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Chapter 8 Solutions

Chapter 8 Solutions. 8.5 Molarity and Dilution. Molarity (M). Molarity (M) is a concentration term for solutions gives the moles of solute in 1 L solution molarity (M) = moles of solute liter of solution. Preparing a 1.0 Molar NaCl Solution.

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Chapter 8 Solutions

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  1. Chapter 8 Solutions 8.5 Molarity and Dilution

  2. Molarity (M) Molarity (M) • is a concentration term for solutions • gives the moles of solute in 1 L solution molarity (M) = moles of solute liter of solution

  3. Preparing a 1.0 Molar NaCl Solution A 1.0 M NaCl solution is prepared • by weighing out 58.5 g of NaCl (1.00 mole) and • adding water to make 1.0 liter of solution

  4. Guide to Calculating Molarity

  5. Example of Calculating Molarity What is the molarity of 0.500 L NaOH solution if it contains 6.00 g of NaOH? STEP 1 Given: 6.00 g of NaOH in 0.500 L solution Need: molarity (mole/L) of NaOH solution STEP 2 Plan: g of NaOH moles of NaOH molarity

  6. Example of Calculating Molarity (continued) STEP 3Write equalitites and conversion factors: 1 mole of NaOH = 40.0 g of NaOH 1 mole NaOH and 40.0 g NaOH 40.0 g NaOH 1 mole NaOH STEP 4 Set up problem to calculate molarity: 6.00 g NaOH x 1 mole NaOH = 0.150 mole of NaOH 40.0 g NaOH 0.150 mole NaOH = 0.300 mole NaOH 0.500 L NaOH solution 1 L NaOH solution = 0.300 M NaOH

  7. Learning Check What is the molarity of a solution if 325 mL of the solution contains 46.8 g of NaHCO3? 1) 0.557 M NaHCO3 2) 1.44 M NaHCO3 3) 1.71 M NaHCO3

  8. Solution 3) 1.71 M STEP 1 Given: 46.8 g of NaHCO3 325 mL (0.325 L) NaHCO3 solution Need: molarity (mole/L) of NaHCO3 solution STEP 2 Plan: g of NaHCO3 moles of NaHCO3 molarity STEP 3Write equalities and conversion factors: 1 mole of NaHCO3 = 84.0 g of NaHCO3 1 mole NaHCO3 and 84.0 g NaHCO3 84.0 g NaHCO3 1 mole NaHCO3

  9. Solution (continued) STEP 4 Setup problem to calculate moles and molarity of NaHCO3: 46.8 g NaHCO3 x 1 mole NaHCO3 84.0 g NaHCO3 = 0.557 mole of NaHCO3 0.557 mole NaHCO3 = 1.71 mole NaHCO3 0.325 L NaHCO3 solution 1 L NaHCO3 solution = 1.71 M NaHCO3

  10. Learning Check What is the molarity of a KNO3 solution if 225 mL of the solution contains 34.8 g of KNO3? 1) 0.344 M 2) 1.53 M 3) 15.5 M

  11. Solution 2) 1.53 M KNO3 STEP 1 Given: 34.8 g of KNO3 225 mL (0.225 L) KNO3 solution Need: molarity (mole/L) of KNO3 solution STEP 2 Plan: g of KNO3 moles of KNO3 molarity STEP 3Write equalities and conversion factors: 1 mole of KNO3 = 101.1 g of KNO3 1 mole KNO3 and 101.1 g KNO3 101.1 g KNO3 1 mole KNO3

  12. Solution (continued) STEP 4 Set up problem to calculate moles and molarity of KNO3: 34.8 g KNO3 x 1 mole KNO3 101.1 g KNO3 = 0.344 mole of KNO3 0.344 mole KNO3 = 1.53 mole KNO3 0.225 L KNO3 solution 1 L KNO3 solution = 1.53 M KNO3 In one setup: 34.8 g KNO3 x 1 mole KNO3x 1 = 1.53 M 101.1 g KNO3 0.225 L

  13. Molarity Conversion Factors The units of molarity are used to write conversion factors for calculations with solutions.

