Examples of Chemistry Laws

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:Concentrationi = Quantityi / Volumesolution 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:Quantityion = Quantityelectorlyte Coefficiention

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 • When 70 ml of 3.0-Molar Na2CO3 is added to 30 ml of 1.0-Molar NaHCO3 the resulting concentration of Na+ is: • 2.0 M • 2.4 M • 4.0 M • 4.5 M • 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.) Volumefinal_solution = Sum of Volumesolutioni So, we have Volumefinal_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). QuantityNa+ = Sum of QuantityNa+i for each original solution i. Need QuantityNa+i • What is QuantityNa+1? Use Law L.2, Concentration of ions in a strong electrolyte solution: QuantityNa+1 = Quantityelectorlyte1 CoefficientNa+electrolyteCoefficientNa+electrolyte = 2, for electrolyte = Na2CO3. Need Quantityelectrolyte1 • What is Quantityelectrolyte1? Use the concentration of solute formula (Law L.1.):Quantityelectrolyte1 = Concentrationelectrolyte1 Volumesolution1 So, we have Quantityelectrolyte1 = 3.0 M.  0.07 lit. = 0.21 moles So, QuantityNa+1 = 0.21 x 2 = 0.42 moles.

5. Answer to Sample Question • What is QuantityNa+2? Use Law L.2, Concentration of ions in a strong electrolyte solution: QuantityNa+2 = Quantityelectorlyte2 CoefficientNa+electrolyteCoefficientNa+electrolyte = 1, for electrolyte = NaHCO3.  Need Quantityelectrolyte2 • What is Quantityelectrolyte2? Use the concentration of solute formula (Law L.1.):Quantityelectrolyte2 = Concentrationelectrolyte2 Volumesolution2 So, we have Quantityelectrolyte2 = 1.0 M.  0.03 lit. = 0.03 moles. So, QuantityNa+2 = 0.03 x 1 = 0.03 moles. So, we have QuantityNa+ in the final solution = 0.42 moles + 0.03 moles = 0.45 moles. So, with QuantityNa+ and Volumefinal_solution, we have:ConcentrationNa+ = 0.45 moles / 0.10 lit. = 4.5 M. So, the correct answer is (d)

6. Sample Question – Correct Answer • When 70 ml of 3.0-Molar Na2CO3 is added to 30 ml of 1.0-Molar NaHCO3 the resulting concentration of Na+ is: • 2.0 M • 2.4 M • 4.0 M • 4.5 M • 7.0 M

7. A Simple Example • When 70 ml of 3.0-Molar Na2CO3 is added to 30 ml of 1.0-Molar NaHCO3 the resulting concentration of Na+ is: • 2.0 M • 2.4 M • 4.0 M • 4.5 M • 7.0 M

8. output Question 26 context result Mix Mixture has-part raw material Aqueous Solution Aqueous Solution Na+ base conc. volume conc. base volume conc. ?? 3.0 M 0.07 lit 1.0 M 0.03 lit Na2CO3 NaHCO3 Question Representation

9. Background Knowledge Chemistry laws: • Concentration of a solute • Composition of strong electrolyte solutions • Conservation of mass • Conservation of volume etc.

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

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

12. Conservation of Mass output input context Chemical result Mix has-part raw-material part-of Chemical … Chemical1 Chemicaln quantity quantity quantity Quantity *moles Quantity *moles Quantity *moles Explanation Template 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 <Chemical> in <Chemical1> is <Quantity1> *moles … The quantity of <Chemical> in <Chemicaln> is <Quantityn> *moles Therefore, the quantity of <Chemical> = X *moles Law 3: Conservation of Mass

13. Conservation of Volume output Mixture result input Mix context volume raw-material Volume *liter … Chemical1 Chemicaln volume volume Volume *liter Volume *liter Explanation Template 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 <Volume1>, …, and <Volumen> = <Volumeresult> *liter Therefore, the volume of <Mixture> = <Volumeresult> *liter Law 4: Conservation of Volume

14. chemical Strong Electrolyte Solution superclass result Mix Mixture superclass Strong Electrolyte has-part raw material superclass Aqueous Solution Aqueous Solution Na+ base conc. volume base volume conc. 3.0 M 0.07 lit 1.0 M 0.03 lit Na2CO3 NaHCO3 Step 1: Reclassify Terms

15. volume ?? *liters quantity Mixture ?? *moles conc. ?? *molar has-part volume conc. Concentration *molar Chemical Volume *liters result Mix quantity Mixture has-part raw material Quantity *moles Aqueous Solution Aqueous Solution Na+ base volume conc. conc. volume base Law 1 3.0 M 0.07 lit 1.0 M 0.03 lit Na2CO3 NaHCO3 Step 2: Use Law 1 to Compute Concentration

