Question

Outline how you would prepare 500 mL of the following solutions:

0.01M Tris, pH 7.5, 100 mM NaCl

0.05 M Tris,pH 7.8, 500 mM NaCl, 20 mM CaCl2

100 mM Tris, pH 8.7, 20 mM EDTA

You have the following available:

Tris base (FW = 121.1 g/n, pKa = 8.21)

NaCl (FW = 58.44 g/n) 100 mM CaCl2 6 M HCl 6 M NaOH 500 mM EDTA

Answer 1

1. 0.01M Tris

F.W = 121.1 g/n

M = w x 1000/M

0.01 = w x 1000/ 121.1

w = 0.001211 g , so 0.01M Tris is prepared by dissolving 0.001211 g in 1L of solution

So 0.001211 g in 1000 mL of solution

so 0.0006055 g in 500 mL of solution

2. 100 mM NaCl = 100 x 10^-3 M

M = w x 1000/M

100 x 10^-3 = w x 1000 / 58.44 = 0.005844 g in 1L of solution = 0.005844 g in 1000 mL of solution = 0.002922 g in 500 mL of solution.

3. 0.05 M Tris

  0.05 = w x 1000/ 121.1

w = 0.0060555 g in 1 L or 1000 mL of solution = 0.00302775 in 500 mL of solution

4. 500 mM NaCl

500 x 10^-3 = w x 1000 / 58.44 = 0.02922 g in 1L of solution = 0.02922 g in 1000 mL of solution = 0.01461 g in 500 mL of solution.