  14. Example of Calculations Using Molarity How many grams of KCl are needed to prepare 125 mL of a 0.720 M KCl solution? STEP 1 Given: 125 mL (0.125 L) of 0.720 M KCl Need: grams of KCl STEP 2 Plan: L of KCl moles of KCl g of KCl

  15. Example of Calculations Using Molarity (continued) STEP 3Write equalities and conversion factors: 1 mole of KCl = 74.6 g of KCl 1 mole KCl and 74.6 g KCl 74.6 g KCl 1 mole KCl 1 L of KCl = 0.720 mole of KCl 1 L and 0.720 mole KCl 0.720 mole KCl 1 L STEP 4 Set up problem to cancel mole KCl: 0.125 L x 0.720 mole KCl x 74.6 g KCl = 6.71 g of KCl 1 L 1 mole KCl

  16. Learning Check How many grams of AlCl3 are needed to prepare 125 mL of a 0.150 M solution? 1) 20.0 g of AlCl3 2) 16.7 g of AlCl3 3) 2.50 g of AlCl3

  17. Solution 3) 2.50 g of AlCl3 STEP 1 Given: 125 mL (0.125 L) of solution 0.150 M AlCl3 solution Need: g of AlCl3 STEP 2 Plan: L of solution moles of AlCl3 g of AlCl3 STEP 3Write equalities and conversion factors: 1 mole of AlCl3 = 133.5 g of AlCl3 1 mole AlCl3 and 133.5 g AlCl3 133.5 g AlCl3 1 mole AlCl3

  18. Solution (continued) STEP 3(continued) 1 L of KCl = 0.150 mole of AlCl3 1 L and 0.150 mole AlCl3 0.150 mole AlCl3 1 L STEP 4 Set up problem: 0.125 L x 0.150 mole AlCl3 x 133.5 g = 2.50 g of AlCl3 1 L 1 mole AlCl3

  19. Learning Check How many milliliters of 2.00 M HNO3 contain24.0 g of HNO3? 1) 12.0 mL of 2.00 M HNO3 2) 83.3 mL of 2.00 M HNO3 3) 190 mL of 2.00 M HNO3

  20. Solution 3) 190 mL of HNO3 STEP 1 Given: 24.0 g of HNO3 2.00 M HNO3 solution Need: mL of HNO3 solution STEP 2 Plan: g of HNO3 moles of HNO3 L of HNO3 solution STEP 3Write equalities and conversion factors: 1 mole of = 63.0 g of HNO3 1 mole HNO3 and 63.0 g HNO3 63.0 g HNO3 1 mole HNO3

  21. Solution (continued) STEP 4 Set up problem to calculate volume, in mL, of HNO3: 24.0 g HNO3 x 1 mole HNO3 x 1000 mL HNO3 63.0 g HNO3 2.00 moles HNO3 = 190 mL of a 2.00 M HNO3 solution

  22. Dilution In a dilution, • water is added • volume increases • concentration of solute decreases

  23. Initial and Diluted Solutions In the initial and diluted solution, • the moles of solute are the same • the concentrations and volumes are related by the following equations: For percent concentration C1V1 = C2V2 Concentrated Diluted solution solution For molarity M1V1 = M2V2 Concentrated Diluted solution solution

  24. Guide to Calculating Dilution Quantities

  25. Example of Dilution Calculations Using Percent Concentration What volume of a 2.00% (m/v) HCl solution can be prepared by diluting 25.0 mL of 14.0% (m/v) HCl solution? STEP 1 Prepare a table: C1 = 14.0% (m/v) V1 = 25.0 mL C2 = 2.00% (m/v) V2 = ? STEP 2 Solve dilution expression for the unknown C1V1 = C2V2 V2 =V1C1 C2 STEP 3Set up the problem using known quantities: V2 =V1C1 = (25.0 mL)(14.0%) = 175 mL C2 2.00%

  26. Learning Check What is the percent (m/v) of a solution prepared by diluting 10.0 mL of 9.00% NaOH to 60.0 mL?

  27. Solution What is the percent (m/v) of a solution prepared by diluting 10.0 mL of 9.00% NaOH to 60.0 mL? STEP 1 Prepare a table: C1 = 9.00% (m/v) V1 = 10.0 mL C2 = ?V2 = 60.0 mL STEP 2 Solve the dilution expression for the unknown: C1V1 = C2V2 C2 =C1V1 V2 STEP 3Set up the problem using known quantities: C2 =C1V1 = (10.0 mL)(9.00%) = 1.50% (m/v) V2 60.0 mL

  28. Example of Dilution Calculations Using Molarity What is the molarity (M) of a solution prepared by diluting 0.180 L of 0.600 M HNO3 to 0.540 L? STEP 1 Prepare a table: M1 = 0.600 M V1 = 0.180 L M2= ?V2 = 0.540 L STEP 2 Solve the dilution expression for the unknown: M1V1 = M2V2 STEP 3Set up the problem using known quantities: M2 =M1V1 = (0.600 M)(0.180 L) = 0.200 M V2 0.540 L

  29. Learning Check What is the final volume (mL) of 15.0 mL of a 1.80 M KOH diluted to give a 0.300 M solution? 1) 27.0 mL of a 1.80 M KOH 2) 60.0 mL of a 1.80 M KOH 3) 90.0 mL of a 1.80 M KOH

