# Chem Lab

Topics: PH, Acid, Acid dissociation constant Pages: 8 (1951 words) Published: December 12, 2012
1. Introduction2

2. Graphs
2.1 HC2H3O2 titration curve3
2.2 H3PO4 titration curve4
2.3 H2A titration curve5

3. Calculations
3.1 HC2H3O2 Calculations
a. Exact molarity of the HC2H3O2 solution6
b. Ka from the initial pH6
c. Ka from the pH at halfway point6
d. Ka from the pH at the end point7
3.2 H3PO4 Calculations
a. Exact molarity of the H3PO4 solution7
b. Ka1 from the initial pH7
c. Ka1 from the pH at the first halfway point8
d. Ka2 from the pH at the first end point8
e. Ka2 from the pH at the second halfway point9
3.3 H2A Calculations
a. Molecular weight of the acid9
b. Ka1 from the initial pH9
c. Ka1 from the pH at the first halfway point10
d. Ka2 from the pH at the first end point10
e. Ka2 from the pH at the second halfway point11
4. Discussion12-14

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-|...