Anionic hydrolysis of strong conjugate base #CH_3COO^-# with its ICE-table
#CH_3COO^"-" +H_2O stackrel(K_h) (rightleftharpoons)CH_3COOH+OH^-#
#I" "" "c" "" "" "" "" "" "" " 0" "" "" "" "" "0#
#C" "-ch" "" "" "" "" "" " ch" "" "" "" "ch#
#E" "c-ch" "" "" "" "" "" " ch" "" "" "" "ch#
Where c is concentration of #CH_3COO^-# or sodium acetate before hydrolysis and h is the degree of hydrolysis.
So #K_h=([CH_3COOH][OH])/([CH_3COO^-])......color(red)((2))#
#=>K_h=((ch)(ch))/(c(1-h))~~ch^2#
#=>h=sqrt(K_h/c) ......color(red)((3))#
Considering ionic equilibrium of dissociation of #CH_3COOH# in aqueous solution.
#CH_3COOH + H_2O stackrel(K_a) rightleftharpoons H_3O^(+) + CH_3COO^-#
#K_a = ([H_3O^(+) ][CH_3COO^-])/([CH_3COOH])
......color(red)((4))#
Considering ionic equilibrium of dissociation of #H_2O#
#H_2O + H_2O stackrel(K_w)rightleftharpoons H_3O^(+)+OH^-#
The ionic product of water
#K_w=[H_3O^+][OH^-]=10^-14" ".....color(red)((5))#
Taking log
#-logK_w=-log(10^-14)=14#
#=>pK_w=-log([H_3O^+][OH^-])=log(10^-14)=14#
#=>pK_w=pH+pOH=-log(10^-14)=14 ......color(red)((6))#
Now by (4) and (5)
#K_w/K_a=([CH_3COOH][OH])/([CH_3COO^-])=K_h#
Again
#[OH^-]=ch=cxxsqrt(K_h/c)=sqrt(K_hc)#
#pOH=-log[OH^-]=-logsqrt(K_hc)#
#=>pOH=-logsqrt(K_hc)#
#=-1/2logK_h-1/2logc#
#=>pOH=1/2(-log(K_w/K_a)-logc)#
#=>pOH=1/2(-logK_w+logK_a-logc)#
#=>pOH=1/2(pK_w-pK_a-logc)#
#=>14-pH=1/2(14-pK_a-logc)#
#=>14-pH=7-1/2(pK_a+logc)#
#=>pH=7+1/2(pK_a+logc)....(7)#
Here #pH=9.48#
#K_a=1.8xx10^-5#
#=>pK_a=-log(1.8xx10^-5)#
#=5-log1.8=4.74#
Inserting the values in (7)
#9.48=7+1/2(4.74+logc)#
#=>1/2logc=9.48-7-2.37=0.11#
#=>logc=0.22#
#c=1.66M#
