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Concentration of Solutions

Concentration of Solutions. Review:. Solutions are made up of Solute - substance dissolved or present in lesser proportion 2) Solvent - substance that is the dissolving medium - what the solute is dissolved in - many times this is water. Solutions can be classified as

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Concentration of Solutions

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  1. Concentration of Solutions

  2. Review: Solutions are made up of • Solute - substance dissolved or present in lesser proportion 2) Solvent - substance that is the dissolving medium - what the solute is dissolved in - many times this is water

  3. Solutions can be classified as • Electrolyte solutions - solutions that conduct electricity (solute is either an ionic compound or forms ions when it dissolves) • Nonelectrolyte solutions - solutions that do not conduct electricity (solute is a molecular compound that does not form ions as it dissolves)

  4. Components that make up mixtures can be • miscible - will form solutions in most any proportions (only liquids or gases) 2) immiscible substances - will not form solutions - example: oil and water

  5. There is a limit to the amount of solute that can be dissolved in a given amount of solvent - this is the "solubility" of the solute. Solubility of a solute can be expressed in the following terms: 1) soluble – The substance mostly dissolves 2) insoluble – Very little of the substance dissolves 3) slightly soluble – In between: some dissolves but it may not be enough to affect the properties of the solution.

  6. Concentration • The concentration of a solution refers to the amount of solute dissolved in the solvent • Qualitative terms used are dilute (not much solute) and concentrated (alot of solute) - concentrated does not mean pure.

  7. Molarity is one way to measure the concentration of a solution. moles of solute Molarity (M) = volume of solution in liters Molarity Molarity is the most used - other units include: molality, normality, formality , mole fraction, % weight, % volume (proof)

  8. Making a Solution…

  9. Sample Problem #1 3.85 g of NaCl is dissolved in enough water to make 81.0 mL of sol’n. Calculate the molarity of the solution. 3.85 g NaCl 1 mol NaCl ------------------ X ------------------- = .812M NaCl .0810 L sol’n 58.5 g NaCl

  10. Sample Problem #2 How many grams of NaCl are required to make 450 mL of .500 M soln? .500 moles NaCl 58.5 g NaCl .45 L soln X ---------------------- X ---------------- = 13 g NaCl 1 L soln 1 mole NaCl

  11. Sample Problem #3 How many mL of a .250 M NaCl soln can be prepared using 7.51 g NaCl? 1 mole NaCl 1 L soln 1000 mL 7.51 g NaCl X-------------- X ----------------X -------- = 514mL NaCl soln58.5 g NaCl .250 mole NaCl 1 L

  12. Titration

  13. Sample Problem #4 25.0 mL of .325 molar hydrochloric acid (HCl) completely neutralizes 35.0 mL of a calcium hydroxide solution. What is the molarity of the calcium hydroxide solution? 25.0 mL 35.0 mL .325 M ? M 2 HCl + Ca(OH)2 2 H2O + CaCl2 .0250 L HCl .325 mole HCl 1 mole Ca(OH)2 ------------------ X ---------------- X ---------------- = .116 M Ca(OH)2 .0350 L Ca(OH)2 1 L HCl 2 mole HCl

  14. Sample Problem #5 How many mL of .525 M nitric acid (HNO3) solution would completely neutralize 22.5 mL of .275 M Ca(OH)2 base solution? .525 M .275 M ? mL 22.5 mL 2 HNO3 + Ca(OH)2 ----> 2 H2O + Ca(NO3)2 .275 mole Ca(OH)2 2 mole HNO3 1 L HNO3 .0225 L Ca(OH)2 X -------------------- X ------------------- X ----------------- 1 L Ca(OH)2 1 mole Ca(OH)2.525 moles HNO3 = .0236 L or 23.6 mL HNO3

  15. Dilutions

  16. Sample Problem #6 20.0 mL of .250 M HCl solution is added to 30.0 mL of .150 M HCl solution. What is the concentration of the resulting solution? total moles total molarity = ------------------ total Liters .0200 L X .250 moles/L = .00500 moles HCl .0300 L X .150 moles/L = .00450 moles HCl ------------ ------------------------- .0500 L .00950 moles HCl total .00950 moles HCl M = -------------------------- = .190 M HCl .0500 L soln.

  17. Sample Problem #7 10.0 mL of .375 M HCl solution is diluted by adding water to a new volume of 50.0 mL of solution. What is the concentration of the resulting solution? total moles total molarity = ------------------- total Liters .0100 L X .375 moles/L = .00375 moles HCl .00375 moles HCl M = ---------------------------- = .0750 M HCl .0500 L soln.

  18. Sample Problem #7 - 2 10.0 mL of .375 M HCl solution is diluted by adding water to a new volume of 50.0 mL of solution. What is the concentration of the resulting solution? M1V1 = M2V2 (.375M)(10.0mL) = M(50.0mL) M = .0750 M

  19. Sample Problem #8 Indicate the concentration of each ion present in the solution formed by mixing 44.0 mL of 0.100 M Na2SO4 and 25.0 mL of 0.150 M KCl. Na2SO4 + KCl X (No reaction!) Total Volume = 44.0mL + 25.0mL = 69.0mL=.0690L

  20. Sample Problem #9 1.00 g of aluminum reacts with 75.0 mL of 0.300 M ZnI2 solution. How many grams of zinc are produced? 0.300 M 1.00 g 75.0 mL x g 2 Al (s) + 3 ZnI2 (aq)  2 AlI3 (aq) + 3 Zn (s) 0.300 mole ZnI2 2 mole Al 27.0 g Al .0750 L soln x ------------------ x --------------- x ----------- = 0.405 g Al 1 L soln 3 mole ZnI2 1 mole Al Al is excess or ZnI2 is limiting 0.300 mole ZnI2 3 mole Zn 65.4 g Zn .0750 L soln x ----------------- x ------------- x ------------- = 1.47 g Zn 1 L soln 3 mole ZnI2 1 mole Zn

  21. Sample Problem #10 75.0 mL of 0.300 M ZnI2soln is added to 125 mL of 0.450 M AgNO3 soln. How many grams of the precipitate AgI are produced? 0.300 M 0.450 M 75.0 mL 125 mL x g ZnI2 (aq) + 2 AgNO3 (aq)  2 AgI (s) + Zn(NO3)2 (aq) 0.300 mole ZnI2 2 mole AgNO3 1 L soln 0.0750 L soln x --------------- x --------------- x ------------ = 0.100 L AgNO3 1 L soln 1 mole ZnI2 0.450 mole AgNO3 excess AgNO3 or ZnI2 is limiting 0.300 mole ZnI2 2 mole AgI 235 g AgI 0.0750 L soln x ----------------- x --------------- x -------------= 10.6 g AgI 1 L soln 1 mole ZnI2 1 mole AgI

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