1. Calculation for Exp# 8
By,
Kinjan Patel.

2. Calculation for Experiment 8
Goal: Find out mass and wt/wt% of NaHCO3 and Na2CO3 from the Unknown.
The experiment has three parts:
Part1: Titration of 25 ml of unknown with 0.1 M HCl.
This part include three titration.
Part2: Titration of 25 ml unknown + 50 ml of NaOH + 10 ml of 10 (wt/wt)% BaCl2 with 0.1 M HCl.
Part3: Blank Titration of 25 ml DI water + 50 ml NaOH with 0.1 M HCl.
This part include two titration.

3. What is Unknown contains?
Unknown contain mixture of Na2CO3 and NaHCO3.
Suppose, the total mass of unknown is
gram.
The total volume of unknown is = 250ml.

4. Part 1: Titration of 25ml unknown with 0.1 M HCl.
We did three titration.
Suppose for trial 1, we got ml titration volume.
So moles of unknown reacted with HCl
= ml X 0.1 M / 1000 ml
= moles
We can find out moles of other two titration and then we can find out average of all three titration moles.
Suppose the average of all three titration moles is,
= moles

5. Part 3: Blank titration of 25 ml DI water + 50 ml NaOH with 0.1 M HCl.
We did two titration in this part.
Suppose for trial 1 and for trial 2, we got ml volume.
The reaction between NaOH and HCl is,
NaOH HCl ---- H2O NaCl
1mol mol
So total moles of OH—reacted with H+ =
= 0.1 M X ml/1000 ml= moles
= average total moles of OH_.

6. Part 2: titration of 25 ml of unknown + 50 ml of 0
Part 2: titration of 25 ml of unknown + 50 ml of 0.1 M NaOH + 10 ml wt/wt% BaCl2 with 0.1 M HCl.
We did three titration in this part.
Suppose for trial 1 we got ml titration volume.
When we added 50 ml of 0.1 M NaOH in unknown, the reaction take place between HCO3_ and some OH_,
HCO3_ + OH_  CO3_ H2O
And,
CO3_ Ba  BaCO3
remaining OH_ will react with H+ and make water as in blank titration.
NaOH HCl ---- H2O NaCl
1mol mol

7. Moles of NaHCO3=Moles of HCO3_
From these equation we can says that,
Total moles of HCO3_ = Total moles of OH_ reacted with HCO3_
= (Total moles of OH_ reacted with HCl in
_ part 3 ) – (remaining moles of OH_
reacted with HCl as in part 2.)
For trial 1 in part 2, we got titration value= ml
so remaining moles of OH_ reacted with HCl:
= ml X 0.1 M/1000 ml
= moles.
Total moles of HCO3_ = moles – moles
= moles.

8. Mass and (wt/wt)% of NaHCO3 in Unknown.
For finding mass of NaHCO3, we have to know three things in this experiment.
Molar mass of NaHCO3 = gram/moles.
Moles of NaHCO3 = moles.
We used 25 ml of unknown from 250 ml of unknown, so we have to consider the dilution factor.
Grams of NAHCO3 = moles X molar mass X dilution factor
= moles X g/mol X
(250ml/25ml)
= gram.
Wt/wt% of NaHCO3 = (gram of NaHCO3 / total grams of unknown)X100
= ( gram/ gram) X 100
= 38.81%

9. Moles of Na2CO3

10. Moles of Na2CO3 From the graph we can says that,
Moles of Na2CO3 = total moles of unknown – moles of NaHCO3 from the unknown.
the equation for Na2CO3 reaction with HCl is,
CO3_ H  HCO3_
HCO3_ + H  H2CO3
One can conclude that,
CO3_2 can react with 2 moles of H+ while HCO3_ will react with 1 mole of H+ to make product H2CO3.
So, Moles of Na2CO3
= total moles of unknown – moles of NaHCO3 from the unknown 2
= moles – moles
= moles.

11. Mass and (wt/wt)% of Na2CO3 in Unknown.
For finding mass of Na2CO3, we have to know three things in this experiment.
Molar mass of Na2CO3 = gram/moles.
Moles of Na2CO3 = moles.
We used 25 ml of unknown from 250 ml of unknown, so we have to consider the dilution factor.
Grams of NA2CO3 = moles X molar mass X dilution factor
= moles X g/mol X
(250ml/25ml)
= gram.
Wt/wt% of Na2CO3 = (gram of Na2CO3 / total grams of unknown)X100
= ( gram/ gram) X 100
= 67.13%

12. At last……….
The interesting fact about this lab calculation is that, you will always get almost double mass for Na2CO3 than NaHCO3. This is because Na2CO3 will react with 2 mol of H+ and NaHCO3 will react with 1 mol of H+ in the titration with 0.1 M HCl. Q:
Write the equations for the titration of NaHCO3 and Na2CO3 with HCl. A:
CO3_ H  HCO3_
HCO3_ + H  H2CO3