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If you had to make a buffer with a 250 mM solution of Tris-Base (FW= 121.14)...

If you had to make a buffer with a 250 mM solution of Tris-Base (FW= 121.14) in a volume of 600 mL, how many grams of Tris-Base will you need to weigh out?

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Answer #1

FW = 121.14 g/L

Desired volume of the solution = 600 ml = 0.6 L

Desired molarity of the solution = 250 mM = 0.25 M

Here, 121.14 g in 1 L will have a molarity of 1 M (or 1000 mM)

The equation is

Grams / desired volume of the solution in L = required molarity in mole/L x FW in g/L

Grams = desired volume of the solution in L x  required molarity in mole/L x FW in g/L

Substituting the equation with the values:

Grams = 0.6 x 0.25 x 121.14

= 18.171 g

Therefore, for making 250 mM of Tris-base in a volume of 600 mL, 18.171g of Tris-base is needed.

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