Question

Consider 2.50 L of a buffer solution made from NaOH and 3.50 M H3PO4 at pH 2.25 that has a total concentration of phosphate containing species of 0.170 M.

What volume of 3.50 M H3PO4 was required to make the buffer solution?

What mass of NaOH was required to make the buffer solution?

Answer 1

pH of phosphate buffer = pka1 + log(H2PO4^-/H3PO4)

   pka1 phosphate buffer = 2.16

Total no of mol of buffer = 2.5*0.17 = 0.425 mol

no of mol of H2PO4^-1 = NaOH = x mol

no of mol of H3PO4 = 0.425-x mol

2.25 = 2.16 +log(x/(0.425-x))

x = 0.234 ml

no of mol of NaOH must take = x = 0.234 mol

mass of NaOH must take = 0.234*40 = 9.36 g

volume of H3PO4 MUST TAKE = (0.425-0.234)/3.5 = 0.0546 L

                = 54.6 ml

SO that,take 54.6 ml of 3.50 M H3PO4 and 9.36 g of NaOH then add water up to 2.5 L