Question

	T1	T2	T3
Mass of Oxalic (g)	0.1942	0.2110	0.1990
initial volume of KOH (mL)	2.64	4.77	3.22
Final volume of KOH (mL)	16.78	18.22	17.49

Calculate molarity of KOH for all three trials

Calculate the average of all 3 trials

calculate for the precision of the 3 trials

Answer 1

Oxalic acid --> diprotic acid...

so

2HA + 2KOH = 2H2O + K2A

MW of acid = 90.03 g/mol

mol of oxalic acid = mass/MW = mass/90.03

T1) mol of oxalic acid = mass/MW = 0.1942/90.03 = 0.00215 mol

T2) mol of oxalic acid = mass/MW = 0.2110/90.03 = 0.00234 mol

T3) mol of oxalic acid = mass/MW = 0.1990/90.03 = 0.00221 mol

ratio is 1:2 so

T1) mol of KOH = 2*mol of acid = 2* 0.00215 = 0.0043

T2) mol of KOH = 2*mol of acid = 2*0.00234 = 0.00468

T3) mol of KOH = 2*mol of acid = 2*0.00221 = 0.00442

so.. Molarities

V1 = 16.78-2.64 = 14.14

V2 = 18.22-4.77 = 13.45

V3= 17.49-3.22 = 14.27

T1) M of KOH = mol of KOH/V = 0.0043/(14.14*10^-3) = 0.3041

T2)M of KOH = mol of KOH/V = 0.00468 /(13.45*10^-3) = 0.3479

T3) M of KOH = mol of KOH/V = 0.00442/ (14.27*10^-3) = 0.30974

so..

Averate molarities = (0.3041+0.3479+0.30974)/3= 0.32058 M o fKOH