How would you make up 100 ml of 50 mM Tris-HCl buffer at pH 7.4, using...

How would you make up 100 ml of 50 mM Tris-HCl buffer at pH 7.4, using the 0.5 M Tris base stock solution, and any other required materials?

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Expert Solution

Here the buffer is

Tris + HCl ---- > Tris-HCl

First calculate the number of moles of Trisl+ Tris-HCl = volume * molarty

= 100 ml+50mM

= 5.0 m mole

[Tris] + [Tris-HCl ]= 5.0 m mole ------(1)

The pH of the buffer is given as follows:

pH = pKa + log [Tris]/ [Tris-HCl]

pKa FOR Tris-HCl = 8.072

And pH = 7.4

Therefore;

pH = pKa + log [Tris]/ [Tris-HCl]

7.4= 8.072 + log [Tris]/ [Tris-HCl]

log [Tris]/ [Tris-HCl] = -0.0672

[Tris]/ [Tris-HCl] = 10^-0.0672 = 0.213

Or [Tris]= 0.213 [Tris-HCl] ---- (2)

From equation 1 and 2

[Tris] + [Tris-HCl ]= 5.0 m mole ------(1)

[Tris]= 0.213 [Tris-HCl] ---- (2)

We will get

[Tris-HCl ]= 4.1 m mole

And

[Tris]= 0.9 mmole

Hence 4.1 m mole of HCl to0.5 M tris base and making the final volume of 100 ml will give the pH 7.4

queen_honey_blossom answered 1 year ago

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