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Science league topic 10

Science league topic 10. Solution concentration and calculation. Solution Concentration. qualitatively, solutions are often described as dilute or concentrated dilute solutions have a small amount of solute compared to solvent

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Science league topic 10

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  1. Science league topic 10 Solution concentration and calculation

  2. Solution Concentration • qualitatively, solutions are often described as dilute or concentrated • dilute solutions have a small amount of solute compared to solvent • concentrated solutions have a large amount of solute compared to solvent • quantitatively, the relative amount of solute in the solution is called the concentration Tro, Chemistry: A Molecular Approach

  3. Solution Concentration: Molarity • moles of solute per 1 liter of solution • used because it describes how many molecules of solute in each liter of solution Tro, Chemistry: A Molecular Approach

  4. Preparing 1 L of a 1.00 M NaCl Solution Tro, Chemistry: A Molecular Approach

  5. g KBr mol KBr M L sol’n Example 4.5 – Find the molarity of a solution that has 25.5 g KBr dissolved in 1.75 L of solution • Sort Information Given: Find: 25.5 g KBr, 1.75 L solution Molarity, M • Strategize Concept Plan: Relationships: 1 mol KBr = 119.00 g, M = moles/L • Follow the Concept Plan to Solve the problem Solution: • Check since most solutions are between 0 and 18 M, the answer makes sense Check:

  6. Using Molarity in Calculations • molarity shows the relationship between the moles of solute and liters of solution • If a sugar solution concentration is 2.0 M, then 1 liter of solution contains 2.0 moles of sugar • 2 liters = 4.0 moles sugar • 0.5 liters = 1.0 mole sugar • 1 L solution : 2 moles sugar Tro, Chemistry: A Molecular Approach

  7. mol NaOH L sol’n Example 4.6 – How many liters of 0.125 M NaOH contains 0.255 mol NaOH? • Sort Information Given: Find: 0.125 M NaOH, 0.255 mol NaOH liters, L • Strategize Concept Plan: Relationships: 0.125 mol NaOH = 1 L solution • Follow the Concept Plan to Solve the problem Solution: • Check since each L has only 0.125 mol NaOH, it makes sense that 0.255 mol should require a little more than 2 L Check:

  8. Dilution • often, solutions are stored as concentrated stock solutions • to make solutions of lower concentrations from these stock solutions, more solvent is added • the amount of solute doesn’t change, just the volume of solution moles solute in solution 1 = moles solute in solution 2 • the concentrations and volumes of the stock and new solutions are inversely proportional M1∙V1 = M2∙V2 Tro, Chemistry: A Molecular Approach

  9. V1, M1, M2 V2 Example 4.7 – To what volume should you dilute 0.200 L of 15.0 M NaOH to make 3.00 M NaOH? • Sort Information Given: Find: V1 = 0.200L, M1 = 15.0 M, M2 = 3.00 M V2, L • Strategize Concept Plan: Relationships: M1V1 = M2V2 • Follow the Concept Plan to Solve the problem Solution: • Check since the solution is diluted by a factor of 5, the volume should increase by a factor of 5, and it does Check:

  10. Solution Stoichiometry • since molarity relates the moles of solute to the liters of solution, it can be used to convert between amount of reactants and/or products in a chemical reaction Tro, Chemistry: A Molecular Approach

  11. L Pb(NO3)2 mol Pb(NO3)2 mol KCl L KCl Example 4.8 – What volume of 0.150 M KCl is required to completely react with 0.150 L of 0.175 M Pb(NO3)2 in the reaction 2 KCl(aq) + Pb(NO3)2(aq)  PbCl2(s) + 2 KNO3(aq) • Sort Information Given: Find: 0.150 M KCl, 0.150 L of 0.175 M Pb(NO3)2 L KCl • Strategize Concept Plan: Relationships: 1 L Pb(NO3)2 = 0.175 mol, 1 L KCl = 0.150 mol, 1 mol Pb(NO3)2 = 2 mol KCl • Follow the Concept Plan to Solve the problem Solution: • Check since need 2x moles of KCl as Pb(NO3)2, and the molarity of Pb(NO3)2 > KCl, the volume of KCl should be more than 2x volume Pb(NO3)2 Check:

  12. Titration Tro, Chemistry: A Molecular Approach

  13. Titration The base solution is the titrant in the burette. As the base is added to the acid, the H+ reacts with the OH– to form water. But there is still excess acid present so the color does not change. At the titration’s endpoint, just enough base has been added to neutralize all the acid. At this point the indicator changes color. Tro, Chemistry: A Molecular Approach

  14. Write down the given quantity and its units. Given: 10.00 mL HCl 12.54 mL of 0.100 M NaOH Example 4.14:The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? Tro, Chemistry: A Molecular Approach

  15. Write down the quantity to find, and/or its units. Find: concentration HCl, M Information Given: 10.00 mL HCl 12.54 mL of 0.100 M NaOH Example 4.14:The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? Tro, Chemistry: A Molecular Approach

  16. Collect Needed Equations and Conversion Factors: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) 1 mole HCl = 1 mole NaOH 0.100 M NaOH 0.100 mol NaOH  1 L sol’n Information Given: 10.00 mL HCl 12.54 mL of 0.100 M NaOH Find: M HCl Example 4.14:The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? Tro, Chemistry: A Molecular Approach

  17. Write a Concept Plan: Information Given: 10.00 mL HCl 12.54 mL of 0.100 M NaOH Find: M HCl CF: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L M = mol/L Example 4.14:The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? mL NaOH L NaOH mol NaOH mol HCl mL HCl L HCl Tro, Chemistry: A Molecular Approach

  18. Apply the Solution Map: Information Given: 10.00 mL HCl 12.54 mL of 0.100 M NaOH Find: M HCl CF: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L M = mol/L CP: mL NaOH → L NaOH → mol NaOH → mol HCl; mL HCl → L HCl & mol  M Example:The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? = 1.25 x 10-3 mol HCl Tro, Chemistry: A Molecular Approach

  19. Apply the Concept Plan: Information Given: 10.00 mL HCl 12.54 mL NaOH Find: M HCl CF: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L M = mol/L CP: mL NaOH → L NaOH → mol NaOH → mol HCl; mL HCl → L HCl & mol  M Example:The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? Tro, Chemistry: A Molecular Approach

  20. Check the Solution: Information Given: 10.00 mL HCl 12.54 mL NaOH Find: M HCl CF: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L M = mol/L CP: mL NaOH → L NaOH → mol NaOH → mol HCl; mL HCl → L HCl & mol  M Example:The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? HCl solution = 0.125 M The units of the answer, M, are correct. The magnitude of the answer makes sense since the neutralization takes less HCl solution than NaOH solution, so the HCl should be more concentrated. Tro, Chemistry: A Molecular Approach

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