Question

Triethanolamine is a base. What proportion of a 0.04 M solution of this base should be mixed with 0.06 M hydrochloric acid to give 1 litre of buffer solution of pH 7.2 ?

Answer 1

You would mix 659 ml of 0.04 M TEOA with 341 ml of 0.06 M HCl.

Explanation:

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.