Results and Calculation

A. Molarity of oxalic acidMoles of oxalic acid in standard solution= = 0.0175 mol C2H2O4

Molarity of C2H2O4= = 0.07 M of C2H2O4

B. Standardization of NaOHTitration123

Final (mL)19.919.919.6

Initial (mL)000

Volume used (mL)19.919.919.6

Average (mL)19.8

Table 1: Standardization of NaOHC2H2O4 (aq) + NaOH(aq) C2HO4-Na+ (aq) + H2O(aq)

Number of moles C2H2O4 reacted= () x 0.07= 0.0007 mol

From the equation, mole ratio C2H2O4 of with NaOH is 1:1,Thus, number of moles NaOH reacted= 0.0007 mol

Concentration of NaOH= = 0.0353 M of NaOH

pOH= -log [OH-]= -log(0.0353)= 1.45pH + pOH = 14pH = 14 1.45 = 12.55

C. Molarity of diluted acetic acidTitration123

Final (mL)7.16.86.6

Initial (mL)000

Volume used (mL)7.16.86.6

Average (mL)6.83

Table 2: Titration of Acetic acidC2H4O2 (aq) + NaOH(aq) C2H3O2-Na+ (aq) + H2O(aq)Molarity of NaOH = 0.0353 M

Number of moles NaOH reacted= 0.0353 M x = 2.411 x 10-4 mol NaOH

From the equation,1 mol NaOH = 1 mol C2H4O2= 2.411 x 10-4 mol C2H4O2

Molarity of moles of diluted C2H4O2 solution= = 2.411 x 10-3 mol

Number of mole of C2H4O2 in 10mL (undiluted) = 2.411 x 10-3 molMolarity of acetic acid= = 0.2411 MMolas mass of C2H4O2= 2(12) + 4(1) + 2(16)= 60

Therefore, weight-volume percentage of C2H4O2 in vinegar= = 1.45%

Percentage of acetic acid in vinegar is 1.45%.

Ka of C2H4O2 = 1.74 x 10-51.74 x 10-5 = = 1.74 x 10-5 (0.2411) = 4.195 x 10-6

pH = -log(4.195 x 10-6) = 5.38

ObservationTitrated solutionInitialFinal

Change of colorColorlessLight/Pale pink

Table 3: Physical observation

Graph

Figure 1: Titration curve of oxalic acid

Figure 2: Titration curve of acetic acid in vinegar