The #sf(HCO_3^-)# ion is basic:
#sf(HCO_(3(aq))^(-)+H_2O_((l))rightleftharpoonsH_2CO_(3(aq))+OH_((aq))^-)#
#sf(K_(b)=([H_2CO_3][OH^(-)])/([HCO_3^(-)])=1.99xx10^(-8)color(white)(x)"mol/l")#
We know that #sf([H_2CO_3]=[OH^(-)])#
Because the dissociation is small I will assume that the equilibrium concentration of #sf([HCO_3^-]}# is equal to #sf(0.1color(white)(x)"mol/l")#
So now we can write:
#sf(K_b=[OH^(-)]^(2)/(0.1)=1.99xx10^(-8))#
#:.##sf([OH^-]^(2)=1.99xx10^(-8)xx0.1=1.99xx10^(-9))#
From which:
#sf([OH^-]=4.46x10^(-5)color(white)(x)"mol/l")#
#:.##sf(pOH=-log(4.46xx10^(-5))=4.35)#
#sf(pH+pOH=14)#
#:.##sf(pH=14-4.35=9.65)#
