Question

Titration of Phosporic acid.

Calculate the pH after 25mL, 50mL, 100mL, 150mL, 200mL, 250mL, 300mL, and 350mL of base is added.

Ka1=7.5E-3, Ka2=6.2E-8, Ka3=4.2E-13

Answer 1

Phosphoric acid reacts with sodium hydroxide in a step-wise fassion:

Sodium dihydrogen phosphate formed
NaOH + H3PO4 ? Na[H2PO4] + H2O

Disodium hydrogen phosphate formed
NaOH + Na[H2PO4] ? Na2[HPO4] + H2O

Sodium phosphate (also called trisodium phosphate) formed
NaOH + Na2[HPO4] ? Na3PO4 + H2O

After after 25mL addition

100.0 mL x 0.100 mol phosphoric acid/L = 10.00 mmol acid

25.0 mL x 0.100 mol NaOH/L = 2.5.00 mmol hydroxide

       H3PO4(aq)                            H2PO4- (aq) +    H+ (aq)     Ka1 = 0.0075

I: 10.00 mmol/125.0mL M                   0 M                   0 M

C: -x M                                               +x M                  +x M

E: 0.0800-x M                                     x M                    x M

7.5 x 10-3 = x2/(0.0800 - x) assuming x<<0.0800 leads to x = 2.45x 10-1 M

x = [H+] = 2.45x 10-1 M pH = 0.39

For 50mL

H2PO4- (aq) HPO42- (aq) + H+ (aq)                            Ka2 = 6.2 x 10-8

If we calculate as above 6.2 x 10-8 = x2/(0.0660 - x) assuming x<<0.0660 leads to x = 6.39x 10-3 M

x = [H+] =6.39x 10-3 M pH = 2.41

For 100 mL

HPO42- (aq) PO43- (aq) + H+ (aq) Ka3 = 4.2 x 10-13

pH = 2.41

in similar way we can calculate remaining ph valus