Investigation 1 ANALYSIS OF BAKING SODA

『EQUIPMENT』

    One sample spoon

    One electronic balance

    Three 250ml beakers

    Two 50ml burettes

    One hob

『CHEMICALS』

    1mol/L of HCl

    2g of NaHCO3 sample for 3 shares

    A few drops of phenolphthalein

1mol/L of NaOH

『Design』

1. Use electric balance to measure the mass of a clean dry flask.
2. Put some soda in the flask. Then measure the mass of the flask with soda. Last we can know the mass of soda..
3. Read the first volume number of HCL.
4. Put some HCL solution to the flask that with no foam. Then read the volume number of HCL added.
5. Add 1~2 drops of thymol blue indicator to the flask. Then we can see the mixture turn red.
6. Read the first volume number of NaOH.
7. Add some NaOH in the mixture until the pink color just disappears. Then read the volume number of NaOH added.
8. Repeat step A to step G twice.

『METHOD』

1. We used electric balance to measure the mass of a clean dry flask.
2. Then put some soda in the flask. And measured the mass of the flask with soda. Last we could know the mass of soda.
3. We Reade the first volume number of HCL.
4. And Put some HCL solution to the flask that with no foam. Then read the volume number of HCL added.
5. Added 1~2 drops of thymol blue indicator to the flask. Then we saw the mixture turn red.
6. Read the first volume number of NaOH.
7. Added some NaOH in the mixture until the pink color just disappears. Then read the volume number of NaOH added.
8. Repeat step A to step G twice.

『DATA COLLECTION』

『DATA ANALYSIS』

First time

『Percentage error』

1. HCL + NaOH → NaCL + H2O

NaOH: HCL = 1: 1

nNaOH=C×V=1.00×7.7/1000 = 0.007 mol

nHCL = 0.007 mol

nHCL = C×V =1.00×21.1/1000 = 0.0211 mol

nHCL = 0.0211 – 0.0077 = 0.0134 mol

  nHCL + NaHCO3 → NaCL + H2O + CO2

  HCL: NaHCO3 = 1:1

  nNaHCO3 =0.0134 mol

  m= n×M = 0.0134×84.01 = 1.125734 g

The percentage error: （1.998 – 1.125734）÷ 1.125734=77.48%

2. HCL + NaOH → NaCL + H2O

NaOH: HCL = 1: 1

nNaOH=C×V=1.00×10.1/1000 = 0.0101 mol

nHCL = 0.0101 mol

nHCL = C×V =1.00×23.4/1000 = 0.0234 mol

nHCL = 0.0234 –0.0101 = 0.0133 mol

  nHCL + NaHCO3 → NaCL + H2O + CO2

  HCL: NaHCO3 = 1:1

  nNaHCO3 =0.0133 mol

    m = n × M =0.0133×84.01 =1.117333 g

The percentage error: (2.021-1.117333) ÷ 1.117333=80.88%

3. HCL + NaOH → NaCL + H2O

NaOH: HCL = 1: 1

nNaOH=C×V=1.00×6.2/1000 = 0.0062 mol

nHCL = 0.0062 mol

nHCL = C×V =1.00×19.6/1000 = 0.0196 mol

nHCL = 0.0196 – 0.0062 = 0.0134 mol

  nHCL + NaHCO3 → NaCL + H2O + CO2

  HCL: NaHCO3 = 1:1

  nNaHCO3 =0.0134 mol

m = n × M = 0.0134×84.01=1.125734 g

The percentage error: (2.025-1.125734) ÷ 1.125734=79.88%

『DISCUSSION &CONCLUSIONS』

In investigation one, we could not use the equipment very well. When we added the NaOH, the solutions have been red but we did not stop it. At next investigation we need to more carefulness.

『Reference』

John Green & Sadru Damji

《Chemistry》

For use with the International Baccalaureate Diploma Programme

3rd Edition