In: Chemistry
How many milliliters of 1.00 M KOH should be added to 4.90 g of phosphoric acid (MM= 98.0g/mol) to give a pH of 11.00 when diluted to 1000 ml? (Ka1=7.11 x10-3, Ka2=6.32x10-8, Ka3=4.5x10-13)
molarity of H3PO4 = (4.90 / 98.0) x (1000/1000) = 0.05 M
H3PO4 -----------------> H2PO4- + H+
0.05-x x x
Ka1 = x^2 / 0.05-x
7.11 x 10^-3 = x^2 / 0.05-x
x = 0.0156
H2PO4- = 0.0156 M
pKa2 = -logKa2 = -log(6.32x10-8) = 7.20
H2PO4- + OH- --------------------------> HPO4-2 + H2O
0.0156 V 0 0
0.0156- V 0 V V
pH = pKa2 + log [V / 0.0156 -V]
11.00 = 7.20 + log (V / 0.0156 -V)
V / 0.0156 -V = 6309.6
V = 98.429 - 6309.6 V
V = 0.01556 L
V = 15.6 ml
volume of KOH = 15.6 mL