Investigation 1 ANALYSIS OF BAKING SODA
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Introduction
Investigation 1 ANALYSIS OF BAKING SODA ?Aim? Determine the mass percentage of NaHCO3 in the sample. ?SAFETY? a. Wear safety goggles and a laboratory apron in the laboratory at all time. b. Notify your teacher immediately if any chemicals, especially concentrated acid or base, are spilled. ?BACKGROUND? Titration is usually the case when insoluble solid reagents are used. Back titration is usually employed for quantitative work with substances of this kind. In this investigation we use back titration to determine the mass of baking soda. Then we can calculate the percent error of the experimer. The amount of solute can be calculated from the volume of the solution of known concentration. The amount of the unknown may then be found using the balanced equation. Finally the concentration of the second solution used. ...read more.
Middle
Then read the volume number of NaOH added. h. Repeat step A to step G twice. ?METHOD? a. We used electric balance to measure the mass of a clean dry flask. b. Then put some soda in the flask. And measured the mass of the flask with soda. Last we could know the mass of soda. c. We Reade the first volume number of HCL. d. And Put some HCL solution to the flask that with no foam. Then read the volume number of HCL added. e. Added 1~2 drops of thymol blue indicator to the flask. Then we saw the mixture turn red. f. Read the first volume number of NaOH. g. Added some NaOH in the mixture until the pink color just disappears. Then read the volume number of NaOH added. ...read more.
Conclusion
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 ?? ?? ?? ?? 1 ...read more.
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