The two equilibria in question are:
#sf(CO_3^(2-)+H_2OrightleftharpoonsHCO_3^(-)+OH^-)#
#sf(K_(b1)=([HCO_3^(-)][OH^-])/([CO_3^(2-)]))#
and
#sf(HCO_3^(-)+H_2OrightleftharpoonsH_2CO_3+OH^-)#
#sf(K_(b2)=([H_2CO_3][OH^-])/([HCO_3^-]))#
To find the base dissociation constants we use:
#sf(K_(b1)xxK_(a2)=10^(-14))#
#:.##sf(K_(b1)=(10^(-14))/(4.8xx10^(-11))=2.08xx10^(-4))#
and
#sf(K_(b2)xxK_(a1)=10^(-14))#
#:.##sf(K_(b2)=(10^(-14))/(4.2xx10^(-7))=2.38xx10^(-6))#
Because #sf(K_(b1)# is about 200x greater than #sf(K_(b2)# I will assume that nearly all of the #sf(OH^-)# ions come from the first dissociation.
We know that #sf(pH=11.9)# so we can say:
#sf(pOH=14-pH=14-11.9=2.1)#
#:.##sf([OH^-]=7.94xx10^(-3)=0.00794color(white)(x)"mol/l")#
If we call the initial concentration of #sf(CO_3^(2-)# #sf(Xcolor(white)(x)"mol/l")# we can set up an #sf(ICE)# table in #sf("mol/l"rArr)#
#" "sf(CO_3^(2-)+H_2OrightleftharpoonsHCO_3^(-)+OH^(-))#
#sf(I" " " "X" " " " " " " " " "0" " " " " "0)#
#sf(C" " ""-x" "" " " "+x" " " " " "+x)#
#sf(E""" "(X-0.00794)" "0.00794" "0.00794)#
Assuming #sf((X-0.00794)rArrX)# we can say:
#sf(K_(b1)=2.08xx10^(-4)=((0.00794)^2)/(X))#
#:.##sf(X=(63.043xx10^(-6))/(2.08xx10^(-4))color(white)(x)"mol/l")#
#sf(X=0.30color(white)(x)"mol/l")#
Quick Method:
Expression for pOH of a weak base gives:
#sf(pOH=1/2[pK_b-logb])#
#sf(pOH=1/2[3.682-(logb)]=2.1)#
#sf(logb/2=-0.259)#
#sf(b=0.30color(white)(x)"mol/l")#
