(aq) — ion-product constant of water (水的离 子积常数） KW 25℃： c(H +)= c(OH －)=1.0 ×10-7mol·L-1 100℃： =1.0 ×10-14 KW =5.43 ×10-13 KW T ↑ ， KW ↑ ( H O ) (OH ) 3 + − or K = c c W ( H O ) (OH ) 3 + − = c c c c KW Why?
11 Anwser: Von't Holf equation R 5.2.2 The pH scale (a measure of acidity) pH=-Igic(H,O")} pOH=-1gc(OH-) Since K={c(H,0*)}{c(OH)}=1.0x1014 SO -lgc(H")-lgc(OH)=-1g K=14 pH+pOH=pK=14 pH only for the cases of [H+]or [OH-]<1 mol.1
pOH = −lg{ (OH )}− c pH lg{ } (H O ) 3 = − + c 5.2.2 The pH scale (a measure of acidity) { }{ } (H O ) (OH ) 1.0 10 14 = 3 = × + − − Since KW c c − lg (H ) − lg (OH ) = −lg =14 + − so c c KW ∴ pH + pOH = pKW =14 ⎟⎟⎠⎞ ⎜⎜⎝⎛ − Δ ∴ = 1 2 θ r m θ1θ2 1 1 ln R T T H KK Anwser: Von’t Holf equation pH only for the cases of [H+] or [OH-]≤１mol.l-1
Q?Calculate the pH of a solution of 1.0X10-mol.dm-3 HCH2O。 Answer: pH=-lg0x10-8=8 Proton conservation equation in acid-base theory Basis:the total number of H+obtained by bases the total number of H donated by acids. How to write a proton conservation equation?
Q? Calculate the pH of a solution of 1.0×10-8mol.dm-3 HCl/H2O。 Answer： pH=-lg1.0x10-8 = 8 Proton conservation equation in acid-base theory Basis: the total number of H+ obtained by bases = the total number of H+ donated by acids. How to write a proton conservation equation?
(1)listing all substances involved in H+-transfer processes (2)selecting zero standards:Many of species take part in proton transfer.Generally solvent and solute is involved in proton transfer of molecules or ions. (3)listing products after accepting and donating H+, numbers of H+donated and accepted,respectively( 得、失质子产物及得失质子数与零水准比较) (4)listing the proton conservation equation(根据得、失质子总数相 等的原则，以浓度表示得失电子数目) The concentration of protons at equilibrium will be multiplied by each proton number,then add together,write on the left; The concentration of lost protons at equilibrium will be multiplied by each lost proton number,then add together,write on the right;
(1) listing all substances involved in H+-transfer processes (2) selecting zero standards: Many of species take part in proton transfer. Generally solvent and solute is involved in proton transfer of molecules or ions. (3) listing products after accepting and donating H+, numbers of H+ donated and accepted, respectively( 列出 得、失质子产物及得失质子数:与零水准比较) (4) listing the proton conservation equation (根据得、失质子总数相 等的原则, 以浓度表示得失电子数目) The concentration of protons at equilibrium will be multiplied by each proton number, then add together, write on the left; The concentration of lost protons at equilibrium will be multiplied by each lost proton number, then add together,write on the right;
Example (1)listing all substances involved in H+transfer processes HAc/H,O system solvent:H2O,OH,H3Oi solute:HAc,Ac (2)zero standards H2O HAc (3)listing products after accepting and donating H+, numbers of H+donated and accepted,respectively Proton-accepted zero standards proton-lost products products HAc H+ Ac -H+ HO+ H20 OH (4)Write out H+conservation equation 质子守恒式∑，[得质子产物=∑y,[失质子产物] [H3O*]=[Ac]+[OH-] concentration at equilibria
Example 1: Proton-accepted products proton-lost products zero standards HAc/H2O system H3O+ HAc AcH2O OH- -H+ -H+ +H+ [H3O+] = [Ac-] + [OH-] 质子守恒式 concentration at equilibria (1) listing all substances involved in H+ transfer processes solvent： H2O，OH-, H3O+ solute：HAc, Ac- (2) zero standards ：H2O HAc ∑ [得质子产物] = ∑ [失质子产物] i j v v (3) listing products after accepting and donating H+, numbers of H+ donated and accepted, respectively (4) Write out H+ conservation equation