1. Examples of Chemistry Laws L. 1) Concentration of a solute In a solution, the concentration of a solute i is given by the following formula: Concentration i = Quantity i / Volume solution L. 2) Composition of strong electrolyte solutions When an aqueous solution contains one or more strong electrolytes, each one of them dissociates into their component ions. The quantity (in moles) of each ion in solution is given by the following formula: Quantity ion = Quantity electorlyte  Coefficient ion

2. Examples of Chemistry Laws L.3) Conservation of mass When mixing several solutions, the quantity of each solute in the resulting solution is equal to the sum of the quantities of that solute in each of the original solutions. L.4) Conservation of volume When mixing several solutions, the volume of the resulting solution is equal to the sum of the volumes of the original solutions (within scope).

3. Sample Question 26.When 70 ml of 3.0-Molar Na 2 CO 3 is added to 30 ml of 1.0-Molar NaHCO 3 the resulting concentration of Na + is: a)2.0 M b)2.4 M c)4.0 M d)4.5 M e)7.0 M

4. Answer to Sample Question What is the concentration of Na + in the final solution? Apply Law L.1.  Need both the quantity of Na+ and the final solution’s volume. What is the volume of the resulting solution? Use the conservation of volume law (Law L.4.) Volume final_solution = Sum of Volume solution i So, we have Volume final_solution = 0.07 lit. + 0.03 lit. = 0.10 lit. What is the quantity of Na+ in the final solution? Use the law of conservation of mass (L.3). Quantity Na+ = Sum of Quantity Na+ i for each original solution i.  Need Quantity Na+ i What is Quantity Na+ 1 ? Use Law L.2, Concentration of ions in a strong electrolyte solution: Quantity Na+ 1 = Quantity electorlyte 1  Coefficient Na+ electrolyte Coefficient Na+ electrolyte = 2, for electrolyte = Na 2 CO 3.  Need Quantity electrolyte 1 What is Quantity electrolyte 1 ? Use the concentration of solute formula (Law L.1.): Quantity electrolyte 1 = Concentration electrolyte 1  Volume solution 1 So, we have Quantity electrolyte 1 = 3.0 M.  0.07 lit. = 0.21 moles So, Quantity Na+ 1 = 0.21 x 2 = 0.42 moles.

5. Answer to Sample Question What is Quantity Na+ 2 ? Use Law L.2, Concentration of ions in a strong electrolyte solution: Quantity Na+ 2 = Quantity electorlyte 2  Coefficient Na+ electrolyte Coefficient Na+ electrolyte = 1, for electrolyte = NaHCO 3.  Need Quantity electrolyte 2 What is Quantity electrolyte 2 ? Use the concentration of solute formula (Law L.1.): Quantity electrolyte 2 = Concentration electrolyte 2  Volume solution 2 So, we have Quantity electrolyte 2 = 1.0 M.  0.03 lit. = 0.03 moles. So, Quantity Na+ 2 = 0.03 x 1 = 0.03 moles. So, we have Quantity Na+ in the final solution = 0.42 moles + 0.03 moles = 0.45 moles. So, with Quantity Na+ and Volume final_solution, we have:Concentration Na+ = 0.45 moles / 0.10 lit. = 4.5 M. So, the correct answer is (d)

6. Sample Question – Correct Answer 26.When 70 ml of 3.0-Molar Na 2 CO 3 is added to 30 ml of 1.0-Molar NaHCO 3 the resulting concentration of Na + is: a)2.0 M b)2.4 M c)4.0 M d)4.5 M e)7.0 M

7. A Simple Example When 70 ml of 3.0-Molar Na 2 CO 3 is added to 30 ml of 1.0-Molar NaHCO 3 the resulting concentration of Na + is: a)2.0 M b)2.4 M c)4.0 M d)4.5 M e)7.0 M

8. Question Representation volume Mix Aqueous Solution Mixture Na + raw material Na 2 CO 3 3.0 M0.07 lit NaHCO 3 0.03 lit volume 1.0 M conc.base conc. result has-part conc. Question 26 context ?? output

9. Background Knowledge Chemistry laws: 1.Concentration of a solute 2.Composition of strong electrolyte solutions 3.Conservation of mass 4.Conservation of volume etc.

