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 ?

Mar 1, 2017

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 $\textsf{X N}$.

It acts as a weak base:

$\textsf{X N + {H}_{2} O r i g h t \le f t h a r p \infty n s X N {H}^{+} + O {H}^{-}}$

For which:

$\textsf{{K}_{b} = \frac{\left[X N {H}^{+}\right] \left[O {H}^{-}\right]}{\left[X N\right]} - 5.4 \times {10}^{- 7} \textcolor{w h i t e}{x} \text{mol/l}}$ at $\textsf{{25}^{\circ} C}$

These are equilibrium concentrations which we will approximate to initial concentrations.

$\textsf{p {K}_{b} = - \log \left(5.4 \times {10}^{- 7}\right) = 6.26}$

We need to get the ratio of $\textsf{\left[X N {H}^{+}\right]}$: $\textsf{\left[X N\right]}$ so rearranging:

$\textsf{\left[O {H}^{-}\right] = {K}_{b} \times \frac{\left[X N\right]}{\left[X N {H}^{+}\right]}}$

Taking -ve logs of both sides gives:

sf(pOH=pK_b+log(([XN^+])/([XN]))

The target pH is 7.2 so we can say :

$\textsf{p H + p O H = 14}$

$\therefore$$\textsf{p O H = 14 - p H = 14 - 7.2 = 6.8}$

Putting in the numbers:

$\textsf{6.8 = 6.26 + \log \left(\frac{\left[X N {H}^{+}\right]}{\left[X N\right]}\right)}$

$\therefore$$\textsf{\log \left(\frac{\left[X N {H}^{+}\right]}{\left[X N\right]}\right) = 6.8 - 6.26 = 0.54}$

From which:

$\textsf{\frac{\left[X N {H}^{+}\right]}{\left[X N\right]} = 3.467}$

Since the total volume is common to both we can write the ratios in terms of moles:

$\textsf{\frac{{n}_{X N {H}^{+}}}{{n}_{X N}} = 3.467}$

We need to create a solution with the salt and base in this ratio.

Let $\textsf{{V}_{1}}$ = volume of $\textsf{X N}$

Let $\textsf{{V}_{2}}$ = volume of $\textsf{H C l}$

We know:

$\textsf{{V}_{1} + {V}_{2} = 1}$ lltre

By adding $\textsf{H C l}$ 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 $\textsf{H C l}$ reacts as follows:

$\textsf{X N + {H}^{+} \rightarrow X N {H}^{+}}$

This tells us that for every mole of $\textsf{{H}^{+}}$ added, then 1 mole of $\textsf{X N {H}^{+}}$ will be formed and 1mole of $\textsf{X N}$ will be consumed.

From the equation we can say :

$\textsf{{n}_{X N {H}^{+}} = {n}_{{H}^{+}}}$ which have been added.

and $\textsf{{n}_{X N}}$ = sf(n_(XN_("initial"))-n_(H^+)" "color(red)((1)))

$\textsf{{n}_{{H}^{+}}}$ added =$\textsf{0.06 \times {V}_{2}}$

$\therefore$$\textsf{{n}_{X N {H}^{+}}}$ formed =$\textsf{0.06 \times {V}_{2}}$

$\textsf{{n}_{X {N}_{\text{initial}}} = 0.04 \times {V}_{1}}$

Since $\textsf{{V}_{1} = \left(1 - {V}_{2}\right)}$ we can write this as:

sf(n_(XN_("initial"))=0.04xx(1-V_2)" "color(red)((2)))

Substituting $\textsf{{n}_{X {N}_{\text{initial}}}}$ from sf(color(red)((2)) into $\textsf{\textcolor{red}{\left(1\right)} \Rightarrow}$

$\textsf{{n}_{X N} = 0.04 \times \left(1 - {V}_{2}\right) - 0.6 {V}_{2}}$

$\therefore$$\textsf{{n}_{X N} = 0.04 - 0.04 {V}_{2} - 0.06 {V}_{2}}$

$\therefore$$\textsf{{n}_{X N} = 0.04 - 0.1 {V}_{2} \text{ } \textcolor{red}{\left(3\right)}}$

We know that:

$\textsf{{n}_{X N {H}^{+}} = 0.06 {V}_{2} \text{ } \textcolor{red}{\left(4\right)}}$

Dividing sf(color(red)((4)) by $\textsf{\textcolor{red}{\left(3\right)} \Rightarrow}$

$\textsf{\frac{{n}_{X N {H}^{+}}}{{n}_{X N}} = \frac{0.06 {V}_{2}}{0.04 - 0.1 {V}_{2}} = 3.467}$

$\therefore$$\textsf{0.06 {V}_{2} = 3.467 \left(0.04 - 0.1 {V}_{2}\right)}$

$\textsf{{V}_{2} = \frac{0.13868}{0.4067} = 0.3409 \textcolor{w h i t e}{x} L}$

$\therefore$$\textsf{{V}_{1} = 1 - 0.34098 = 0.65902 \textcolor{w h i t e}{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.