Learning Center
Plans & pricing Sign in
Sign Out

Basic Concepts_ Molarity_ Solutions and Dilutions


									Review of Basic Concepts,
Molarity, Solutions, Dilutions
      and Beer’s Law
           Chapter 4
         Aqueous Solutions
 In Chemistry, many reactions take place in
  water. This is also true for Biological
 Reactions that take place in water are said
  to occur in an aqueous solution.
 Three types of reactions take place in
  aqueous solutions: Precipitation, Acid-
  Base and Redox.
Properties of Aqueous Solutions
   Solution- a homogeneous mixture of two or
    more substances.
   Solute- a substance in a solution that is present
    in the smallest amount.
   Solvent- a substance in a solution that is
    present in the largest amount.
   In an aqueous solution, the solute is a liquid or
    solid and the solvent is always water.
Properties of Aqueous Solutions
   All solutes that dissolve in water fit into one of
    two categories: electrolyte or non-electrolyte.
   Electrolyte- a substance that when dissolved in
    water conducts electricity
   Non-electrolyte- a substance that when
    dissolved in water does not conduct electricity.
   To have an electrolyte, ions must be present in
        Electrolytic Properties of
           Aqueous Solutions
   NaCl in water.
     What happens?
     NaCl(s) → Na+(aq) + Cl–(aq)
     Completely dissociates
    Strong vs. Weak Electrolytes
   How do you know when an electrolyte is
    strong or weak?

   Take a look at how HCl dissociates in
       HCl(s) →   H+(aq) + Cl–(aq)
Electrolytic Properties of
   Aqueous Solutions
Electrolytic Properties of Aqueous
Hydrated Ions
         Electrolytic Properties of
            Aqueous Solutions
 What about weak electrolytes?
 What makes them weak?
       Ionization of acetic acid

          CH3COOH(aq)   ↔ CH3COO–(aq) + H+(aq)
Electrolytic Solutions
          Precipitation Reactions
 Precipitation Reaction- a reaction that
  results in the formation of an insoluble
 These reactions usually involve ionic
 Formation of PbI2:
       Pb(NO3)2(aq) + 2KI(aq) →   PbI2(s) + 2KNO3(aq)
         Precipitation Reactions
   How do you know whether or not a precipitate
    will form when a compound is added to a
   By knowing the solubility of the solute!
   Solubility- The maximum amount of solute that
    will dissolve in a given quantity of solvent at a
    specific temperature.
   Three levels of solubility: Soluble, slightly soluble
    or insoluble.
Precipitation Reactions
         Determining Solubility
   Determine the level of solubility for the
    (1) Ag2SO4
    (2) CaCO3
    (3) Na3PO4
           Acid-Base Reactions
   Acids- generally have a sour taste, change
    litmus from blue to red, can react with certain
    metals to produce gas, conduct electricity.
   Bases- generally have a bitter taste, change
    litmus from red to blue, feel slippery, conduct
   BrØnstead Acid- proton donor
   BrØnstead Base- proton acceptor
            Acid-Base Reactions
   Acid or Base?
       HCl(aq) + H2O(l) → H3O+(aq) + Cl–(aq)

       NH3(aq) + H2O(l) → NH4+(aq) + OH–(aq)
         Acid-Base Reactions
   Look at the following compounds and
    decide whether they are a BrØnstead Acid
    or a BrØnstead Base.
     HBr
     NO2–
     HCO3–
Acid-Base Reactions
Oxidation-Reduction Reactions
 Can also be called Redox reactions.
 Considered electron-transfer reactions.
 Occur in steps called half-reactions.
       Half-Reactions- Two parts to a redox
        reaction that explicitly show the electrons
        involved and where they are transferred.
Oxidation Reduction Reactions
   Oxidation Reaction- refers to the half-reaction
    that involves the loss of electrons.
   Reduction Reaction- refers to the half-reaction
    that involves the gain of electrons.
   Oxidizing agent- the compound or ion in a
    redox reaction that donates electrons.
   Reducing agent- the compound or ion in a
    redox reaction that accepts electrons.
Oxidation-Reduction Reactions
        Concentration of Solutions
   Concentration of a Solution- amount of solute
    present in a given quantity of solvent or solution.
   We will use Molarity, M for measurement.
    Molarity can also be called Molar Concentration.
   Molarity– the number of moles of solute per liter
    of solution.
       Molarity- moles of solutes/ liters of solution
            Or n/v
       Moles- grams of sample/ molecular weight of sample
            Or g/ mw
    Concentration of Solutions
How many moles are there in 24.0g of C?
    moles = g/mw
    moles = 24.0g C/ 12.0g C
    moles = 2.0
There are 2.0 moles of C in 24.0g of C.
      Concentration of Solutions
   How many grams are in 2.0 moles of Boron?
       moles= g/MW
       2.0 moles = grams/ 10.81g Boron
       2.0 moles x 10.81g Boron = grams
       Grams = 21.62
   There are 21.62 g of Boron in 2.0 moles of
      Concentration of Solutions
   What is the Molarity of a 1L solution
    containing 9.0g HCl?
       9.00g HCl x 1 mol HCl/ 18.00g HCl
                    = 0.5 mol HCl
       M = n/v
       M = 0.5 mol HCl/ 1L
       M = 0.5
       The concentration of the solution is 0.5M.
    Preparation of Solutions
 Now that you know how to calculate M, n
  and v, what does that mean?
 You can make your own solutions!
 What are the steps in making a proper
Preparation of Solutions
       Concentration of Solutions
   How many grams of Potassium Dichromate, K2Cr2O7, are required to
    prepare a 250mL solution with a concentration of 2.16M?

