Explain The Fact Molarity Of H3o2 Titration

1. Introduction 2

2. Graphs 2.1 HC2H3O2 titration curve 3 2.2 H3PO4 titration curve 4 2.3 H2A titration curve 5

3. Calculations 3.1 HC2H3O2 Calculations a. Exact molarity of the HC2H3O2 solution 6 b. Ka from the initial pH 6 c. Ka from the pH at halfway point 6 d. Ka from the pH at the end point 7 3.2 H3PO4 Calculations a. Exact molarity of the H3PO4 solution 7 b. Ka1 from the initial pH 7 c. Ka1 from the pH at the first halfway point 8 d. Ka2 from the pH at the first end point 8 e. Ka2 from the pH at the second halfway point 9 3.3 H2A Calculations a. Molecular weight of the acid 9
Ka1 from the pH at the first halfway point
Reaction:
| H3PO4 + OH-  H2PO4- + H2O | initial mmol | 3.88 | 1.939 | 0 | n/a | Δ mmol | -1.939 | -1.939 | +1.939 | n/a | final mmol | 1.94 | 0 | 1.939 | n/a | final M | 0.03907 | 0 | 0.03905 | n/a |
Total volume = 40.00 mL + 9.65 mL= 49.65 mL Equilibrium: | H3PO4  H+ + H2PO4- | initial M | 0.03907 | 0 | 0.03905 | Δ M | -5.0×10-3 | +5.0×10-3 | +5.0×10-3 | final M | 0.03407 | 5.0×10-3 | 0.04405 |

pH= 2.30
6.5×10-3
[H+]= 10-2.30= 5.0×10-3
Ka1= [H+][H2PO4-]/[H3PO4]= (5.0×10-3)(0.04405)/(0.03407)=

d. Ka2 from the pH at the first end point
Simultaneous equilibria:
H2PO4-  H+ + HPO42-
H2PO4- + H2O  H3PO4 + OH- pH= 4.40
[H+]= 10-4.40 = 4.0×10-5
[H+]= √Ka1×Ka2
2.4×10-7
Ka2 = (4.0×10-5)2/6.5×10-3 = e. Ka2 from the pH at the second halfway point

Equilibrium: H2PO4-  H+ + HPO42-
Ka2= [H+][HPO42-]/[H2PO4-]
[H2PO4-]=[HPO42-] so, Ka2= [H+] pH= 6.80
1.6×10-7
[H+]= 10-6.80= 1.6×10-7
Ka2=

3.3 Unknown diprotic acid H2A Calculations

a. Molecular weight of the
A monoprotic weak acid (HC2H3O2) titration curve has a single steep inflection point, polyprotic acid (H3PO4) has multiple less steep inflection points, and the unknown diprotic acid (H2A) yielded a single steep inflection point.
The HC2H3O2 titration curve is expected to start out fairly acidic (pH= 2.80) because the initial specie is the HC2H3O2 itself. As more base is being added, the solution reaches its buffer region wherein the acid and base neutralize each other. It can be identified on the titration curve where the pH does not change abruptly. Here, the main species are HC2H3O2 and C2H3O2- wherein any small addition of base, the OH- reacts with the H+ to form water. This will lower the [C2H3O2-] so the equilibrium will shift to the right to form more H+ and C2H3O2- which will accommodate some more OH- additions until all of the acid will react with the base. The end point signals that an equal amount of mole of acid and base had been added to the solution. The end point can be located by looking at halfway the steep rise on the graph. The main specie in this region is the C2H3O2- wherein the H+ had been “stripped off” which causes the basic pH (pH=8.80). After the end point, the excess OH- and remaining C2H3O2- determines the basic