Question

What is the concentration of borate buffer (61.8 g boric acid H3BO3 and 8 g NaOH in 1000 mL of distilled water)? How much boric acid (g) and NaOH (g) do I need to add to make a 50ml volume?

Answer 1

Part 1.

The no. of moles of boric acid = mass of boric acid/its molar mass = 61.8 g/(61.8 g/mol) = 1 mol

The no. of moles of NaOH = mass of NaOH/(its molar mass) = 8 g/(40 g/mol) = 0.2 mol

Now, the composition of borate buffer = 0.2 mol NaH₂BO₃ + 0.8 mol H₃BO₃

Therefore, the concentration of borate buffer = 1 mol/1000 mL = 1 mol/L = 1 M

Part 2.

50 mL = 50/1000 L = 0.05 L

i.e. The mass of boric acid required = 0.05 L * 1 mol/L = 0.05 mol * 61.8 g/mol = 3.09 g

And the mass of NaOH required = 0.05 L * 0.2 mol/L = 0.01 mol = 0.01*40 g/mol = 0.4 g