5.3.2 Equilibrium of ionization for polyprotic acids H,CO3H++HCO3 three coexisting K82,1=4.2x10-7 equilibria HCO3H++CO32 K81 K9a,2=4.7x10-11 K82 H20= H++OH Kow=1.0x10-14
HCO3 - H+ + CO3 2- H2O H+ + OHH2CO3 H+ + HCO3 - Ｋ a, 1 = 4.2 x 10-7 Ｋ a, 2 = 4.7 x 10-11 Ｋ w = 1.0 x 10-14 Ｋa, 1 Ｋa, 2 >> 5.3.2 Equilibrium of ionization for polyprotic acids three coexisting equilibria
In general, 1st step:H2CO;(aq)+H2O(1)-H;O"(aq)+HCO3 (aq) kH,co,)-k,0)kIc0-42x10-7 Rc(H,CO:) 2nd step:HCO3 (aq)+H2O(1)-H3O(aq)+CO(aq) k8H,c0,)-k,o)c0}-47x10 c(HCO3 when Ka.>103,the H+comes mostly from the first step of dissociation reaction
{ }{ } { } 7 2 3 3 3 a1 2 3 4.2 10 (H CO ) (H O ) (HCO ) (H CO ) - + = = c c c K － { }{ } { } 1 1 3 2 3 3 a2 2 3 4.7 10 (HCO ) (H O ) (CO ) (H CO ) - + = = － － c c c K In general, H CO (aq) H O(l) H O (aq) HCO (aq) 2 3 2 3 3 － 1 st step： + + + HCO (aq) H O(l) H O (aq) CO (aq) 2 3 2 3( 3 － － 2 nd step: + + + when Ka,1 /Ka,2 > 103 , the H+ comes mostly from the first step of dissociation reaction
Example:Calculate the concentrations of H3O,H2CO3, HCO3,CO32-,OH-ions and the pH in a 0.010 mol-L-1 H2CO3 solution.It is known that Ka=4.2 x 107,=4.7 x 10-11 Answer: H2CO;(aq)+H2O(D=H;O*(aq)+HCO3 (aq)
Example：Calculate the concentrations of H3O+ , H2CO3 , HCO3 －, CO3 2－, OH－ ions and the pH in a 0.010 mol·L-1 H2CO3 solution. It is known that Ka ,1 = 4.2 x 10-7 ,Ｋa ,2 = 4.7 x 10-11 Answer H C O (aq) H O(l) H O (aq) HCO (aq) － 2 3 + 2 3 + 3 ： +
H2CO3(aq)+H2O(I)-H3O(aq)+HCO3 (aq) Equi./(mol.L)0.010-x X X K9(H,C03)=4.2×107 [c(H;O)IIc(HCO3 ) x2 C(H2CO3) 0.010-x 0.010-x≈0.010 x=6.5×10-5 c(H,0*)=c(HC03)=6.5×10-5mol·L1 c(H2C03)=0.010mol.L1
5 0.010 0.010 6.5 10 － － x x = 5 1 (H3 O ) (HCO3 ) 6.5 10 mol L － － － = = + c c 1 (H2 CO3 ) 0.010mol L － c = Equi. / (mol L ) 0.010 1 －x x x - H CO (aq) H O(l) H O (aq) HCO (aq) － 2 3 2 3 3 + + + 7 a1 (H2 CO3 ) 4.2 10- K = [ [ ] x x c c [c － － (H CO ) 0.010 (H O )] (HCO )] 2 2 3 3 3 = = +
Considering the second stage of ionization to calculate c(C032-) HC03(aq)+H,00=H,0*(aq)+C0?(aq) cg/(mol·L)6.5×10-5-y 6.5×10-5+yy K8(H2C03)=4.7×101" =[c(H30[c(C03】=(6.5x105+y)y [c(HCO3] 6.5x105-y 6.5×105±y≈6.5×105，y=K8=4.7×10-" c(C03)=K3mol.L=4.7×10-11mol·L
/(mol L ) 6.5 10 6.5 10 HCO (aq) H O(l) H O (aq) C O (aq) 1 5 5 eq 2 3 2 3 3 c y + y y + + - + － － － － － 1 1 a2 (H2 CO3 ) 4.7 10- K = 1 1 a2 5 5 6.5 10 6.5 10 , 4.7 10 - - - y y = K = 1 1 1 1 a2 2 (C O3 ) mol L 4.7 10 mol L - - - c = K = － Considering the second stage of ionization to calculate c(CO3 2- ) y y y c c c - + = = - + - 5 5 3 2 3 3 6.5x 10 (6.5x 10 ) [ (HCO )] [ (H O )][ (CO )] － －