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The application of common ion to purify hard water
 

The application of common ion to purify hard water

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    The application of common ion to purify hard water The application of common ion to purify hard water Document Transcript

    • The Application ofCommon Ion to Purify Hard Water Chemistry- Experiment Rolla Tyas Amalia Grade 11 Science
    • Aim: To vanish the hardness of natural water. Materials: 1. 4 Pippets 11. 2grm Ca(OH)2 2. 8 tube reactions 12. NaOH 0,025 M and 0,1, 50 ml 3. 8 conical flasks 13. 25 ml Hcl, 0,1 M 4. A funnel 14. Distilled water 5. A stative 6. A tube rack 7. Paper Filter 8. A beaker glass 250 ml 9. A gratitude glass 10. A burette Procedures: 1. Prepare 4 conical flasks, put 2 grm of Ca(OH)2 powder into each flask. 2. Label the flasks with number 1-4. 3. Add 50 ml water into flask 1; 50 ml NaOH 0,025 M into flask 2; 50 ml NaOH 0,05 M into flask 3; and 50 ml NaOH 0,1 M into flask 4. 4. Plug flask in, then shake and let it for one night. 5. After one night, separate the solutions at each flask so you can get the filtrate. 6. Put 10 ml of filtrate 1 into conical flask and add 3 drops of methyl indicator. Titrate the solution with HCL 0,1 M until there is change in solution color! Make a note on the volume of used HCL. 7. Determine the concentration of OH- in the filtrate using equation V1 M1 = V2 M2 8. Record the observation in a table. 9. Do a calculation and record the data in the following table: Saturated Used (OH-) in (OH-) in (Ca2+) in Ksp= (Ca2+) solution of volume of solution Ca(OH)2 solution (OH-) Ca(OH)2 HCL 0.1 M 8.7 ml 0.087 M 0.174 M 0.087 M 2.6 x 10^-3In water 13.2 ml 0.132 M 0.264 M 0.132 M 9.1 x 10^-3In NaOH 0.025 M 7.3 ml 0.073 M 0.146 M 0.073 M 7.78 x 10^-3In NaOH 0.05 M 15.3 ml 0.153 M 0.292 M 0.153 M 0.024In NaOH 0.1 M
    • Evaluating Data From the data of the experiment, constanta of solubility of Ca(OH)2 inwater is smaller than the constanta of solubility of Ca(OH)2 in NaOH with 0.025M and 0.05 M. This means that Ca(OH)2 is more soluble in water than in NaOHwith 0.025 M and 0.05 M. The effect of NaOH solution in the solubility of Ca(OH)2 is reducing thesolubility of insoluble substances, that we known as the common ion effect. Wecould see that the solubility of Ca(OH)2 in NaOH is decreasing as theconcentration of NaOH is increasing. My conclusion is Ca(OH)2 is more soluble in water than NaOH. The NaOHwith bigger concentration will give the common ion effect to the solubility ofCa(OH)2 in NaOH. The bigger the concentration of NaOH, the smaller thesolubility of the Ca(OH)2, which make the Ca(OH)2 can be more soluble.Communicating The effect of common ion to the solubility is reducing the solubility of theinsoluble substance, so that it might be soluble. The effect of common ion to thevalue of Ksp is decreasing the value of the Ksp. From the data of the experiment,the Ksp of Ca(OH)2 in NaOH with 0.1 M is smaller than the Ksp of Ca(OH)2 inwater. This means that salt is more soluble in a solution with common ion thanin water.