Question

How many grams of Na3PO4 must be added to 1.0 L of 0.20 M Na2HPO4 to produce a buffer solution with pH = 11.7?

Answer 1

Solution :-

How many grams of Na3PO4 must be added to 1.0 L of 0.20 M Na2HPO4 to produce a buffer solution with pH = 11.7

Solution :-

HPO4^2-   and PO4^3- makes the buffer solution

So using the pka3 of the phosphoric acid we can find the moles of the Na3PO4 needed to make the solution with pH 11.7

ka3 = 4.2*10^-13

pka3 = -log ka3

       = -log [4.2*10^-13]

      = 12.37675

pH= pka3 + log ([PO4^3-]/[HPO4^2-])

moles of HPO4^2- = 0.20 mol per L * 1.00 L = 0.20 mol

pH= pka3 + log ([PO4^3-]/[HPO4^2-])

11.7 = 12.37675+ log ([PO4^3-]/[0.20])

11.7 –12.37675= log ([PO4^3-]/[0.20])

-0.67675= log ([PO4^3-]/[0.20])

Antilog [-0.67675] =[PO4^3-]/[0.20]

0.2105 = [PO4^3-]/[0.20]

0.2105 * 0.20 mol = [PO4^3-]

0.0421 mol = [PO4^3-]

1 mol PO4^3- = 1 mol Na3PO4 so the moles of Na3PO4 needed = 0.0421 mol

Now lets convert the moles of Na3PO4 to its mass

Mass of Na3PO4 = moles * molar mass

                               = 0.0421 mol * 163.941 g per mol

                               = 6.90 g Na3PO4

So the amount of Na3PO4 needed = 6.90 g