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
b. Ka1 from the initial pH 9 c. Ka1 from the pH at the first halfway point 10 d. Ka2 from the pH at the first end point 10 e. Ka2 from the pH at the second halfway point 11 4. Discussion 12-14

1. Introduction
In this experiment, NaOH(aq) will be used as the titrant to HC2H3O2(aq), H3PO4(aq), and an unknown acid. Using the data obtained for each titration, graph of pH vs. volume of NaOH can be plotted. These graphs can then be analyzed to find the volume of NaOH needed to reach the respective end points for each titration. Finally, the exact molarities of the acidic solutions can be determined using the known volumes and molarity of the base and the acid. These molarities will be used to calculate the respective Ka values of these solutions. This potentiometric titration experiment aims to evaluate the characteristics of a weak acid (HC2H3O2), polyprotic acid (H3PO4), and an unknown diprotic acid solution (H2A) titration curves using NaOH as the titrant; determine the experimental Ka values of the acidic solutions; and to identify the unknown acid.

3. CALCULATIONS

3.1 HC2H3O2 Calculations a. Exact molarity of the HC2H3O2 solution Reaction: HC2H3O2 + OH- C2H3O2- + H2O 0.107 M HC2H3O2
0.209 mol OH-/L base solution × 25.70 mL base solution × 1 mol HC2H3O2/1 mol OH- × 1/50.00 mL acid solution = b. Ka from the initial pH
Equilibrium: HC2H3O2  H+ + C2H3O2- pH= 2.80
2.4×10-5
[H+]= 10-2.80 = 1.6×10-3
Ka= [H+][C2H3O2-]/[HC2H3O2]= (1.6×10-3)(1.6×10-3)/(0.107-1.6×10-3)=

c. Ka from the pH at halfway point
Reaction:
| HC2H3O2 + OH-  C2H3O2- + H2O | initial mmol | 5.35 | 2.686 | 0 | n/a | Δ mmol | -2.686 | -2.686 | +2.686 | n/a | final mmol | 2.66 | 0 | 2.686 | n/a | final M | 0.0424 | 0 | 0.0427 | n/a |
Total volume = 50.00 mL + 12.85 mL= 62.85 mL
Equilibrium: HC2H3O2  H+ + C2H3O2- pH= 4.65
[H+]= 10-4.65 = 2.2×10-5
2.2×10-5
Ka= [H+][C2H3O2-]/[HC2H3O2]= (2.2×10-5)(0.0427)/(0.0424)=

d. Ka from the pH at the end point
Equilibrium: C2H3O2- + H2O  HC2H3O2 + OH-
Kb= [HC2H3O2][OH-]/[C2H3O2-] pH= 8.80 pOH= 14.00-8.80= 5.20
[OH-]= 10-5.20= 6.3×10-6
[C2H3O2-]= 5.35 mmol HC2H3O2/75.70 mL solution= 0.07067 M C2H3O2- Kb= (6.3×10-6)(6.3×10-6)/(0.07067)= 5.6×10-10
1.8×10-5
Ka= Kw/Kb= 1.0×10-14/5.6×10-10 =

3.2 H3PO4 Calculations

a. Exact molarity of the H3PO4 solution Reaction: H3PO4 + OH-  H2PO4- + H2O

0.201 mol OH-/L base solution × 19.30 mL base solution × 1 mol H3PO4/1 mol OH- ×
0.0970 M H3PO4
1/40.00 mL acid solution =

b. Ka1 from the initial pH
Equilibrium: H3PO4  H+ + H2PO4-
Ka1= [H+][H2PO4-]/[H3PO4] pH= 1.64
7.1×10-3
[H+]= 10-1.64= 0.023 Ka1= (0.023)(0.023)/(0.0970-0.023)=

c.
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
acid
Reaction: H2A + OH-  HA- + H2O

120 g/mol H2A
0.201 mol OH-/L base solution × 0.01678 L base solution × 1 mol H2A/1 mol OH- =3.37×10-3 mol H2A
Molecular Weight= 0.4031 g H2A/3.37×10-3 mol H2A=

b. Ka1 from the initial pH Equilibrium: H2A  H+ + HA- Ka1= [H+][HA-]/[ H2A] pH= 2.68 [H+]= 10-2.68= 2.1×10-3
6.7×10-5
[H2A]= 3.37×10-3 mol H2A/0.05000 L solution= 0.0674 M H2A
Ka1= (2.1×10-3)(2.1×10-3)/(0.0674-2.1×10-3)=

