Question

Back titration. Help, please.
Calculate the mass of the calcium carbonate dissolved in hydrochloric acid.

Info from experiment is as follows:

NaOH (poured into burette) had a concentration of 0.5 moldm3 and a volume (average titre) of 11.27ml.
The concentration of the HCl (which was standardized in a 250ml volumetric flask) was also 0.5moldm3 and the volume of the HCl which the CaCO2 was dissolved in was 50ml. In addition, 15ml of that solution was pipetted into a conical flask and titrated.
Thanks in advance.

Answer 1

@Cuanchi

Answer 2

typo *_* It's CaCO3, sorry, my bad.

Answer 3

1. calculate the number of moles of NaOH in the 11.25ml of 0.5M NaOH that you used to titrate the 15 ml of solution (mix of HCl + CaCO3)

moles= molarity x volume (L) = 0.5 M x 0.01125 L = "A" moles NaOH

2. because the stoiquimetry of the titration is one to one

NaOH + HCl -> NaCl + H2O

the xxx moles of NaOH is going to be the same than the moles of HCl extra from the mix of HCl and CaCO3. Then "B" moles of HCl = "A' moles of NaOH

3. Calculate the "left over" moles of HCl in the 50mL

"B" moles of HCl x 50 mL / 15mL = "C" moles of HCl

4. Calculate the "Total moles" of HCl in the 50mL 0.5 M HCl that you add to the CaCO3

moles= molarity x volume (L) = 0.5 M x 0.050 L = "D" moles HCl

5. subtract from the Total moles of HCl the left over HCl moles to find the moles of HCl that reacted with the CaCO3

"Total" - "left over" = "reactants"

"D" - "C" = "E" moles of HCl used to react with the CaCO3

6. Now the stoiquiometry of this other reaction is different than the titration

2HCl + CaCO3 - > CaCl2 + H2CO3 -> CaCl2(aq) + CO2(g) + H2O(l)

Then for each mole of CaCO3 you will need 2 moles of HCl

"E" moles of HCl x 1 mol of CaCO3 / 2 moles of HCl = "F" moles of CaCO3

7. Finally with the "F" moles of CaCO3 you can calculate the mass of CaCO3 if you know the molecular mass of CaCO3

"F" moles of CaCO3 x molecular mass of CaCO3 / 1 mol CaCO3 = "G" mass of CaCO3"

Answer 4

thank you so much!