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
