Question

Tris is a molecule that can be used to prepare buffers for biochemical experiments. It exists in two forms: Tris (a base) and TrisH (an acid). The MW of Tris base is 121.14 g/mol; the MW of TrisH is 157.6 g/mol (the extra weight is due to the Cl- counterion that is present in the acid). The Ka of the acid is 8.32 X 10-9. Assume that you have TrisH in solid form (a powder), unlimited 1M HCl, 1 M NaOH and distilled water. How would you prepare 1 L of a 0.02 M Tris Buffer, pH?

Answer 1

Solution :

For the reaction :

$\text{TrisH}^+ + H_2O \rightarrow \text{Trish}^- + H_3O^+$

we have

$Ka = \frac{[\text{Tris}^- \times H_3O]}{\text{Tris}^+}$

   $=\frac{x^2}{0.02 -x}$

  $=8.32 \times 10^{-9}$

Clearing $x$, we have $x = 1.29 \times 10^{-5} \text{ moles of acid}$

So to reach $\text{pH} = 7.8 (\text{pOH}= 14-7.8=6.2)$, one must have the $\text{OH}^-$ concentration of the :

$\text{[OH}^-]=10^{-pOH} = 6.31 \times 10^{-7} \text{ moles of base}$

So we can add enough of 1 M NaOH in order to neutralize the acid that is calculated above and also adding the calculated base.

$\text {n NaOH}=1.29 \times 10^{-5}+6.31 \times 10^{-7}$

            $= 1.35 \times 10^{-5} \text{ moles}$

Volume NaOH $= 1.35 \times 10^{-5} \text{ moles} \times \frac{1000 \text{ mL}}{1 \text{ mol}} = 0.0135 \text{ mL}$

Tris mass $H^+ = 0.02 \text{ mol} \times 157.6 \text{ g/mol}=3.152 \text{ g}$

Now to prepare the said solution we must mix:

$3.152 \text{ g Tris H} + 0.0135 \text{ mL NaOH} \ 1 M$ gauge to 1000 mL with water.