Calculation for Exp

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.
• This part include three titration.
• 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
• 2.5600 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 43.71 ml titration
  volume.
• So moles of unknown reacted with HCl
• 43.71 ml X 0.1 M / 1000 ml
• 0.004371 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,
• 0.004372 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 54.00
  ml volume.
• The reaction between NaOH and HCl is,
• NaOH HCl ----? H2O NaCl
• 1mol 1mol
• So total moles of OHreacted with H
• 0.1 M X 54.00 ml/1000 ml 0.005400 moles
• average total moles of OH_.

6
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 42.71 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_2 H2O
• And,
• CO3_2 Ba2 -------? BaCO3
• remaining OH_ will react with H and make water
  as in blank titration.
• NaOH HCl ----? H2O NaCl
• 1mol 1mol

7
Moles of NaHCO3Moles 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
  42.71 ml
• so remaining moles of OH_ reacted with HCl
• 42.71 ml X 0.1 M/1000 ml
• 0.004271 moles.
• Total moles of HCO3_ 0.005400 moles 0.004271
  moles
• 0.001129 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 84.0066gram/moles.
• Moles of NaHCO3 0.001129 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
• 0.001129 moles X
  84.0066 g/mol X
• 
  (250ml/25ml)
• 0.9936 gram.
• Wt/wt of NaHCO3 (gram of NaHCO3 / total grams
  of unknown)X100
• (0.9936 gram/
  2.5600 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_2 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
• 0.004372 moles 0.001129 moles
• 2
• 0.001621 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 105.988 gram/moles.
• Moles of Na2CO3 0.001621 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
• 0.001621 moles X
  105.988 g/mol X
• 
  (250ml/25ml)
• 1.7186 gram.
• Wt/wt of Na2CO3 (gram of Na2CO3 / total grams
  of unknown)X100
• (1.7186 gram/
  2.5600 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_2 H --------? HCO3_
• HCO3_ H --------? H2CO3