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Solution Stoichiometry (Lecture 2)

Solution Stoichiometry (Lecture 2). Mass concentration; Dilution; and volumetric analysis. What will I learn?. What is mass concentration? (and how is it related to molar concentration?) What is dilution? What is volumetric analysis? (or titration?)

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Solution Stoichiometry (Lecture 2)

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  1. Solution Stoichiometry(Lecture 2) Mass concentration; Dilution; and volumetric analysis

  2. What will I learn? • What is mass concentration?(and how is it related to molar concentration?) • What is dilution? • What is volumetric analysis? (or titration?) • How to calculate an unknown concentration in a volumetric analysis problem?

  3. Other Concentration Units • Sometimes, concentration can be expressed in mass per unit volume

  4. Other Concentration Units • Sometimes, concentration can be expressed in mass per unit volume • Units: gdm-3

  5. Other Concentration Units • Sometimes, concentration can be expressed in mass per unit volume • Concentration (in gdm-3) can be converted to molar concentration via the equation:

  6. Example 1 • 2.00 g of NaOH is dissolved in water to give a final volume of 150.0 cm3. • Calculate the concentration (in gdm-3) of the NaOH solution formed

  7. Example 1 • 2.00 g of NaOH is dissolved in water to give a final volume of 150.0 cm3. • Calculate the concentration (in moldm-3) of the solution above

  8. Example 1 • 2.00 g of NaOH is dissolved in water to give a final volume of 150.0 cm3. • Calculate the concentration (in moldm-3) of the solution above

  9. Example 1 • 2.00 g of NaOH is dissolved in water to give a final volume of 150.0 cm3. • Calculate the concentration (in moldm-3) of the solution above

  10. Example 1 • 2.00 g of NaOH is dissolved in water to give a final volume of 150.0 cm3. • Calculate the concentration (in moldm-3) of the solution above

  11. 4 Dilution • Dilution is the process of adding more solvent to a solution Remains constant decreases increases

  12. Example 1 • 10.0 cm3 of a NaOH solution of concentration 1.500 moldm-3 was diluted to 250 cm3 using distilled water. Calculate the concentration of NaOH after dilution. Rearranging,

  13. Example 1 • 10.0 cm3 of a NaOH solution of concentration 1.500 moldm-3 was diluted to 250 cm3 using distilled water. Calculate the concentration of NaOH after dilution.

  14. Example 1 • 10.0 cm3 of a NaOH solution of concentration 1.500 moldm-3 was diluted to 250 cm3 using distilled water. Calculate the concentration of NaOH after dilution.

  15. Example 1 • 10.0 cm3 of a NaOH solution of concentration 1.500 moldm-3 was diluted to 250 cm3 using distilled water. Calculate the concentration of NaOH after dilution.

  16. Example 1 • 10.0 cm3 of a NaOH solution of concentration 1.500 moldm-3 was diluted to 250 cm3 using distilled water. Calculate the concentration of NaOH after dilution.

  17. Example 1 • 10.0 cm3 of a NaOH solution of concentration 1.500 moldm-3 was diluted to 250 cm3 using distilled water. Calculate the concentration of NaOH after dilution.

  18. Example 1 • 10.0 cm3 of a NaOH solution of concentration 1.500 moldm-3 was diluted to 250 cm3 using distilled water. Calculate the concentration of NaOH after dilution.

  19. Example 2 • A concentrated solution of H2SO4 has a concentration of 3.35 moldm-3. What volume of this acid is required to prepare 250.0 cm3 of 0.130 moldm-3 H2SO4 solution?

  20. Example 2 • A concentrated solution of H2SO4 has a concentration of 3.35 moldm-3. What volume of this acid is required to prepare 250.0 cm3 of 0.130 moldm-3 H2SO4 solution?

  21. Example 2 • A concentrated solution of H2SO4 has a concentration of 3.35 moldm-3. What volume of this acid is required to prepare 250.0 cm3 of 0.130 moldm-3 H2SO4 solution?

  22. Example 2 • A concentrated solution of H2SO4 has a concentration of 3.35 moldm-3. What volume of this acid is required to prepare 250.0 cm3 of 0.130 moldm-3 H2SO4 solution? Rearranging,

  23. Example 2 • A concentrated solution of H2SO4 has a concentration of 3.35 moldm-3. What volume of this acid is required to prepare 250.0 cm3 of 0.130 moldm-3 H2SO4 solution?

  24. Example 2 • A concentrated solution of H2SO4 has a concentration of 3.35 moldm-3. What volume of this acid is required to prepare 250.0 cm3 of 0.130 moldm-3 H2SO4 solution?

  25. Example 2 • A concentrated solution of H2SO4 has a concentration of 3.35 moldm-3. What volume of this acid is required to prepare 250.0 cm3 of 0.130 moldm-3 H2SO4 solution?

  26. Example 2 • A concentrated solution of H2SO4 has a concentration of 3.35 moldm-3. What volume of this acid is required to prepare 250.0 cm3 of 0.130 moldm-3 H2SO4 solution?