  30. Solution What is the final volume (mL) of 15.0 mL of a 1.80 M KOH diluted to give a 0.300 M solution? STEP 1 Prepare a table: M1 = 1.80 M V1 = 15.0 mL M2 = 0.300 MV2 = ? STEP 2 Solve the dilution expression for the unknown: M1V1 = M2V2 STEP 3Set up the problem using known quantities: V2=M1V1 = (1.80 M)(15.0 mL) = 90.0 mL M2 0.300 M

  31. Molarity in Chemical Reactions In a chemical reaction, • the volume and molarity of a solution are used to determine the moles of a reactant or product molarity ( mole ) x volume (L) = moles 1 L • if molarity (mole/L) and moles are given, the volume (L) can be determined moles x 1 L = volume (L) moles

  32. Guide to Calculations Involving Solutions in Chemical Reactions

  33. Using Molarity of Reactants How many mL of 3.00 M HCl are needed to completely react with 4.85 g of CaCO3? 2HCl(aq) + CaCO3(s) CaCl2(aq) + CO2(g) + H2O(l) STEP 1 Given: 3.00 M HCl; 4.85 g of CaCO3 Need: volume of HCl in mL STEP 2Write a plan: g of CaCO3 moles of CaCO3 moles of HCl mL of HCl STEP 3Write equalities and conversion factors: 1 mole of CaCO3 = 100.1 g 1 mole CaCO3 and 100.1 g CaCO3 100.1 g CaCO3 1 mole CaCO3

  34. Using Molarity of Reactants (continued) STEP 3(continued) 1 mole of CaCO3 = 2 moles of HCl 1 mole of CaCO3 and 2 moles of HCl 2 moles of HCl 1 mole of CaCO3 1000 mL of HCl = 3.00 moles of HCl 1000 mL of HCland 3.00 moles of HCl 3.00 moles of HCl 1000 mL of HCl STEP 4 Set up problem to calculate mL of HCl: 4.85 g CaCO3 x 1 mole CaCO3 x 2 moles HCl x 1000 mL HCl 100.1 g CaCO3 1 mole CaCO3 3.00 moles HCl = 32.3 mL of HCl required

  35. Learning Check If 22.8 mL of 0.100 M MgCl2 is needed to completely react 15.0 mL of AgNO3 solution, what is the molarity of the AgNO3 solution? MgCl2(aq) + 2AgNO3(aq) 2AgCl(s) + Mg(NO3)2(aq) 1) 0.0760 M 2) 0.152 M 3) 0.304 M

  36. Solution 3) 0.304 M AgNO3 STEP 1 Given: 22.8 mL (0.228 L) of 0.100 M MgCl2 Need: molarity of AgNO3 STEP 2Write a plan to calculate molarity: mL of MgCl2 moles of MgCl2 moles of AgNO3 molarity of AgNO3 STEP 3Write equalities and conversion factors: 0.100 mole of MgCl2 = 1 L of MgCl2 0.100 mole MgCl2 and 1 L MgCl2 1 L MgCl2 0.100 mole MgCl2

  37. Solution (continued) STEP 3 (continued) 1 mole of MgCl2 = 2 moles AgNO3 2 moles AgNO3 and 1 mole MgCl2 1 mole MgCl2 2 moles AgNO3 STEP 4 Set up problem to calculate molarity of AgNO3: 0.0228 L x 0.100 mole MgCl2 x 2 moles AgNO3 x 1____ 1 L 1 mole MgCl2 0.0150 L = 0.304 mole/L = 0.304 M AgNO3

  38. Learning Check How many liters of H2 gas at STP are produced when125 mL of 6.00 M HCl reacts with sufficient Zn? Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g) 1) 4.20 L of H2 2) 8.40 L of H2 3) 16.8 L of H2

  39. Solution 2) 8.40 L of H2 gas STEP 1 Given: 125 mL (0.125 L) of 6.00 M HCl Need: liters of H2 at STP STEP 2Write a plan to calculate liters of H2: L of HCl moles of HCl moles of H2 liters of H2 STEP 3Write equalities and conversion factors: 6.00 moles of HCl = 1 L of HCl 6.00 moles HCland 1 L HCl 1 L HCl 6.00 moles HCl 22.4 L of H2 = 1 mole of H2 at STP 22.4 L H2 and 1 mole H2 1 mole H2 22.4 L H2

  40. Solution (continued) STEP 3 (continued) 2 moles HCl = 1 mole H2 2 moles HCland 1 mole H2 1 mole H2 2 moles HCl STEP 4 Set up problem to calculate liters of H2: 0.125 L x 6.00 moles HCl x 1 mole H2 x 22.4 L 1 L 2 moles HCl 1 mole H2 = 8.40 L of H2 gas

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