16. 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

17. volume ?? *liters .1 quantity ?? *moles conc. Chemical result Mix ?? *molar volume raw-material Volume *liter … result Chemical Chemical Mix Mixture volume has-part volume raw material Law 4 Aqueous Solution Aqueous Solution Na+ Volume *liter Volume *liter base volume conc. conc. volume base 3.0 M 0.07 lit 1.0 M 0.03 lit Na2CO3 NaHCO3 Step 3: Use Law 4 to Compute Volume

18. volume .1 *liters Mix Mixture result raw material has-part Chemical result Mix quantity Aqueous Solution Aqueous Solution Na+ has-part base base conc. raw-material volume volume conc. conc. ?? *moles part-of Na2CO3 NaHCO3 0.03 liters ?? *molar Chemical 0.07 liters … Chemical Chemical 1.0 M 3.0 M quantity quantity quantity has-part ?? *moles Quantity *moles Quantity *moles Na+ Na+ Law 3 quantity ?? *moles ?? *moles Step 4: Use Law 3 to Compute Quantity

19. volume .1 *liters Mix Mixture result raw material has-part quantity Aqueous Solution Aqueous Solution Strong Electrolyte Na+ base base conc. volume volume conc. conc. ?? *moles quantity Na2CO3 NaHCO3 0.03 liters ?? *molar 0.07 liters has-part Quantity *moles 1.0 M 3.0 M Cation Anion has-part Law 2 quantity Na+ Na+ quantity quantity quantity Quantity *moles Quantity *moles ?? *moles ?? *moles ?? *moles Step 5: Use Law 2 to Compute Quantity of Ionic Parts

20. volume .1 *liters Mix Mixture result raw material has-part quantity Aqueous Solution Aqueous Solution Na+ base base conc. volume volume conc. conc. ?? *moles Na2CO3 NaHCO3 0.03 liters ?? *molar 0.07 liters 1.0 M 3.0 M Mixture has-part quantity Na+ Na+ has-part volume quantity conc. Concentration *molar ?? *moles Chemical ?? *moles ?? *moles Volume *liters quantity .21 Law 1’ Quantity *moles Step 6: Use Law 1’ to Compute Quantity

21. volume .1 *liters Mix Mixture result raw material has-part quantity Aqueous Solution Aqueous Solution Strong Electrolyte Na+ base base conc. volume volume conc. conc. ?? *moles quantity Na2CO3 NaHCO3 0.03 liters ?? *molar 0.07 liters has-part Quantity *moles 1.0 M 3.0 M Cation Anion has-part Law 2 quantity Na+ Na+ quantity quantity quantity Quantity *moles Quantity *moles .42 .21 *moles ?? *moles ?? *moles Step 7: Wind out of Law 2 from step 5

22. volume .1 *liters Mix Mixture result raw material has-part quantity Aqueous Solution Aqueous Solution Na+ base base conc. volume volume conc. conc. ?? *moles Na2CO3 NaHCO3 0.03 liters ?? *molar 0.07 liters 1.0 M 3.0 M has-part quantity Na+ Na+ quantity .03 .21 *moles .42 *moles ?? *moles Step 8-10: Similar to steps 5-7

23. volume .1 *liters Mix Mixture result raw material has-part Chemical result Mix quantity Aqueous Solution Aqueous Solution Na+ .45 has-part base base conc. raw-material volume volume conc. conc. ?? *moles part-of Na2CO3 NaHCO3 0.03 liters ?? *molar Chemical 0.07 liters … Chemical Chemical 1.0 M 3.0 M quantity quantity quantity has-part ?? *moles Quantity *moles quantity Quantity *moles Na+ Na+ Law 3 quantity .21 *moles .42 *moles .03 *moles Step 11: Wind out of Law 3 from Step 4

24. volume .1 *liters Mix Mixture result raw material has-part quantity Aqueous Solution Aqueous Solution Na+ Mixture base conc. volume volume conc. conc. .45 *moles Na2CO3 0.03 liters ?? *molar 0.07 liters base has-part volume NaHCO3 4.5 1.0 M 3.0 M conc. Concentration *molar Chemical has-part Volume *liters quantity quantity Na+ Na+ Quantity *moles quantity .21 *moles .42 *moles .03 *moles Law 1 Step 12: Wind out of Law 1 from Step 2

25. 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