Question

Phosphoric acid is a triprotic acid with the pK values pK = 2.148, pK2 = 7.198, and pK3 12.375. You wish to prepare 1.000 L of a 0.0300 M phosphate buffer at pH 7.400. To do this, you choose to mix the two salt forms involved in the second ionization, NaH2 POą and Na, HPO4, in a 1.000 L volumetric flask and add water to the mark. What mass of each salt will you add to the mixture? mass NaH, PO4 0.803 3.31 mass Na,HPO4 What other combination of phosphoric acid and its salts could be mixed to prepare this buffer? H3PO and Na2HPO Na, HPO, and Na, РО, 4 HPO and NaH,PO HPO and Na,PO 4 NaH,PO, and Na,РО, 4

Answer 1

The Ka values for the phosphoric acid are The phosphoric acid has the following ionizations H3PO4 = H2PO4 + H+ pka2 = 2.148 H2PO4 = HPO4-2 + H+ pka = 7.198 HPO2-2 = PO2-3 + H+ pka, = 12.375 Since the given pH (7.400) is nearer to second ionization. The phosphate buffer must be prepared by H2PO4-1 and HPO4-2 [base] pH = pK, +log [acid] = 7.400 = 7.198+log Woo [HPO,?] 0% [H,PO, 1 - [HPO,” ] = 0.202 10% [H,PO, ] [HPO41 -100.202 = 1.592209 [H,PO, -] [HPO, ?]=1.592209x[H,PO, ] Let the conc. of buffer = 0.0300M That is [HPO, ?]+[H,PO, ]=0.03 1.592209x[H,PO, -]+[H,PO, ]=0.03 2.592209x [H2PO4 ]=0.03 0.03 [H,PO, 1= 2.592209 = 0 =0.011573 [HPO, ?]=1.592209x [H,PO, ]=1.592209x0.011573 = 0.018427 [H,PO, ]=0.011573M [HPO, 2]=0.018427M Volume of the buffer solution = 1.000 L Number of moles of H,PO, present in the solution = M*V = 0.011573M x1.000 L =0.011573 mol Number of moles of HPO, - present in the solution = M*V =0.018427Mx1.000 L = 0.018427 mol Molecular weight NaH,PO, = 119.98 g/mol Mass of NaH,PO. = molesx mol.wt. = 0.011573 molx119.98 g/mol =1.39 g Mass of NaH,PO = 1.39 g| Molecular weight of dibasic potassium phosphate (Na2HPO.) = 141.96 g/mol Mass of Na, HPO, = moles x mol.wt. = 0.018427 molx141.96 g/mol = 2.62 g Mass of Na HPO4 = 2.62 g

To obtain this pH, we need H2PO4 and HPO4-2 present in the solution. Best combination is NaH2PO4 and Na2HPO4. Other are a) H3PO, and NaH, PO, do not generate the above combination. b) H3PO4 and Na2HPO4 generate the above combination by taking H3PO4 in excess, as H3PO4 reacts with Na2HPO4 to get NaH2PO4 c) H3PO4 and Na3PO, do not generate the above combination. d) NaH PO, and Na3PO, can generate the required combination by taking excess of NaH2PO4. e) Na2HPO4 and Na3PO4 do not generate the required combination.