When the acid is added to the carbonate solution, the #sf(CO_3^(2-)# ions are first protonated:
#sf(H^(+)+CO_3^(2-)rarrHCO_3^(-))#
This means we are interested in the 2nd dissociation of carbonic acid:
#sf(HCO_3^(-)rightleftharpoonsH^(+)+CO_3^(2-))#
#sf(K_(a2)=([H^+][CO_3^(2-)])/([HCO_3^(-)])=4.69xx10^(-11)#
Rearranging:
#sf([H^+]=K_(a2)xx([HCO_3^-])/([CO_3^(2-)]))#
#sf(pH=9.45)#
#:.##sf(-log[H^+]=9.45)#
#sf([H^+]=3.55xx10^(-10)color(white)(x)"mol/l")#
#:.##sf(3.55xx10^(-10)=4.69xx10^(-11)xx([HCO_3^-])/([CO_3^(2-)]))#
#sf(([HCO_3^-])/([CO_3^(2-)])=(3.55xx10^(-10))/(4.69xx10^(-11))=7.569" "color(red)((1)))#
This is the ratio of acid : co-base we need to achieve to get the target pH of 9.45.
A certain volume of HCl which I will call V must be added such that the total volume of buffer created is 1 Litre.
From the original equation you can see that if n moles of HCl is added, then n moles of #sf(HCO_3^-)# is formed.
Since #sf(n=cxxv)# we can say that the no. moles #sf(H^+)# added is given by:
#sf(n_(H^+)=0.200xxV)#
So we can say that the no. moles #sf(HCO_3^-)# formed is given by:
#sf(n_(HCO_3^(-))=0.200xxV)#
Since the total volume must be 1 Litre we can say that the initial moles of #sf(CO_3^(2-))# must be #sf(0.3xx(1-V))#.
So the no. moles of #sf(CO_3^(2-))# remaining is given by:
#sf(n_(CO_3^(2-))=0.300xx(1-V)-0.200xxV)#
Since the total volume is common, we can put these into #sf(color(red)((1))rArr)#
#sf((0.2V)/(0.3(1-V)-0.2V)=7.569)#
#sf((0.2V)/(0.3-0.3V-0.2V)=7.569)#
#sf((0.2V)/(0.3-0.5V)=7.569)#
#sf(0.2V=7.569(0.3-0.5V))#
#sf(0.2V=2.2707-3.7845V)#
#sf(3.9845V=2.2707)#
#sf(V=2.2707/3.9845=0.5699color(white)(x)L)#
#sf(V=569.9color(white)(x)ml)#
This is the volume of #sf(0.200M)# HCl required.
The volume of #sf(0.300M)# #sf(Na_2CO_3)# required will be #sf(1-0.5699=0.4301color(white)(x)L)#
#sf(=430.1color(white)(x)ml)#
Iteration check:
#sf([H^+]=K_(a2)xx([HCO_3^-])/([CO_3^(2-)]))#
#sf([H^+]=4.69xx10^(-11)xx(0.2V)/(0.3(1-V)-0.2V))#
#sf([H^+]=4.69xx10^(-11)xx(0.2xx0.5699)/((0.3xx0.4301)-0.2xx0.5699)#
#sf([H^+]=0.35517xx10^(-9)color(white)(x)"mol/l")#
#sf(pH=-log(0.35517xx10^(-9))=9.45)#
So that's all good.
