Question

You made a 1 M KCl stock solution and a 1 M MES solution, pH 6.15 (pKa=6.15). The final buffer that you wish to make is 50 mM MES (pH 6.15) containing 100 mM of added KCl. How much of each solution will you need to add to make 0.2 mL (200 uL) of this buffer?_____ uL 1 M KCl stock ______uL 1 M MES (pH 6.15) stock ______uL. What is the total ionic strength of the buffer you made in Question mM 2 (in)

Answer 1

For the first part,

pH = pKa + log([base]/acid])

6.15 = 6.15 + log([base]/acid])

[base] = [acid]

Given,

[base] + [acid] = 50 mM x 0.2 ml = 10 mmol

[acid] + [acid] = 10 mmol

[acid] = 5 mmol

[base] = 5 mmol

So we would need for preparation of buffer,

[acid] = 0.1 ml x 5 mmol = 0.5 ml (500 ul) of 1 M KCl stock solution

[base] = 0.1 ml x 5 mmol = 0.5 ml (500 ul) of 1 M MES stock solution

concentration of buffer = 50 mM

Total Ionic strength of buffer = 1/2(50 x 1^2 + 50 x 1^2) = 50 mM