<ul><li>You can now complete Lab 19 </li></ul>
<ul><li>A  soluble salt  is an ionic compound that dissolves in water.  </li></ul><ul><li>An  insoluble salt  is an ionic ...
<ul><li>A soluble salt dissolves in water.  </li></ul><ul><ul><li>Breaks apart into ions </li></ul></ul><ul><li>Insoluble ...
<ul><li>The solubility rules predict whether a salt is soluble or insoluble in water. Table 8.8 </li></ul>
<ul><li>When solutions of salts are mixed, a solid forms when ions of an insoluble salt combine. </li></ul>
<ul><li>Precipitate- an insoluble compound that forms as a result a reaction </li></ul><ul><li>The precipitate forms when ...
<ul><li>Pb(NO 3 ) 2  (aq)  + NaBr (aq)    ? (s) + ? (aq) </li></ul><ul><li>Approach:  Consider all possible ion combinati...
<ul><li>You should be able to identify when a precipitate will form based on the solubility of the products.  </li></ul><u...
<ul><li>You can now complete Worksheet 1 </li></ul>
<ul><li>The concentration of a solution is the amount of solute dissolved in a specific amount of solution (solutions = so...
<ul><li>The mass percent (%m/m)- concentration of the solution that describes the mass of solute in every 100 g of solutio...
<ul><li>  grams of solute   + grams of solvent </li></ul><ul><li>50.0 g KCl solution </li></ul>
<ul><li>Mass percent (%m/m) is calculated from the grams of solute (g KCl) and the grams of solution (g KCl solution). </l...
<ul><li>Mass Percent:  Is the g of solute in 100 g of solution. </li></ul><ul><li>mass percent  =  g of  solute    100 g o...
<ul><li>The mass/volume percent (%m/v)  </li></ul><ul><li>Concentration is the ratio of the mass in grams (g) of solute in...
<ul><li>A percent mass/volume solution is prepared by weighing out the grams of solute (g) and adding water to give the fi...
<ul><li>Mass/volume percent (%m/v)  is calculated from the grams of solute (g KCl) and the volume of solution (mL KCl solu...
<ul><li>Is the g of solute in 100 mL of solution. </li></ul><ul><li>mass/volume %  =  g of  solute      100 mL of solution...
<ul><li>The volume percent (%v/v)  </li></ul><ul><li>Concentration  is the percent volume (mL)  of solute (liquid) to volu...
<ul><li>Is the mL of solute in 100 mL of solution. </li></ul><ul><li>volume %  (v/v) =  mL of  solute    100 mL of solutio...
<ul><li>Two conversion factors can be written for any type of % value.  Table 8.9 </li></ul>
<ul><li>How many grams of NaCl are needed to prepare </li></ul><ul><li>250 g of a 10.0% (m/m) NaCl solution? </li></ul><ul...
<ul><li>How much (in g) of KI do you need to make a 2% (m/v) if the final volume is 250 mL?  </li></ul>
<ul><li>Molarity is a concentration unit for the moles of solute in the liters (L) of solution. </li></ul><ul><li>Molarity...
<ul><li>A 1.0 M NaCl solution is prepared by weighing out 58.5 g NaCl ( 1.0 mole) and adding water to make 1.0 liter of so...
<ul><li>What is the molarity of a NaOH solution prepared by adding 4.0 g of solid NaOH to water to make 0.50 L of solution...
<ul><li>What is the molarity of a solution made by dissolving 12.5 g NaCl in water to make 500. mL of solution? </li></ul>...
<ul><li>The units in molarity can be used to write  conversion factors.  Table 8.10 </li></ul>
<ul><li>How many liters of a 2.00M NaCl solution do you need to provide 67.3 g of NaCl </li></ul><ul><li>1. Determine the ...
<ul><li>How many grams of KCl is needed to prepare 4.0L of of a 0.20 M KCl solution?  </li></ul>
<ul><li>Diluting a solution </li></ul><ul><li>Is the addition of water.  </li></ul><ul><li>Decreases concentration. </li><...
