lab 13g Pacific Academy, Surrey CHEM 12

Lab 13G

Analysis of Results:

Part 1)
a) Mass of oxalic acid dihydrate
H​2​C​2​O​4​ * 2H​2​O = 126g/mol

b) Number of mol of oxalic acid measured
1 mol
0.58g x 126g =4.6x10​-3​ mol


c) Concentration of oxalic dihydrate
4.6x10−3
Conc = .1L = 4.6x10​-2

Part 2)
a) Number of mol of oxalic acid
4.6x10​-2​M x 0.025L = 1.2x10​-3​ mol

b) Diprotic so 2x the number of mol of acid.
1.2x10​-3​mol x 2 = 2.4x10​-3​ mol NaOH

c) Average mL of NaOH used = 23.70mL
2,4x10−3 mol of N aOH
Conc = 0.02370L = 0.10M NaOH

Part 3)
a) mol of NaOH used for each titration.
Titration #1: 0.10M NaOH x 0.03220L = 3.22x10​-3​ =3.2x10​-3
Titration #2: 0.10M NaOH x 0.03195L = 3.195x10​-3​ = 3.2x10​-3

b) mol of acid = mol of base because the acid is monoprotic.
Titration #1: 3.2x10​-3​ mol of unknown acid
Titration #2: 3.2x10​-3​ mol of unknown acid

c)molar mass
Because titration of #1 and #2 is the same, and the same amount of acid was used.
0.30g
Molar mass = 3.2x10−3 = 94g/mol




Part 4)
a) Measured pH for both trial is 2.33