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.