<ul><li>In a dilution, the amount (moles or grams) of solute does not change. </li></ul><ul><li>Amount solute (initial) =A...
<ul><li>Add water to the 3.0 M solution to lower its concentration to 0.50 M  </li></ul><ul><li>Dilute the solution!   </l...
But how much water  do we add?
<ul><li>How much water is added? </li></ul><ul><li>The important point is that ---> </li></ul>PROBLEM:  You have 50.0 mL o...
<ul><li>Amount of NaOH in original solution =  </li></ul><ul><li>M • V   =  </li></ul><ul><li>(3.0 mol/L)(0.050 L)  =  0.1...
<ul><li>Conclusion: </li></ul><ul><li>add 250 mL of water  to 50.0 mL of 3.0 M NaOH to make 300 mL of 0.50 M NaOH.  </li><...
<ul><li>What is the molarity of an HCl solution prepared by </li></ul><ul><li>adding 0.50 L of water to 0.10 L of a 12 M H...
<ul><li>What is the molarity of an HCl solution prepared by </li></ul><ul><li>adding 0.50 L of water to 0.10 L of a 12 M H...
<ul><li>What is the molarity of an HCl solution prepared by </li></ul><ul><li>adding 0.50 L of water to 0.10 L of a 12 M H...
<ul><li>The dilution calculation can be expressed as an equation. </li></ul><ul><li>C 1 V 1   =  C 2 V 2 C = concentration...
<ul><li>A shortcut </li></ul><ul><li>C initial  • V initial   =  C final  • V final </li></ul>
<ul><li>How many milliliters of 6.0 M NaOH solution are needed to prepare 1.0 L of a 0.15 M NaOH solution? </li></ul><ul><...
<ul><li>How many milliliters of 6.0 M NaOH solution are needed to prepare 1.0 L of a 0.15 M NaOH solution? </li></ul><ul><...
<ul><li>Describe how to make a 0.12 M solution CuCl from 10mL of a 0.29 M solution of CuCl.  </li></ul>
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  • don’t need to know Hg2+.
  • When you put soluble ions in solution together, they are all bumping around, and when the right ones find each other, they form the precipitate- In this case, lead iodide is insoluble (see exceptions of soluble compounds), and when they find each other, they form a solid.
  • ______ PbBr 2 (s)+ ______ NaNO 3 (aq)
  • FeCl3 (aq) + 3NaOH(aq) = Fe(OH)3 (s) + 3NaCl (aq)
  • solute takes up space
  • .427 M NaCl .335 M KCl
  • 60. g KCl
  • you want .15 mols in a 0.50 mols/L soln, so you need 0.30 L
  • Çözelti Hazırlama

    1. 1. <ul><li>You can now complete Lab 19 </li></ul>
    2. 2. <ul><li>A soluble salt is an ionic compound that dissolves in water. </li></ul><ul><li>An insoluble salt is an ionic compound that does not dissolve in water. </li></ul>
    3. 3. <ul><li>A soluble salt dissolves in water. </li></ul><ul><ul><li>Breaks apart into ions </li></ul></ul><ul><li>Insoluble salts do not dissolve in water. </li></ul><ul><ul><li>Attraction between ions is to strong for water to break apart </li></ul></ul>
    4. 4. <ul><li>The solubility rules predict whether a salt is soluble or insoluble in water. Table 8.8 </li></ul>
    5. 5. <ul><li>When solutions of salts are mixed, a solid forms when ions of an insoluble salt combine. </li></ul>
    6. 6. <ul><li>Precipitate- an insoluble compound that forms as a result a reaction </li></ul><ul><li>The precipitate forms when ions in solution recombine in a way that makes an insoluble compound. </li></ul><ul><li>Pb(NO 3 ) 2 (aq) + 2 KI(aq) ---> </li></ul><ul><li>2 KNO 3 (aq) + PbI 2 (s) </li></ul>
