Lab 3
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chem lab 3- solubility and equilibrium

chem lab 3- solubility and equilibrium

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Lab 3 Lab 3 Presentation Transcript

  • Lab 3 Solubility
  • Concentration and Conductivity
    • Demonstration
    • Sodium chloride was added to 300 mL of water in small increments
      • 1 st scoop dissolves
        • and solution barely conducts electricity (dim light)
      • 2 nd scoop dissolves and light gets brighter
      • 3 rd scoop dissolves and lights brighter
      • 4 th scoop dissolves but light gets only a bit brighter
      • 5 th scoop does not appear to dissolve
  • Increasing Conductivity
    • Suggests increasing number of ions in same volume of water to allow solution to conduct a higher voltage
    + + + + + + + - - - - - - - - - - - - + + + + + Least concentrated Most concentrated Compare the number of ions in the same volume
  • Graph to compare conductivity and concentration conductivity concentration Increasing concentration — Increasing conductivity Conductivity no longer increases
  • Eventually Constant Conductivity
    • Conductivity no longer changes
    • This suggests that the concentration is no longer changing
    • Solution must have reached a maximum concentration
    • Undissolved solid remains at the bottom
    • The solution is said to be SATURATED.
  • Lab Procedures
    • A saturated solutions of several ionic compounds were prepared
      • Salt was added until solid remained at the bottom
    • The solution was decanted
      • The solution was poured down a stirring rod leaving the solid in behind in the beaker
      • The goal was to isolate the saturated solution
    • 10.0 mL of the saturated solution heated until all the water was removed
      • This allowed us to measure the amount of salt that was dissolved in the 10.0 mL of solution
  • Calculating the concentration of the saturated solution
    • Use the data of # grams in 10.0 mL of solution
    • Convert grams to moles
    • Convert mL to L
    MOLARITY moles of solute liter of solution
  • Sample Data
  • EOS The attractions of water dipoles for ions pulls the ions out of the crystalline lattice and into aqueous solution
  • Equilibrium in a Saturated Solution
    • When the solution is saturated, there is no net change- conditions remain constant
    • The solution and solid are in equilibrium
    • Both reactions (dissolving and crystallizing) are occurring at equal rates
    • Example
      • NaCl(s)  Na + (aq) + Cl - (aq)
      • The ions are dissociating as fast as the ions crystallize
  • Dynamic Nature of Equilibrium When a system reaches equilibrium , the forward and reverse reactions continue to occur … but at equal rates. We are usually concerned with the situation after equilibrium is reached. After equilibrium the concentrations of reactants and products remain constant.
  • Dynamic Equilibrium Illustrated NaCl dissolves and recrystallizes continuously. After a time, the solution contains radioactive Na + … NaCl containing radioactive Na + is added to a saturated NaCl solution. … and the added salt now contains some stable Na + .
  • Calculating an Equilibrium Constant
    • Since conditions remain constant at equilibrium, an equilibrium constant (Keq) can be calculated
    • Keq is calculated
      • as a ratio of concentrations of products compared to reactants at equilibrium
      • the coefficients in the equation become the exponents in the Keq
    • Since the concentrations of solids and liquids are constant, they are both left out of the Keq
  • Ksp- the equilibrium constant for saturated solutions of ionic compounds
      • NaCl(s)  Na + (aq) + Cl - (aq)
    • Ksp = [Na + ][Cl - ]
      • If the concentration of dissolved NaCl in a saturated solution is 5 M
      • Then the concentration of sodium ions must be 5 M and the concentration of chloride ions must be 5 M
