Question

Molarity of NaOH and molecular weight of the unknown acid

Part A	Run 1	Run 2
Mass of H2C2O4*2H20 used in grams	0.2127	0.2124
Moles of H2C2O4*2H2O (mol)	1.69 *10^-5	1.69*10^-5
Number of protons available for reaction with OH-	2H	2H
Moles of OH- which reacted (mol)	2	2
Volume of NaOH solution used (mL)	18.71	19.15
Molarity of NaOH soultion (M)	.12	.12
Average molarity of NaOH (M)
Part B with unkown sample	Run 1	Run2
Mass of unknown acid used for titration (g)	.25	.25
Volume of NaOH solution used (mL)	26.94	28.30
Moles of NaOH which reacted (mol)	2	2
Moles of NaOH which reacted (mol)	3.23*10^-3	3.40*10^-3
Moles of unknown acid (mol)	77.44	73.61

Molecular weight of unknown acid ( g/mol)	77.44	73.61
Average molecular weight of unknown acid (g/mol)	75.52	75.52

1)If the oxalic acid I accidentally used was a dehydrated solid H2C2O4 & not the dihydrate H2C2O4*2H2O I actually used, how would this change the experimental weight of the unknown acid? (increase decrease or no change)

2) Calculate the density of your standardized solution of NaOH in units of gram per milliliter.

3)In a time teperature data for an acid base and redox reaction experiment... Why is it necessary to correct for the heat capacity of your calorimeter?

Answer 1

Q1.

if the oxalic acid was dehydrated ... the same mass will have less H+ ions... therefore the concentration of the acid will be actually higher

since less amount of acid reacts with the same amout of base

Q2.

M of NaOH = 0.12 mol / L

assume 1 Liter solution so:

mol of NaOH = 0.12

mass of NaOH = 0.12*40 = 4.8 g of NaOH

mass of water = 1000 g approx

so

D = mass/V = (1000+4.8)/1000 = 1.0048 g/mL

Q3.

We need to corret it since heat cpaacity is dependent of temperature, at higher T, the heat capacity

typically increases