    7. 7. <ul><li>Pb(NO 3 ) 2 (aq) + NaBr (aq)  ? (s) + ? (aq) </li></ul><ul><li>Approach: Consider all possible ion combinations and determine which is insoluble (ie, which is the solid) based on solubility rules. </li></ul>
    8. 8. <ul><li>You should be able to identify when a precipitate will form based on the solubility of the products. </li></ul><ul><li>When iron (III) chloride reacts with sodium hydroxide in water, what are the products? Write a balanced equation and indicate if there are any precipitates. </li></ul>
    9. 9. <ul><li>You can now complete Worksheet 1 </li></ul>
    10. 10. <ul><li>The concentration of a solution is the amount of solute dissolved in a specific amount of solution (solutions = solute + solvent). </li></ul><ul><li>amount of solute </li></ul><ul><li>amount of solution </li></ul><ul><li>The percent concentration describes the amount of solute that is dissolved in 100 parts of solution. </li></ul><ul><li>amount of solute </li></ul><ul><li>100 parts solution </li></ul>
    11. 11. <ul><li>The mass percent (%m/m)- concentration of the solution that describes the mass of solute in every 100 g of solution. </li></ul><ul><li>Concentration is the percent by mass of solute in a solution. </li></ul><ul><li>mass percent = g of solute x 100% g of solution </li></ul>
    12. 12. <ul><li> grams of solute + grams of solvent </li></ul><ul><li>50.0 g KCl solution </li></ul>
    13. 13. <ul><li>Mass percent (%m/m) is calculated from the grams of solute (g KCl) and the grams of solution (g KCl solution). </li></ul><ul><li>g of KCl = 8.0 g </li></ul><ul><li>g of solvent (water) = 42.0 g </li></ul><ul><li>g of KCl solution = 50.0 g </li></ul><ul><li>8.0 g KCl (solute) x 100 = 16% (m/m) KCl </li></ul><ul><li>50.0 g KCl solution </li></ul>
    14. 14. <ul><li>Mass Percent: Is the g of solute in 100 g of solution. </li></ul><ul><li>mass percent = g of solute 100 g of solution </li></ul><ul><li>10% (m/m) = 10 g of solute in 100g of solution </li></ul><ul><li>15% (m/m) = 15 g of solute in 100 g of solution. </li></ul>
    15. 15. <ul><li>The mass/volume percent (%m/v) </li></ul><ul><li>Concentration is the ratio of the mass in grams (g) of solute in a volume (mL) of solution. </li></ul><ul><li>mass/volume % = g of solute x 100% mL of solution </li></ul>
    16. 16. <ul><li>A percent mass/volume solution is prepared by weighing out the grams of solute (g) and adding water to give the final volume of the solution. </li></ul>
    17. 17. <ul><li>Mass/volume percent (%m/v) is calculated from the grams of solute (g KCl) and the volume of solution (mL KCl solution). </li></ul><ul><li>g of KI = 5.0 g KI </li></ul><ul><li>mL of KI solution = 250.0 mL </li></ul><ul><li>5.0 g KI (solute) x 100 = 2.0%(m/v) KI </li></ul><ul><li>250.0 mL KI solution </li></ul>
    18. 18. <ul><li>Is the g of solute in 100 mL of solution. </li></ul><ul><li>mass/volume % = g of solute 100 mL of solution </li></ul><ul><li>10% (m/v) = 10 g of solute in 100 mL of solution </li></ul><ul><li>15% (m/v) = 15 g of solute in 100 mL of solution </li></ul>
    19. 19. <ul><li>The volume percent (%v/v) </li></ul><ul><li>Concentration is the percent volume (mL) of solute (liquid) to volume (mL) of solution. </li></ul><ul><li>volume % (v/v) = mL of solute x100% mL of solution </li></ul>
    20. 20. <ul><li>Is the mL of solute in 100 mL of solution. </li></ul><ul><li>volume % (v/v) = mL of solute 100 mL of solution </li></ul><ul><li>10% (v/v) = 10 mL of solute in 100 mL of solution </li></ul><ul><li>15% (v/v) = 15 mL of solute in 100 mL of solution </li></ul>
