As an FDA physiologist, you need 0.625 L of phosphoric acid acid / dihydrogen phosphate (H3PO4 (aq) / H2PO4 - (aq) ) buffer with a pH of 2.75. How do you prepare this solution from stock solutions of 1.0 M H3PO4 (formic acid) and 1.5 M NaOH

0.4058L of 1.0M H3PO4

0.2192L of 1.5M NaOH

Explanation:

The pKa of the H3PO4 / H2PO4- buffer is 2.12

To solve this question we must use H-H equation for this system:

pH = pKa + log [H2PO4-] / [H3PO4]

2.75 = 2.12 +  log [H2PO4-] / [H3PO4]

0.63 = log [H2PO4-] / [H3PO4]

4.2658 = [H2PO4-] / [H3PO4] (1)

Where [] could be taken as the moles of each reactant

As you have H3PO4 solution, the reaction with NaOH is:

H3PO4 + NaOH → H2PO4- + Na+ + H2O

As you can see, both H3PO4 and H2PO4- comes from the same 1.0M H3PO4 solution

The moles of H3PO4 are:

[H3PO4] = Moles H3PO4 - Moles NaOH

And for H2PO4-:

[H2PO4-] = Moles NaOH added

Replacing in (1):

4.2658 = [Moles NaOH] / [Moles H3PO4 - Moles NaOH]

4.2658 Moles H3PO4 - 4.2658 moles NaOH = Moles NaOH

4.2658 Moles H3PO4 = 5.2658 moles NaOH (1)

In volume:

0.625L = Moles H3PO4 / 1.0M + Moles NaOH / 1.5M

0.625 = Mol H3PO4 + 0.6667 Moles NaOH (2)

Replacing (2) in (1):

4.2658 Moles H3PO4 = 5.2658 (0.625 - Mol H3PO4 / 0.6667)

4.2658 Moles H3PO4 = 5.2658 (0.625 - Mol H3PO4) / 0.6667

4.2658 Moles H3PO4 = 5.2658*(0.9375 - 1.5 mol H3PO4)

4.2658 Moles H3PO4 = 4.9367 -7.8983 mol H3PO4

12.1641 mol H3PO4 = 4.9367

Mol H3PO4 = 0.4058moles * (1L / 1.0moles) =

0.4058L of 1.0M H3PO4

And:

0.625L - 0.4058L =

0.2192L of 1.5M NaOH

the electrical  conductivity  of asolution  of an electrolyte is measured by determining the resistance of thesolution  between two flat or cylindrical electrodes separated by a fixed distance.