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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, unlimited 1 M NaOH and an unlimited supply of distilled water. How would you prepare 1 L of a 0.02 M Tris Buffer, pH

Sagot :

Solution :

The reaction :

[tex]$\text{TrisH}^+ + \text{H}_2\text{O} \rightarrow \text{Tris}^- +\text{H}_3\text{O}^+$[/tex]

We have

[tex]$K_a = \frac{[\text{Tris}^-]\times[\text{H}_3\text{O}^-]}{[\text{TrisH}^+]}$[/tex]

     [tex]$=\frac{x^2}{0.02-x}$[/tex]

     [tex]$= 8.32 \times 10^{-9}$[/tex]

Clearing x, we have [tex]$x=1.29 \times 10^{-5}$[/tex] moles of acid

Now to reach pH = [tex]$7.8 (\text{ pOH} = 14-7.8 = 6.2)$[/tex], we must have an [tex]$OH^-$[/tex] concentration of

[tex]$[OH^-] = 10^{-pOH}$[/tex]

          [tex]$=10^{-6.2}$[/tex]

         [tex]$=6.31 \times 10^{-7}$[/tex] moles of base

We must add enough NaOH of 1 M to neutralize the acid calculated above and also add the calculated base.

[tex]$n \ NaOH = 1.29 \times 10^{-5} + 6.31 \times 10^{-7}$[/tex]

               [tex]$=1.35 \times 10^{-5}$[/tex] moles

Vol [tex]$NaOH = 1.35 \times 10^{-5} \text{ moles} \times \frac{1000 \ mL}{1 \ mol}$[/tex]

                 = 0.0135 L

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

                        = 3.152 g

To prepare the said solution we must mix

-- [tex]$3.152 \ g \text{ TrisH}^+$[/tex]

-- [tex]$0.0135 \ mL \ NaOH \ 1M$[/tex]

-- [tex]$\text{Gauge to 1000 mL with H}_2\text{O}$[/tex]

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