Question

An aqueous solution is made by dissolving 11.24 g of sodium phosphate (tribasic) anhydrous salt in water to make 250 mL of solution. What is the expected pH of this solution? The pKa values for phosphoric acid are 2.15, 7.2, and 12.38.

Answer 1

molarity of NaCl = (w/M)*(1000/V)

w = weight of Na3PO4 = 11.24 g

M = molarmass of Na3PO4 = 164 g/mol

V = volume of Na3PO4 solution = 250 ml

molarity of Na3PO4 = (11.24/164)*(1000/250)

                   = 0.274 M

Na3PO4 salt of weak acid,strong base = basic

pH = 7+1/2(pka+logC)

pka3 of H3PO4 = 12.38

     = 7+1/2(12.38+log0.274)

     = 12.91