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

For the reaction :

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

we have

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

`$=(x^2)/(0.02 -x)$`

`$=8.32 * 10^(-9)$`

Clearing
`$x$`, we have
`$x = 1.29 * 10^(-5) \text{ moles of acid}$`

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

`$\text{[OH}^-]=10^(-pOH) = 6.31 * 10^(-7) \text{ moles of base}$`

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.

`$\text {n NaOH}=1.29 * 10^(-5)+6.31 * 10^(-7)$`

`$= 1.35 * 10^(-5) \text{ moles}$`

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

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

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.