         250mL x 1L/ 1000mL = .250L

         M= n/v
         n= M x v
         n= 2.16M x .250L
         n= 0.54 mol

         moles = g/MW
         Grams = moles x MW
         Grams = 0.54 mol K2Cr2O7 x 294.2 g K2Cr2O7
         Grams = 159

         159 grams of K2Cr2O7 are needed to prepare the requested solution.
       Concentration of Solutions
   In a biochemical assay, a chemist needs to add 0.381g of glucose to a
    reaction mixture. Calculate the volume in millimeters of a 2.53M glucose
    solution that she should use for this addition.

         moles = g/MW
         moles = 0.381g C6H12O6/ 180.2g C6H12O6
         moles = 2.114 x 10 –2 mol C6H12O6

         M = n/v
         v = n/M
         v = 2.114 x 10 –2 mol C6H12O6 / 2.53M C6H12O6
         v = 8.36mL

         She should use 8.36mL of the 2.53M glucose solution.
        Preparation of Solutions
   Explain the process of creating 1L of 3.0M KCl.
       M = n/v
       n = 3.0M x 1L
       n = 4.0 mol of KCl needed

       moles= g/MW
       Grams = moles x MW
       Grams = 4.0 mol KCl x 36.0g KCl
       Grams = 144g KCl

       Weigh out 144g of KCl. Put in a 1L flask. Add enough dH20 to
        dissolve KCl. Fill flask to 1L meniscus.
            Dilution of Solutions
   Dilution- the procedure for preparing a less
    concentrated solution from a more concentrated
   Dilutions can be made in increments of 10, 20,
    50 or any other value.
   Serial Dilution- the process of diluting a solution
    by removing part of it, placing this in a new flask
    and adding water to a known volume in the new
          Dilution of Solutions
   When you want to dilute a solution, what
    happens to the number of moles present
    in the solution?
     Do they increase?
     Decrease?
     Stay the same?
Dilution of Solutions
Dilution of Solutions
            Dilution of solutions
   Since moles are constant before and after
    dilution, we can use the following formula
    for calculations.

        MiVi   = MfVf
              Dilution of Solutions
   Describe how you would prepare 800mL of a 2.0M H2SO4 solution,
    starting with a 6.0M stock solution of .

        800mL x 1L/ 1000mL
          = 0.800L

        MiVi = MfVf
        6.0M x Vi = 2.0M x 0.800L
        6.0M x Vi = 1.6M x L
        Vi = 1.6M x L/ 6.0M
        Vi = 0.26L

        0.26L of the 6.0M H2SO4 solution should be diluted to give a
          final volume of 800mL.
       Concentration of Solutions
   There are several ways to determine the concentration
    of a solution.
   In this week’s lab, we will be using spectroscopy to help
    us identify the molar concentration of an unknown
   Spectroscopy is helpful because it gives us:
      Amount of light transmitted through a solution
      Amount of light absorbed by a solution
      Beer’s Law- a relationship between proportionality
        constant, path length of radiation going through
        solution and concentration of the solution.
   A = abc
   A = -log(T)
                      Beer’s Law
   If we know the value of T, we can solve the
    previous equation and figure out what A is.
   Absorbance vs. wavelength
   Absorbance vs. concentration
       Use excel and be sure to add in your equation for the
       Calculation of A or C:
            A = mc
Beer’s Law

To top