    21. 21. <ul><li>Two conversion factors can be written for any type of % value. Table 8.9 </li></ul>
    22. 22. <ul><li>How many grams of NaCl are needed to prepare </li></ul><ul><li>250 g of a 10.0% (m/m) NaCl solution? </li></ul><ul><li>1. Write the 10.0 %(m/m) as conversion factors. </li></ul><ul><li>10.0 g NaCl and 100 g solution </li></ul><ul><li>100 g solution 10.0 g NaCl </li></ul><ul><li>2. Use the factor that cancels given (g solution). </li></ul><ul><li>250 g solution x 10.0 g NaCl = 25 g NaCl 100 g solution </li></ul>
    23. 23. <ul><li>How much (in g) of KI do you need to make a 2% (m/v) if the final volume is 250 mL? </li></ul>
    24. 24. <ul><li>Molarity is a concentration unit for the moles of solute in the liters (L) of solution. </li></ul><ul><li>Molarity (M) = moles of solute = moles </li></ul><ul><li> liter of solution L </li></ul><ul><li>Examples: </li></ul><ul><li>2.0 M HCl = 2.0 moles HCl </li></ul><ul><li> 1 L </li></ul><ul><li>6.0 M HCl = 6.0 moles HCl </li></ul><ul><li> 1 L </li></ul>
    25. 25. <ul><li>A 1.0 M NaCl solution is prepared by weighing out 58.5 g NaCl ( 1.0 mole) and adding water to make 1.0 liter of solution. </li></ul>
    26. 26. <ul><li>What is the molarity of a NaOH solution prepared by adding 4.0 g of solid NaOH to water to make 0.50 L of solution ? </li></ul><ul><li>1. Determine the moles of solute. </li></ul><ul><li>4.0 g NaOH x 1 mole NaOH = 0.10 mole </li></ul><ul><li>40.0 g NaOH </li></ul><ul><li>2. Calculate molarity. </li></ul><ul><li>0.10 mole = 0.20 mole = 0.20 M NaOH </li></ul><ul><li> 0.50 L 1 L </li></ul><ul><li> </li></ul>
    27. 27. <ul><li>What is the molarity of a solution made by dissolving 12.5 g NaCl in water to make 500. mL of solution? </li></ul><ul><li>What is the molarity of a solution made by dissolving 12.5 g KCl in water to make 500. mL of solution? </li></ul>
    28. 28. <ul><li>The units in molarity can be used to write conversion factors. Table 8.10 </li></ul>
    29. 29. <ul><li>How many liters of a 2.00M NaCl solution do you need to provide 67.3 g of NaCl </li></ul><ul><li>1. Determine the moles of solute desired. </li></ul><ul><li>67.3 g NaCl x 1 mole NaCl = 1.15 mole </li></ul><ul><li>58.5 g NaCl </li></ul><ul><li>2. Calculate volume. </li></ul><ul><li>1.15 mole x 1L = 0.575 L NaCl </li></ul><ul><li> 2.00 mole </li></ul>
    30. 30. <ul><li>How many grams of KCl is needed to prepare 4.0L of of a 0.20 M KCl solution? </li></ul>
    31. 31. <ul><li>Diluting a solution </li></ul><ul><li>Is the addition of water. </li></ul><ul><li>Decreases concentration. </li></ul><ul><li>Concentrated Diluted </li></ul><ul><li>Solution Solution </li></ul>
    32. 32. <ul><li>In a dilution, the amount (moles or grams) of solute does not change. </li></ul><ul><li>Amount solute (initial) =Amount solute (dilute) </li></ul><ul><li>A dilution increases the volume of the solution. </li></ul><ul><li>Amount solute Amount solute </li></ul><ul><li>Volume (initial) Volume (increased) </li></ul><ul><li>As a result, the concentration of the solution decreases. </li></ul>
    33. 33. <ul><li>Add water to the 3.0 M solution to lower its concentration to 0.50 M </li></ul><ul><li>Dilute the solution! </li></ul>
    34. 34. But how much water do we add?
