Determination of Na2Co3 in an Unknown.

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Purpose: The purpose of this was to determine the concentration of sodium carbonate in an unknown sample by titration. The solution of hydrochloric acid was prepared and standardized using Na2CO3.

Observations:

Week 1: Standardizing hydrochloric acid using sodium carbonate with bromocresol green indicator

Table 1: Titration of sodium carbonate using hydrochloric acid

Trial| Mass of Na2CO3 (g)| Burette Reading (mL)| Final Volume of HCl (mL)| | | Initial | Final | |
1| 0.2123| 0.00| 42.34| 42.34|
2| 0.2195| 0.00| 47.24| 47.24|
3| 0.2049| 0.00| 26.65| 46.65|
After Boiling|
1| | 31.52| 32.22| 0.70|
2| | 29.34| 30.61| 1.27|
3| | 30.61| 31.52| 0.91|

Week 2: Determination of Sodium Carbonate in the unknown via titrating using bromocresol indicator

Sample number: SC 8

Table 2: Titration of the unknown using hydrochloric acid

Trial | Mass of unknown (g)| Burette reading (before boiling)| Final Volume of HCl (mL)| | | Initial| Final| |
1| 0.5065| 0.00| 30.25| 30.25|
2| 0.5007| 0.00| 28.72| 28.72|
3| 0.5026| 0.00| 29.01| 29.01|
After Boiling|
1| 29.01| 29.49| 0.48|
2| 29.49| 29.91| 0.42|
3| 29.91| 30.21| 0.30|
Calculations:

Mass % of Na2CO3 in the unknown for trial 1

Na2CO3 + 2HCl ---> 2NaCl + H2O + CO2

Molar mass Na2CO3= 105.99g/mol

Mass of Na2CO3= 0.2123g

Moles of Na2CO3= 0.2123g/105.99g/mol

Moles of Na2CO3= 2.003 x 10-3 moles

Mole-to-Mole Ratio

1 Na2CO3: 2 HCl

Moles of HCl= 2 x 2.003 x 10-3moles

Moles of HCl= 4.006 x 10-3

Molarity of HCl= (4.006 x 10-3)/0.04304
Molarity of HCl= 0.09308M

Moles of HCl= 0.09308M x 0.03073L
Moles of HCl= 2.8601 x 10-3 moles

Moles of Na2CO3 = (2.860 x 10-3 moles) / 2
Moles of Na2CO3 = 1.4301 x 10-3

Mass of Na2CO3== 1.4301 x10-3 x 105.99g/mol
Mass of Na2CO3= 0.152g

Mass % Na2CO3= (0.1523g/0.5065) x100
Mass % Na2CO3= 30.07%

Results:

Trials | % Mass of Na2CO3 in unknown (%)|
1| 30.07|
2| 26.38|
3| 25.12|
Mean = 27.19%Standard Deviation= 2.100Relative standard deviation= 7.725%|

Mean = (30.07+26.38+25.12/3)
Mean = 27.19 %

S2=∑(x- μ)2n

S2= ∑ (13.23)/ 3

S= 2.100

Relative standard deviation= (2.100/ 27.19) x100
Relative standard deviation= 7.725 %

Discussion

In this experiment titration was used to determine analyte in moles that is in the solution. This was done by slowly adding a standard solution of known concentration; until the titration is determine to be complete. This occurs after the titration has passed an equivalence point (fade green color appears).

The equivalence point is not something that is observed because around the equivalence point, one drop before and there was no change in pH and the next drop changes the pH sometimes by 3-4 units. The equivalence point is between these two drops and the closer these two-drop are to each other the better the quantification of the analyte.

In this experiment the solution of HCl was standardized and percent purity of Na2CO3 from the unknown was determined by method of standardization by performing titration. According to the chemical equation the number of moles of Na2CO3 needed to neutralize HCl was 1:2 so the molar concentration of the Na2CO3 in the unknown was calculated by using the moles of HCl.

Conclusion:

The sample # SC 8 contains 27.19% sodium carbonate with the relative standard deviation of 7.725%.

Sources of Error

Error| Impact | Recommendation |
The actual equivalence point wasn’t observed because there is no change in pH the drop before the equivalence point after the next drop the pH changes by 3-4 units | The Concentration of the HCl will be effected because it will be past the equivalence point | Right before the equivalence point use a disposable pipette to add final drops of the HCl| Not Reading the burette accurately | It will effect the volume due to personal...
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