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, 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

Answer 1

To prepare 1 L of a 0.02 M Tris buffer at pH 7.8, you need to start with the conjugate base form of Tris and add HCl to adjust the pH. Calculate the moles of Tris using the formula moles = Molarity x volume. Convert moles to grams using the formula mass = moles x molar mass. Use the Henderson-Hasselbalch equation to calculate the ratio of Tris to TrisH and adjust the pH to 7.8. Finally, calculate the volume of 6 M HCl needed to adjust the pH.

Step-by-step explanation:

To prepare 1 L of a 0.02 M Tris buffer at pH 7.8, you need to start with the conjugate base form of Tris and add HCl to adjust the pH.

The first step is to calculate the number of moles of Tris you need. Using the formula Molarity (M) = moles/volume (L), rearrange the formula to calculate moles = Molarity x volume. Since you need 0.02 moles of Tris and the volume is 1 L, you will need 0.02 moles of Tris.

To convert moles to grams, you can use the formula moles = mass/molar mass. Rearranging the formula, mass = moles x molar mass. The molar mass of Tris is 121.14 g/mol, so the mass of Tris needed is 0.02 moles x 121.14 g/mol = 2.42 g Tris.

To adjust the pH to 7.8, you need to add HCl. The pKa of Tris is 8.1, so you can use the Henderson-Hasselbalch equation pH = pKa + log([A-]/[HA]). Rearranging the equation, [A-]/[HA] = 10^(pH - pKa).

Plugging in the values, [A-]/[HA] = 10^(7.8 - 8.1) = 5.62 x 10^(-1).

You will need a 1:4 ratio of Tris to TrisH. Therefore, [A-]/[HA] = 5.62 x 10^(-1) = (0.02 - x)/(x/4), where x is the moles of TrisH you need to add. Solving this equation, x = 0.0073 moles of TrisH.

To calculate the volume of 6 M HCl needed, you can use the equation moles = Molarity x volume. Rearranging the equation, volume = moles/Molarity. The moles of HCl needed is 0.0073 moles x 6 M = 0.044 L = 44 mL. Therefore, you will need to add 44 mL of 6 M HCl to the Tris to adjust the pH to 7.8.

Answer 2

The reaction :

`$\text{TrisH}^+ + \text{H}_2\text{O} \rightarrow \text{Tris}^- +\text{H}_3\text{O}^+$`

We have

`$K_a = \frac{[\text{Tris}^-]*[\text{H}_3\text{O}^-]}{[\text{TrisH}^+]}$`

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

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

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

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

`$[OH^-] = 10^(-pOH)$`

`$=10^(-6.2)$`

`$=6.31 * 10^(-7)$` moles of base

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

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

`$=1.35 * 10^(-5)$` moles

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

= 0.0135 L

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

= 3.152 g

To prepare the said solution we must mix

--
`$3.152 \ g \text{ TrisH}^+$`

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`$0.0135 \ mL \ NaOH \ 1M$`

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`$\text{Gauge to 1000 mL with H}_2\text{O}$`