      • Ksp = [Na + ][Cl - ] = (5 M)(5M) = 25
  • Calculating Concentration given the Ksp
      • AgCl(s)  Ag + (aq) + Cl - (aq)
    • Ksp = [Ag + ][Cl - ]
      • The concentration is unknown
        • Let x = [AgCl], then x also x = [Ag + ] and x = [Cl - ]
      • Ksp = [Ag + ][Cl - ] = (x)(x) = x 2 = 1.6 x 10 -10
        • So x = 1.3 x 10 -5 M
    • Much less silver chloride is present in saturated solution than sodium chloride
      • The lower Ksp shows the compound is less soluble
  • Given the Concentration calculate the Ksp
      • Ag 2 S(s)  2Ag + (aq) + S 2- (aq)
    • Ksp = [Ag + ] 2 [S 2- ]
    • The concentration of silver sulfide in a saturated solution is 3.1 x 10 -17 M = [Ag 2 S]
    • The concentration of sulfide ions must be
    • 3.1 x 10 -17 M = [S 2- ]
    • The concentration of silver ions must be
    • 2 x 3.1 x 10 -17 M = 6.2 x10 -17 M = [Ag + ]
    • Ksp = [Ag + ] 2 [S 2- ]
        • = (6.2 x10 -17 ) 2 (3.1 x 10 -17 )
        • = 3.2 x 10 -49
        • A very small Ksp indicates that this compound is barely soluble
  • Ksp and solubility
    • Ksp is calculated using the concentrations of dissolved ions in the saturated solution, thus Ksp indicates the relative solubility of compounds
    • A Ksp greater than 0.1 indicates a compound is soluble
    • A compound is designated as insoluble is the Ksp is less than 0.1
  •  
  • Calculating the Concentration of Calcium fluoride in a saturated solution
    • Ksp = 5.3 x 10-11
      • Reversible reaction at equilibrium
      • CaF 2 (s)  Ca 2+ (aq) + 2F - (aq)
      • Ksp = [Ca 2+ ][F - ] 2
    • If x = [CaF 2 ], then x = [Ca 2+ ] and 2x = [F - ]
      • Ksp = [Ca 2+ ][F - ] 2 = 5.3 x 10 -11 = (x)(2x) 2
      • (x)(4x 2 ) = 4x 3 = 5.3 x 10 -11
      • Divide by 4 and then cube root
  • Preparing a solution of known molarity
    • Note the final volume of the solution is exactly 250mL.
    • Calculate the molarity of the is solution
  • Again, a molar solution The volume is exactly 1.0 L Note that not all of the initial 1.0 L of water was used to prepare the solution
  • Dilution of a solution
    • Calculate the molarity of the new solution using M 1 V 1 = M 2 V 2
  • Other Units of Concentrations
    • Molality
      • Moles of solute per kg of solvent
    • Mole fraction
      • Moles of solute per total # moles
    • Percent by Mass
    • mass of solute/mass of solution x 100
  • Mole Fraction and Mole Percent The mole fraction ( x i ) of a solution component i is the fraction of all the molecules in the solution that are molecules of i EOS The mole percent of a solution component is its mole fraction multiplied by 100 Sum of x i must equal 1
  • Calculating other concentrations
    • 1.0L of the 0.100M CuSO 4 solution was prepared by adding 25.0g or 0.100 moles of solid to 992mL of water.
    • Calculate the molality of the solution
      • Looking for moles of CuSO 4 per kg of water
    • Calculate the mole fraction of the solution.
      • Looking for moles of CuSO 4 per total moles
      • Total moles = moles of CuSO 4 + moles of water
    • Calculate the percent by mass of the solution.
      • Looking for mass of CuSO 4 per total mass
  • Dilution The desired molarity solutions are often prepared from concentrated stock solutions (routinely used solutions prepared in concentrated form) by adding water. This process is called dilution. Moles of solute before dilution = moles of solute after dilution M 1 V 1 = M 2 V 2
  • Figure 4.11 (a) A Measuring Pipet (b) A Volumetric (transfer) Pipet
  • Example: What volume of 16 M sulfuric acid must be used to prepare 1.5 L of 0.10 M H 2 SO 4 solution? V 1 = Volume before dilution = ? M 1 = Concentration before dilution = 16 M V 2 = Volume after dilution = 1.5 L M 2 = Concentration after dilution = 0.10 M M 1 V 1 = M 2 V 2 V 1 =M 2 V 2 / M 1 = (0.10 M x 1.5 L)/ 16 M = 0.0094 L = 9.4 mL
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