Question

For 50mL of a 0.400M borate buffer at pH 8.5 (boric acid Ka = 5.8x10^-10), Describe how you would make this using sodium borate (NaH2BO3) and 2.50M HCl.

Answer 1

total moles of buffer = 50 x 0.4 / 1000 = 0.02

H3BO3 + NaH2BO3 = 0.02

pH = pKa + log [salt / acid]

8.5 = 9.24 + log [H3BO3 / NaH2BO3]

[H3BO3 / NaH2BO3] = 0.182

[H3BO3 + NaH2BO3] = 0.02

from these 2 equations:

moles of H3BO3 = 0.0169

moles of NaH2BO3 = 3.08 x 10^-3

moles of H3BO3 = moles of HCl = 0.0169

volume of HCl = 0.0169 / 2.50 = 6.76 x 10^-3 L

volume of HCl = 6.76 mL

mass of NaH2BO3 = moles x molar mass

                              = 3.08 x 10^-3 x 83.81

                              = 0.258 g

0.259 g NaH2BO3 and 6.76 mL of HCl required to make this buffer.