    35. 35. <ul><li>How much water is added? </li></ul><ul><li>The important point is that ---> </li></ul>PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? moles of NaOH in ORIGINAL solution = moles of NaOH in FINAL solution
    36. 36. <ul><li>Amount of NaOH in original solution = </li></ul><ul><li>M • V = </li></ul><ul><li>(3.0 mol/L)(0.050 L) = 0.15 mol NaOH </li></ul><ul><li>Amount of NaOH in final solution must also = 0.15 mol NaOH </li></ul><ul><li>Volume of final solution = </li></ul><ul><li>(0.15 mol NaOH)(1 L/0.50 mol) = 0.30 L </li></ul>
    37. 37. <ul><li>Conclusion: </li></ul><ul><li>add 250 mL of water to 50.0 mL of 3.0 M NaOH to make 300 mL of 0.50 M NaOH. </li></ul>
    38. 38. <ul><li>What is the molarity of an HCl solution prepared by </li></ul><ul><li>adding 0.50 L of water to 0.10 L of a 12 M HCl. </li></ul><ul><li>1. Calculate the moles of HCl. </li></ul><ul><li> 0.10 L x 12 moles HCl = 1.2 moles HCl </li></ul><ul><li> 1 L </li></ul>
    39. 39. <ul><li>What is the molarity of an HCl solution prepared by </li></ul><ul><li>adding 0.50 L of water to 0.10 L of a 12 M HCl. </li></ul><ul><li>1. Calculate the moles of HCl. </li></ul><ul><li> 0.10 L x 12 moles HCl = 1.2 moles HCl </li></ul><ul><li> 1 L </li></ul><ul><li>2. Determine the volume after dilution. </li></ul><ul><li> 0.10 L + 0.50 L = 0.60 L solution </li></ul>
    40. 40. <ul><li>What is the molarity of an HCl solution prepared by </li></ul><ul><li>adding 0.50 L of water to 0.10 L of a 12 M HCl. </li></ul><ul><li>1. Calculate the moles of HCl. </li></ul><ul><li> 0.10 L x 12 moles HCl = 1.2 moles HCl </li></ul><ul><li> 1 L </li></ul><ul><li>2. Determine the volume after dilution. </li></ul><ul><li> 0.10 L + 0.50 L = 0.60 L solution </li></ul><ul><li>3. Calculate the molarity of the diluted HCl . </li></ul><ul><li> 1.2 moles HCl = 2.0 M HCl </li></ul><ul><li> 0.60 L </li></ul>
    41. 41. <ul><li>The dilution calculation can be expressed as an equation. </li></ul><ul><li>C 1 V 1 = C 2 V 2 C = concentration </li></ul><ul><li>For molar concentration, the equation is M 1 V 1 = M 2 V 2 </li></ul><ul><li>(moles) = (moles after dilution ) </li></ul><ul><li>For percent concentration, </li></ul><ul><li>% 1 V 1 = % 2 V 2 </li></ul><ul><li>(grams) = (grams after dilution ) </li></ul>
    42. 42. <ul><li>A shortcut </li></ul><ul><li>C initial • V initial = C final • V final </li></ul>
    43. 43. <ul><li>How many milliliters of 6.0 M NaOH solution are needed to prepare 1.0 L of a 0.15 M NaOH solution? </li></ul><ul><li>M 1 = 6.0 M M 2 = .15 M </li></ul><ul><li>V 1 = ??? V 2 = 1.0 L </li></ul><ul><li>Rearrange the dilution equation for V 1 </li></ul>
    44. 44. <ul><li>How many milliliters of 6.0 M NaOH solution are needed to prepare 1.0 L of a 0.15 M NaOH solution? </li></ul><ul><li>M 1 = 6.0 M M 2 = .15 M </li></ul><ul><li>V 1 = ??? V 2 = 1.0 L </li></ul><ul><li>Rearrange the dilution equation for V 1 </li></ul><ul><li>V 1 = M 2 V 2 = 0.15 M x 1.0 L </li></ul><ul><li> M 1 6.0 M </li></ul><ul><li>= 0.025 L x 1000 mL = 25 mL </li></ul><ul><li> 1 L </li></ul>
    45. 45. <ul><li>Describe how to make a 0.12 M solution CuCl from 10mL of a 0.29 M solution of CuCl. </li></ul>

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