Question

moles of oxalix acid

moles/molarity of NaOH

average molarity

volume/molarity of hcl used

average molarity

NAME SECTION DATA SHEET LOCKER A CTOR Write a balanced equation for the reaction between H.CO. and NaOH: 90.33/mol 0.878 0836 TRIAL NUMBER a. Mass of oxalic acid used (g) b. Final Buret reading (ml.] c. Initial Buret reading (ml] d. Volume of NaOH used (b-c) Moles of oxalic acid used f. Moles of NaOH used (2 xe) g. Molarity of NaOH (f/d/1000) Average Molarity 14.50 4.20 6.45 14.50 13.85 13.20 0.0 blocos la.2.0.0.01.29 11.00 LDL Write a balanced equation for the reaction between HCl and NaOH: TRIAL NUMBER 1 23 15.15 3 .20 10.15 15.15 15.00 15.05 i. Final Buret reading j. Initial Buret reading k. Volume NaOH used (1-j) (1. Volume of HCI used Molarity of HCI (h x k/l) Average Molarity Concentration of NOCH ? and HCl? * Conclusion? 88 LAB 8: VOLUMETRIC ANALYSIS: AN ACID-BASE TITRATION

Answer 1

For calculating the moles of oxalic acid, the molar mass of oxalic acid is required. Checking the literature, the molar mass of oxalic acid is 90.03 g/mol

For trial 1

Mass of oxalic acid used = 0.878 g

Moles of oxalic acid = mass / molar mass = 0.878 g/ 90.03 g/mol = 0.00975 mol

Oxalic acid has 2 carboxylic acid groups. So 1 mole of oxalic acid will react with 2 moles of NaOH.

So the moles of NaOH = 2 x moles of oxalic acid = 2 x 0.00975 mol = 0.0195 mol

Molarity of NaOH = moles of NaOH / volume of NaOH solution in liters = 0.0195 mol / (13.85/1000)L = 1.41 mol/ L or 1.41 M

For trial 2

Mass of oxalic acid = 0.876 g

Moles of oxalic acid = 0.876 g/ 90.03 g/mol = 0.00973mol

Moles of NaOH = 2 x 0.00973mol = 0.01946 mol

Molarity of NaOH = 0.01946 mol /(13.70 /1000) L = 1.42 g /mol = 1.42 M

Average molarity = (1.41 M + 1.42 M) /2 = 1.415 M

Volume of HCl is the volume that you have taken in the flask for titration. This will be measured experimentally. Let this volume be V

The Molarity of HCl can be calculated by using the following formula

Molarity of HCl = average Molarity of NaOH x volume of NaOH solution used for titration/ volume of HCl taken in the flask for titration = 1.415 M x 15.00 / V

Please substitute the value of X and calculate.

Similar calculations can be done for trial 2