  27. 5 Dilution • Useful formulae for dilution:

  28. Dilution • Useful formulae for dilution:

  29. Volumetric analysis • A quantitative analysis (vs qualitative analysis) • Analysis => calculation / manipulation of the results to obtain meaningful data • Volumetric analysis => accurate measurement of volume is required

  30. Volumetric analysis

  31. Volumetric analysis

  32. Volumetric analysis

  33. Volumetric analysis • End point is the point in the titration when the indicator undergoes a sharp colour change. • Equivalence point is the point when stoichiometric amounts / volume of reactants have been mixed • The two points are not the same • However, an ideal indicator is one where the equivalence point and end point is very close to each other • (and a sharp colour change occurs when a small extra amount of reactant has been added)

  34. 6 Calculations in VA • Similar to that encountered before (in normal stoichiometry calculations) • Difference: concentration and reacting volumes are now involved

  35. Number of moles Titration volume mass (m) number of moles (n) number of particles (N) volume (V)

  36. Calculations in VA • Given: since

  37. Example 120.0 cm3 of a solution of barium hydroxide, Ba(OH)2, of unknown concentration was placed in a conical flask and titrated with a solution of hydrochloric acid which had a concentration of 0.0600 moldm-3. The volume of the acid required was 25.0 cm3. Calculate the molar concentration of the barium hydroxide. Balanced equation Given c and V c?? Given V

  38. Example 120.0 cm3 of a solution of barium hydroxide, Ba(OH)2, of unknown concentration was placed in a conical flask and titrated with a solution of hydrochloric acid which had a concentration of 0.0600 moldm-3. The volume of the acid required was 25.0 cm3. Calculate the molar concentration of the barium hydroxide. Balanced equation

  39. Example 120.0 cm3 of a solution of barium hydroxide, Ba(OH)2, of unknown concentration was placed in a conical flask and titrated with a solution of hydrochloric acid which had a concentration of 0.0600 moldm-3. The volume of the acid required was 25.0 cm3. Calculate the molar concentration of the barium hydroxide. Balanced equation

  40. Example 120.0 cm3 of a solution of barium hydroxide, Ba(OH)2, of unknown concentration was placed in a conical flask and titrated with a solution of hydrochloric acid which had a concentration of 0.0600 moldm-3. The volume of the acid required was 25.0 cm3. Calculate the molar concentration of the barium hydroxide. Balanced equation From equation:

  41. Example 120.0 cm3 of a solution of barium hydroxide, Ba(OH)2, of unknown concentration was placed in a conical flask and titrated with a solution of hydrochloric acid which had a concentration of 0.0600 moldm-3. The volume of the acid required was 25.0 cm3. Calculate the molar concentration of the barium hydroxide. Balanced equation

  42. Example 120.0 cm3 of a solution of barium hydroxide, Ba(OH)2, of unknown concentration was placed in a conical flask and titrated with a solution of hydrochloric acid which had a concentration of 0.0600 moldm-3. The volume of the acid required was 25.0 cm3. Calculate the molar concentration of the barium hydroxide. Balanced equation

  43. Example 120.0 cm3 of a solution of barium hydroxide, Ba(OH)2, of unknown concentration was placed in a conical flask and titrated with a solution of hydrochloric acid which had a concentration of 0.0600 moldm-3. The volume of the acid required was 25.0 cm3. Calculate the molar concentration of the barium hydroxide. Balanced equation

  44. Example 120.0 cm3 of a solution of barium hydroxide, Ba(OH)2, of unknown concentration was placed in a conical flask and titrated with a solution of hydrochloric acid which had a concentration of 0.0600 moldm-3. The volume of the acid required was 25.0 cm3. Calculate the molar concentration of the barium hydroxide. Balanced equation

  45. Example 120.0 cm3 of a solution of barium hydroxide, Ba(OH)2, of unknown concentration was placed in a conical flask and titrated with a solution of hydrochloric acid which had a concentration of 0.0600 moldm-3. The volume of the acid required was 25.0 cm3. Calculate the molar concentration of the barium hydroxide. Balanced equation

  46. Example 120.0 cm3 of a solution of barium hydroxide, Ba(OH)2, of unknown concentration was placed in a conical flask and titrated with a solution of hydrochloric acid which had a concentration of 0.0600 moldm-3. The volume of the acid required was 25.0 cm3. Calculate the molar concentration of the barium hydroxide. Balanced equation ALTERNATIVELY…

  47. Example 120.0 cm3 of a solution of barium hydroxide, Ba(OH)2, of unknown concentration was placed in a conical flask and titrated with a solution of hydrochloric acid which had a concentration of 0.0600 moldm-3. The volume of the acid required was 25.0 cm3. Calculate the molar concentration of the barium hydroxide. Balanced equation 

  48. Example 120.0 cm3 of a solution of barium hydroxide, Ba(OH)2, of unknown concentration was placed in a conical flask and titrated with a solution of hydrochloric acid which had a concentration of 0.0600 moldm-3. The volume of the acid required was 25.0 cm3. Calculate the molar concentration of the barium hydroxide. Balanced equation 

  49. Example 120.0 cm3 of a solution of barium hydroxide, Ba(OH)2, of unknown concentration was placed in a conical flask and titrated with a solution of hydrochloric acid which had a concentration of 0.0600 moldm-3. The volume of the acid required was 25.0 cm3. Calculate the molar concentration of the barium hydroxide. Balanced equation 

  50. Example 120.0 cm3 of a solution of barium hydroxide, Ba(OH)2, of unknown concentration was placed in a conical flask and titrated with a solution of hydrochloric acid which had a concentration of 0.0600 moldm-3. The volume of the acid required was 25.0 cm3. Calculate the molar concentration of the barium hydroxide. Balanced equation 

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