c. Ka1 from the pH at the first halfway point

Reaction: | H2A + OH- HA- + H2O | initial mmol | 3.37 | 1.686 | 0 | n/a | Δ mmol | -1.686 | -1.686 | +1.686 | n/a | final mmol | 1.684 | 0 | 1.686 | n/a | final M | 0.0288 | 0 | 0.0288 | n/a |
Total volume= 50.00 mL + 8.39 mL = 58.39 mL
Equilibrium:
| H2A  H+ + HA- | initial M | 0.0288 | 0 | 0.0288 | Δ M | -8.9×10-5 | +8.9×10-5 | +8.9×10-5 | final M | 0.0287 | 8.9×10-5 | 0.0289 |

Ka1= [H+][HA-]/[H2A] pH= 4.05 9.0×10-5
[H+]= 10-4.05= 8.9×10-5 Ka1= (8.9×10-5)(0.0289)/(0.0287)=

d. Ka2 from the pH at the first end point Simultaneous equilibria: HA-  H+ + A2- HA- + H2O  H2A + OH- pH= 4.75
[H+]= 10-475 = 1.8×10-5
[H+]= √Ka1×Ka2 3.5×10-6
Ka2 = (1.8×10-5)2/9.0×10-5=

e. Ka2 from the pH at the second halfway point
Equilibrium: HA-  H+ + A2-
Ka2= [H+][A2-]/[HA-]
[HA-]=[A2-] so, Ka2= [H+] pH= 5.40
4.0×10-6
[H+]= 10-5.40= 4.0×10-6 Ka2=

4. Discussion

Summary table of calculated Ka values HC2H3O2 | pH | Calculated Ka | % error | Initial | 2.4×10-5 | -2.6% | Halfway point | 2.2×10-5 | -1.8% | End point | 1.8×10-5 | 0% | Accepted Ka= 1.8×10-5 |

H3PO4 | pH | Calculated Ka1 | % error | Calculated Ka2 | % error | Initial | 7.1×10-3 | 1.1% | | | 1st halfway point | 6.5×10-3 | 2.9% | | | 1st end point | | | 2.4×10-7 | -8.2% | 2nd halfway point | | | 1.6×10-7 | -5.7% | | Accepted Ka1= 7.5×10-3 | Accepted Ka2= 6.2×10-8 |

| Unknown acid H2A | pH | Calculated Ka1 | % error | Calculated Ka2 | % error | Initial | 6.7×10-5 | 0.46% | | | 1st halfway point | 9.0×10-5 | -2.6% | | | 1st end point | | | 3.5×10-6 | -1.2% | 2nd halfway point | | | 4.0×10-6 | -2.3% | | Accepted Ka1= 7×10-5 | Accepted Ka2= 3×10-6 |

Dominant characteristics of acids can be seen just by looking at their respective titration curves.
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
pH.
The H3PO4 titration curve has two inflection points due to the singly ionization of its protons. Since this acid has three H+’s, it is expected that it will have a lower pH compared to HC2H3O2. The acid itself determines its initial pH (pH= 2.68). The pH at the first buffer region is still low (pH= 2.30) because the two main species in this region is H3PO4 and H2PO4-. The acidic hydrogen attached to these species determines the pH in this buffer region because the extra hydrogen ion increases the amount of ionization. The first hydrogen ion is taken off from the acid at the first end point. The principal specie in this region is H2PO4-, which is amphiprotic. It hydrolyzes and causes simultaneous equilibria since this specie can act as an acid or a base. The pH is still quite acidic because of the presence of the two hydrogen ions. At the second halfway point, the principal species are H2PO4- and HPO42-. Since these species are equal in moles in this region, the stripped H+ from the buffer solution determines the pH. The pH in the second end point is expected to be acidic because the main specie in this region is HPO42-. Since essentially all of the acidic hydrogen is gone, the ionization in this region is low. After this end point, the pH is determined by the excess OH-.
The second endpoint is only the visible mark in the graph of the unknown acid. The initial principal specie is the acid itself, H2A. The specie at the second end point is A2-. Since all of the H+ had been removed, the pH is expected to be basic (pH= 8.80). After the end point, the pH is determined by the excess OH-.
The unknown acid is succinic acid. The calculated molecular weight (MW= 120 g/mol H2A) and Ka values (Ka1= 6.7×10-5 Ka2= 3.5×10-6) are close to the acceptable values of succinic acid listed. The distinct characteristic of the titration curve, which is the inflection of the second end point, only means that the Ka values are too close in value as manifested in the calculated Ka values.