Triethanolamine is a tertiary amine which, for simplicity, I will call #sf(XN)#.
It acts as a weak base:
#sf(XN+H_2OrightleftharpoonsXNH^++OH^-)#
For which:
#sf(K_b=([XNH^+][OH^-])/([XN])-5.4xx10^(-7)color(white)(x)"mol/l")# at #sf(25^@C)#
These are equilibrium concentrations which we will approximate to initial concentrations.
#sf(pK_b=-log(5.4xx10^(-7))=6.26)#
We need to get the ratio of #sf([XNH^+])#: #sf([XN])# so rearranging:
#sf([OH^-]=K_bxx([XN])/([XNH^+]))#
Taking -ve logs of both sides gives:
#sf(pOH=pK_b+log(([XN^+])/([XN]))#
The target pH is 7.2 so we can say :
#sf(pH+pOH=14)#
#:.##sf(pOH=14-pH=14-7.2=6.8)#
Putting in the numbers:
#sf(6.8=6.26+log(([XNH^+])/([XN])))#
#:.##sf(log(([XNH^+])/([XN]))=6.8-6.26=0.54)#
From which:
#sf(([XNH^+])/([XN])=3.467)#
Since the total volume is common to both we can write the ratios in terms of moles:
#sf((n_(XNH^+))/(n_(XN))=3.467)#
We need to create a solution with the salt and base in this ratio.
Let #sf(V_1)# = volume of #sf(XN)#
Let #sf(V_2)# = volume of #sf(HCl)#
We know:
#sf(V_1+V_2=1)# lltre
By adding #sf(HCl)# the base will be converted to the salt. The question is how much to add to get this ratio, keeping the total volume to 1 litre?
The #sf(HCl)# reacts as follows:
#sf(XN+H^+rarrXNH^+)#
This tells us that for every mole of #sf(H^+)# added, then 1 mole of #sf(XNH^+)# will be formed and 1mole of #sf(XN)# will be consumed.
From the equation we can say :
#sf(n_(XNH^+)=n_(H^+))# which have been added.
and #sf(n_(XN))# = #sf(n_(XN_("initial"))-n_(H^+)" "color(red)((1)))#
#sf(n_(H^+))# added =#sf(0.06xxV_2)#
#:.##sf(n_(XNH^+))# formed =#sf(0.06xxV_2)#
#sf(n_(XN_("initial"))=0.04xxV_1)#
Since #sf(V_1=(1-V_2))# we can write this as:
#sf(n_(XN_("initial"))=0.04xx(1-V_2)" "color(red)((2)))#
Substituting #sf(n_(XN_("initial")))# from #sf(color(red)((2))# into #sf(color(red)((1))rArr)#
#sf(n_(XN)=0.04xx(1-V_2)-0.6V_2)#
#:.##sf(n_(XN)=0.04-0.04V_2-0.06V_2)#
#:.##sf(n_(XN)=0.04-0.1V_2" "color(red)((3)))#
We know that:
#sf(n_(XNH^+)=0.06V_2" "color(red)((4)))#
Dividing #sf(color(red)((4))# by #sf(color(red)((3))rArr)#
#sf((n_(XNH^+))/(n_(XN))=(0.06V_2)/(0.04-0.1V_2)=3.467)#
#:.##sf(0.06V_2=3.467(0.04-0.1V_2))#
#sf(V_2=0.13868/0.4067=0.3409color(white)(x)L)#
#:.##sf(V_1=1-0.34098=0.65902color(white)(x)L)#
This means we mix 659 ml of 0.04 M TEOA with 341 ml of 0.06 M HCl to get the target pH of 7.2.
You might get this by dilution but this would not have buffering action.
