Çözelti Hazırlama
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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 Çözelti Hazırlama Presentation Transcript

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