Question

Given that you only have reagents listed below calculate and explain how you will prepare a 250 mL solution containing the following final concentrations: 50 mM sucrose, 100 mM potassium chloride, 50 mM Tris Buffer pH 8 at 30° C.

Available Reagents/Stock Solutions: Purified water, 1M sucrose, solid KCl (74.55g/mole) 250 mM Tris-Base, and 250 mM Tris-HCl (pKa of Tris is 8.0 at 30° C )

Answer 1

Available Reagents/Stock Solutions: Purified water, 1M sucrose, solid KCl (74.55g/mole) 250 mM Tris-Base, and 250 mM Tris-HCl (pKa of Tris is 8.0 at 30° C )

Sucrose:

Sucrose conc. = 50 mM = 50 x 10^-3 M

Volume = 250 mL

Stock soln. available = 1 M

So the amount of sucrose required from stock soln. is

M₁V₁ = M₂V₂

1M x V₁ = 50 x 10^-3 M x 250

V₁ = (50 x 10^-3 M x 250 ) / 1 M

V₁ = 12.5 mL

KCl

KCl = 100 mM = 100 x 10^-3 M

Volume = 250 mL

Molecular weight of KCl = 74.55 g /mol

Amount of KCl required = MEV/1000 = (100 x 10^-3 M) x (74.55 g/mol) x 250 / 1000 = 1.86 g

Amount of KCl required = 1.86 g

50 mM Tris Buffer pH 8

pH = pKa + log[A^-]/[HA] (Handerson Equation)

Stock soln. of Tris base = 250 mM

250 mM x V = 50mM x 250

V = ( 50mM x 250) / (250 mM) = 50 mL

V = 50 mL

Similarly for Tris-HCl

Stock soln. of Tris base = 250 mM

250 mM x V = 50mM x 250

V = ( 50mM x 250) / (250 mM) = 50 mL

V = 50 mL

Using Handerson's equation

pH = pKa + log[A^-]/[HA]

pKa = 8 (at 30° C )

pH = 8 + log[50/50]

pH = 8 + log 1 (log 1 = 0)

So take 12.5 mL of sucrose, 1.86g of KCl, 50 mL of Tris-base and 50 mL of Tris-HCl in a 250 mL standard flask (Type A) and add purified water till the mark. This gives the required solution