10. Law 1: Concentration of a Solute The concentration of a chemical in a mixture is the quantity of the chemical divided by the volume of the mixture. Divide the quantity by the volume: / = X *molar Therefore, the concentration of in = X *molar Explanation Template Mixture volume conc. Volume *liters Concentration *molar has-part Chemical Quantity *moles quantity Compute-Concentration Method context input output Note: when this law is applied, the quantities are automatically converted to the units- of-measurement specified here

11. Law 1’: Quantity of a Solute Law 1 (on the previous slide) computed: Concentration = quantity / volume Of course, a slight variant computes: Quantity = concentration * volume Currently, we code this variant as a separate law (call it 1’) because it has a slightly different explanation template

12. Law 2: Composition of Strong Electrolytes Strong Electrolyte Anion has-part Quantity *moles quantity Quantity *moles quantity Cation Quantity *moles quantity Compute-Ions-in-Strong-Electrolyte context input output

13. Law 3: Conservation of Mass Conservation of Mass context input output Mix Chemical 1 Chemical n Chemical raw-material result … Quantity *moles Quantity *moles quantity Chemical has-part Quantity *moles quantity part-of By the Law of Conservation of Mass, the quantity of a chemical in a mixture is the sum of the quantities of that chemical in the parts of the mix. The quantity of in is *moles … The quantity of in is *moles Therefore, the quantity of = X *moles Explanation Template

14. Law 4: Conservation of Volume Mix Chemical 1 Chemical n Mixture raw-material result … Volume *liter Volume *liter volume Volume *liter volume Conservation of Volume context input output By the Law of Conservation of Volume, the volume of a mixture is the sum of the volumes of the parts mixed. The sum of, …, and = *liter Therefore, the volume of = *liter Explanation Template

15. Step 1: Reclassify Terms volume Mix Aqueous Solution Mixture Na + raw material Na 2 CO 3 3.0 M0.07 lit NaHCO 3 0.03 lit volume 1.0 M conc.base conc. result has-part Strong Electrolyte Solution superclass Strong Electrolyte superclass chemical superclass

16. Step 2: Use Law 1 to Compute Concentration Mixture volume conc. Volume *liters Concentration *molar has-part Chemical Quantity *moles quantity Law 1 conc. ?? *molar volume Mix Aqueous Solution Mixture Na + raw material Na 2 CO 3 3.0 M0.07 lit NaHCO 3 0.03 lit volume 1.0 M conc. base conc. result has-part ?? *liters volume ?? *moles quantity

17. The Search is non-deterministic Multiple laws might be used to compute a value for any property. For example, here’s another way to compute concentration:  pH = - log [H + ], where [H + ] is the concentration of H + Since this applies only to H +, this search path ends quickly

18. Step 3: Use Law 4 to Compute Volume Mix Chemical raw-material result … Volume *liter Volume *liter volume Volume *liter volume Law 4.1 conc. ?? *molar volume Mix Aqueous Solution Mixture Na + raw material Na 2 CO 3 3.0 M0.07 lit NaHCO 3 0.03 lit volume 1.0 M conc. base conc. result has-part ?? *liters volume ?? *moles quantity

19. Step 4: Use Law 3 to Compute Quantity volume Mix Aqueous Solution Mixture Na + raw material Na 2 CO 3 3.0 M 0.07 liters 0.03 liters volume 1.0 M conc. base NaHCO 3 base conc. result has-part conc. ?? *molar.1 * liters volume ?? *moles quantity Mix Chemical raw-material result … Quantity *moles Quantity *moles quantity Chemical has-part ?? *moles quantity part-of Law 3 Na + ?? *moles ?? *moles has-part quantity

20. Step 5: Use Law 2 to Compute Quantity of Ionic Parts ?? *moles quantity Strong Electrolyte Anion has-part Quantity *moles quantity Quantity *moles quantity Cation Quantity *moles quantity Law 2 volume Mix Aqueous Solution Mixture Na + raw material Na 2 CO 3 3.0 M 0.07 liters 0.03 liters volume 1.0 M conc. base NaHCO 3 base conc. result has-part conc. ?? *molar.1 * liters volume ?? *moles quantity Na + ?? *moles ?? *moles has-part quantity

