How many milliliters of 1.00 M KOH should be added to 4.90 g of phosphoric acid...

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)

Expert Solution

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

queen_honey_blossom answered 11 months ago

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How many milliliters of 1.00 M KOH should be added to 4.90 g of phosphoric acid...

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