a) After OmL means only NaNO_(2) is present.{:[NaNO_(2)+H_(2)OlongrightarrowHNO_(2)+OH^(-)],[{:[I,0.02M,-,-],[C,-x,+xM,+xM],[E,0.02-x,x,x]:}],[K_(b)=([HNO_(2)][OH^(-)])/([NO_(2)^(-)])],[Kb_(b)" of "NO_(2)^(-)=((K omega))/(Ka)],[=(1xx10^(-14))/(5xx10^(-4))],[k_(b)=2.0 xx10^(-11)],[k_(b)=(x^(2))/(0.02-x)],[k_(b)=(x^(2))/(0.02)quad0.02>>x],[x=sqrt(2xx10^(-11)xx0.02)],[x=6.32 xx10^(-7)M],[[OH^(-)]=x=6.32 xx10^(-7)M],[pOH=-log[OH^(-)]=-log(6.32 xx10^(-7))],[POH=6.2],[pH=14-6.2=7.8.]:}b) First find volume of HCe added at equivalence point.Moles of HCl= Moles of NaNO_(2).{:[" MHCe VHCl "=M_(N^(2NO))V_(N_(N)O_(2))],[V_(HCl)=(0.02Mxx60(mL))/(0.04M)],[V_(HCl)=30mL.]:}(2)/(3) Ve means (2)/(3)xx30mL=20mL.Moles of HCl added =0.04Mxx20mL =0.8mmel.Moles of NaNO_(2)=0.02Mxx60mL.=1.2mmelNow draw ICE table.{:[NaNO_(2)+HCllongrightarrowHNO_( ... See the full answer