21. Step 6: Use Law 1’ to Compute Quantity ?? *moles quantity Mixture volume conc. Volume *liters Concentration *molar has-part Chemical Quantity *moles quantity Law 1’.21 volume Mix Aqueous Solution Mixture Na + raw material Na 2 CO 3 3.0 M 0.07 liters 0.03 liters volume 1.0 M conc. base NaHCO 3 base conc. result has-part conc. ?? *molar.1 * liters volume ?? *moles quantity Na + ?? *moles ?? *moles has-part quantity

22. Step 7: Wind out of Law 2 from step 5 Strong Electrolyte Anion has-part Quantity *moles quantity Quantity *moles quantity Cation Quantity *moles quantity Law 2.42.21 *moles quantity volume Mix Aqueous Solution Mixture Na + raw material Na 2 CO 3 3.0 M 0.07 liters 0.03 liters volume 1.0 M conc. base NaHCO 3 base conc. result has-part conc. ?? *molar.1 * liters volume ?? *moles quantity Na + ?? *moles ?? *moles has-part quantity

23. Step 8-10: Similar to steps 5-7.03.21 *moles quantity volume Mix Aqueous Solution Mixture Na + raw material Na 2 CO 3 3.0 M 0.07 liters 0.03 liters volume 1.0 M conc. base NaHCO 3 base conc. result has-part conc. ?? *molar.1 * liters volume ?? *moles quantity Na + ?? *moles.42 *moles has-part quantity

24. Step 11: Wind out of Law 3 from Step 4 Mix Chemical raw-material result … Quantity *moles Quantity *moles quantity Chemical has-part ?? *moles quantity part-of Law 3.45.21 *moles quantity volume Mix Aqueous Solution Mixture Na + raw material Na 2 CO 3 3.0 M 0.07 liters 0.03 liters volume 1.0 M conc. base NaHCO 3 base conc. result has-part conc. ?? *molar.1 * liters volume ?? *moles quantity Na +.03 *moles.42 *moles has-part quantity

25. Step 12: Wind out of Law 1 from Step 2 Mixture volume conc. Volume *liters Concentration *molar has-part Chemical Quantity *moles quantity Law 1.21 *moles quantity volume Mix Aqueous Solution Mixture Na + raw material Na 2 CO 3 3.0 M 0.07 liters 0.03 liters volume 1.0 M conc. base NaHCO 3 base conc. result has-part conc. ?? *molar.1 * liters volume.45 *moles quantity Na +.03 *moles.42 *moles has-part quantity 4.5

26. Answer and Explanation When 70 ml of 3.0-Molar Na2CO3 is added to 30 ml of 1.0-Molar NaHCO3, what is the resulting concentration of Na+?. The concentration of a chemical in a mixture is the quantity of the chemical divided by the volume of the mixture. By the Law of Conservation of Mass, the quantity of a chemical in a mixture is the sum of the quantities of that chemical in the parts of the mix. In the Na2CO3 strong-electrolyte-solution and the NaHCO3 strong-electrolyte-solution : In the Na2CO3 : Multiply the concentration and the volume: 3 molar * 70 milliliter = 0.21 mole. The quantity of Na+ in the NA2CO3 solution is 0.42 mole. In the NaHCO3 : Multiply the concentration and the volume: 1 molar * 30 milliliter = 0.03 mole. The quantity of Na+ in the Na2CO3 strong-electrolyte-solution and the NaHCO3 strong-electrolyte-solution is 0.45 mole. Therefore, the quantity of Na+ = 0.45 mole. By the Law of Conservation of Volume, the volume of a mixture is the sum of the volumes of the parts mixed. The sum of 70 milliliter and 30 milliliter = 0.10 liter. Therefore, the volume of the strong-electrolyte-solution strong-electrolyte-solution mixture = 0.10 liter. Divide the quantity by the volume:. 0.45 mole / 0.10 liter = 4.50 molar. Therefore, the concentration of Na+ in the Na2CO3 and NaHCO3 mixture = 4.50 molar. When 70 ml of 3.0-Molar Na2CO3 is added to 30 ml of 1.0-Molar NaHCO3, the resulting concentration of Na+ is 4.50 molar