Question

Sample Data Sheet: TITRATION AND MOLARITY

Part I: Preparing the Oxalic Acid solution

1.         Mass of oxalic acid + weighing paper.                     _____1.5765_____________ g

2.         Mass of weighing paper.                                            _____n/a_____________ g

3.         Volume of oxalic acid solution.                                                 250       mL

4.         Concentration of oxalic acid=___________        (show calculation above)

Part III: Completing the Neutralization

  

	Trial #1	Trial #2	Trial #3
Volume of Oxalic Acid	15ml	15ml	15ml
Final Buret Reading of NaOH (mL)	15.87	15.74	19.43
Initial Buret Reading of NaOH (mL)	0.1ml	0.4ml	5.0ml
Volume of NaOH used	15.77ml	15.3ml	14.4ml

CALCULATIONS AND RESULTS

7.        Mass of oxalic acid (H₂C₂O₄^• 2 H₂0).                                           ___________________g

8.        Molecular weight of oxalic acid.                                                     ______________g/mole

  

9.        Number of moles of oxalic acid.                                                    _______________moles

10.       Molarity of oxalic acid.                                                                     _______________M

11.       Number of moles of oxalic acid used in each trial.

Trial 1                         Trial 2                         Trial 3

_____ mol                _____ mol                _____ mol

12.       Number of moles of NaOH at equivalence point:

_____ mol                _____ mol                _____ mol

13.       Molarity of NaOH:

_____ M                     _____ M                     _____ M

14.       Average of 2 closest trials: M NaOH =

Answer 1

Solution-

Balance reaction between oxalic acid & NaOH

H​​​​​​₂C₂O₄ (s)_​​. + 2NaOH (aq) Na₂C₂O₄( aq) + 2H₂O(l)

MW 90.03 40 134. 18

Part A

#Mass of oxalic acid = 1.5765 g

# Total volume of oxalic solution=250 mL ~ 0.25 Liter

Concentration of oxalic acid solution(Molarity)=( number of moles /Volume of solution in liter)

First we have to calculate moles of oxalic acid= (mass/ molar mass of oxalic acid)

=(1.5765/90.03)

=0.01751 g mol of oxalic acid

Now we can calculate Molarity of oxalic solution

=(0.01751/0.25)

=0.07M

Result-

7)mass of oxalic acid= 1.5765g

8)Molar mass of oxalic acid= 90.03g/mol

9)moles of oxalic acid=0.01751g mol

10) Molarity of oxalic acid=0.07M

#11)Number of moles of oxalic acid in each trial

Sr.No	Trial1	trial2	Trial 3
Volume of oxalic acid	15mL	15mL	15mL
Moles of oxalic acid solution	0.00105gmol	0.00105gmol	0.00105gmol

Volume of oxalic acid solution=15mL~0.015 liter

We know that Molarity(M)= number of moles /Volume of solution in liter

We know Molarity of oxalic acid=0.07M

#By putting values, number of moles of oxalic acid=Molarity x volume of oxalic solution in liter

=0.07x 0.015

=0.00105 gmol of oxalic acid

12)number of moles of NaOH at equivalence point

Sr.No	Trial1	Trial 2	Trial3
Volume of NaOH used	15.77mL ~0.01577liter	15.3ml ~0.0153Liter	14.4 ml ~0.0144Liter
Moles of NaOH at equivalence point are same because same amount of oxalic acid used	0.0021gmol	0.0021gmol	0.0021 gmol

From balance reaction it is clear that one mole of oxalic acid solution required 2 moles of NaOH,

So moles of oxalic acid in 0.015 Liter (15mL)=0.00105 gmol

Moles of NaOH required= 2x moles of oxalic acid

=2 x 0.00105

=0.0021 gmol of NaOH required

13)Molarity of NaOH at each trial

Sr.No	Trial 1	Trial 2	trial 3
Volume of NaOH solution in liter	0.01577 Liter	0.0153 Liter	0.0144 Liter
Number of moles of NaOH used	0.0021 gmol	0.0021 gmol	0.0021 gmol
Molarity of NaOH solution	0.133M	0.137M	0.145M
Average Molarity of NaOH	0.138M

#Molarity of trial 1 NaOH= number of moles/Volume of solution in liter

=0.0021/0.01577

=0.133M

Molarity of trial 2 NaOH=0.0021/0.0153

=0.137M

#Molarity of trial 3, NaOH solution= 0.0021/0.0144

=0.145M

14) average Molarity of NaOH=(0.133+0.137+0.145)/3

=0.138M