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