Ella Li's List: All my chemistry bookmarks for 109H

• equation: 
  
     
  
• Polyprotic Acids
   In contrast to a simple monoprotic acid like acetic acid, with only one equilibrium between the acid and conjugate base, a polyprotic acid contains more than one acidic hydrogen. For a polyprotic acid, n acidic hydrogens will exist in solution in equilibrium with n conjugate base forms (for a total of n+1 species). For example, when phosphoric acid (n = 3, a triprotic acid) is dissolved in solution, the following equilibria are established among the four species H₃PO₄ (phosphoric acid itself), H₂PO₄^-1 (dihydrogen phosphate anion), HPO₄^-2 (hydrogen phosphate anion), and PO₄^-3 (phosphate anion):
• Note that each successive pair of species is linked in an independent equilibrium with H₃O⁺ and that each link has a unique equilibrium constant. If we want to understand how polyprotic acid solutions respond to chemical changes (such as addition of OH^- or H₃O⁺), we need to learn how to predict the concentrations of the four phosphate species, as well as H₃O⁺ and OH^-, under a variety of conditions. To account for all six species, we will use relationships such as chemical equilibrium reactions and associated equilibrium constant equations (Eq. (7)-(9)), the mass balance, electroneutrality and water auto-ionization equilibrium relationships (Eq.'s (10) - (12)): 
  
• • This is relate to our HW 12 last several questions, calculate PH, calculate after adding certain amount of acid and base, the buffer is still effective or not.
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• A theoretical titration curve for the titration of 10 mL of 0.1 M phosphoric acid with 0.1 M NaOH is shown below. Because the initial acid and the titrating base concentrations are equal and the initial acid volume was 10 mL, the first equivalence point should appear at 10 mL added base. The graph shows this point clearly, along with the clear second equivalence point at 20 mL added base. But the third equivalence point (at 30 mL added base) is not distinct. This point is washed out because K_a3 is close to K_w, the water dissociation constant. Phosphate ion is a fairly strong base and competes with OH^- for hydrogen ions, at high pH. Note that the half-equivalence points are found in each buffer region, midway between adjacent equivalence points.

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• • The equivalence point is the place with the most rapid change. Do not confuse with the slowest change place in HW which is the buffer's EP.
• • this is similar to our lab -two EP

• • Change the numbers of the acid and base concentration and volume. it illustrate the whole titration process and the calculation.

• CalculateH,S, andG for the above reaction to determine whether  the reaction is spontaneous or not.  

  First let's calculateH_f.  Note that in the above reaction, one mole of  NH₄NO₃ dissociates in water to give one mole each of  NH₄⁺ and NO₃^-:  

   

  

   

  Next, let's calculateS:  

   

  
• Sample free energy calculation

    

  
  

                         

  Compound		Hf		S
  NH4NO3(s)		-365.56		151.08
  NH4+(aq)		-132.51		113.4
  NO3-(aq)		-205.0		146.4

   

  CalculateH,S, andG for the above reaction to determine whether  the reaction is spontaneous or not.  

   First let's calculate  H_f  .    Note  that in the above reaction, one mole of   NH₄NO₃ dissociates in  water to give one mole each of   NH₄⁺ and  NO₃^-: 

• Entropy and probability

   

  As was explained in the preceding lesson, the  distribution of thermal energy in a system is characterized by the number of  quantized microstates that are accessible (i.e., among which energy can be  shared); the more of these there are, the greater the entropy of the system.  This is the basis of an alternative definition of entropy
  

   

  S = k ln Ω (2-2)

   

  in which k is the Boltzmann constant (the gas constant per molecule,  1.3810^–23 J K^–1) and Ω (omega) is the  number of microstates that correspond to a given macrostate of the system. The  more such microstates, the greater is the probability of the system being in the  corresponding macrostate. For any physically realizable macrostate, the quantity  Ω is an unimaginably large number, typically around for one mole. By comparison, the  number of atoms that make up the earth is about 10⁵⁰. But even though  it is beyond human comprehension to compare numbers that seem to verge on  infinity, the thermal energy contained in actual physical systems manages to  discover the largest of these quantities with no difficulty at all, quickly  settling in to the most probable macrostate for a given set of conditions.

   

   

   The reason  S depends on the logarithm of Ω is easy to   understand.  Suppose we have two systems (containers of gas, say) with   S₁,  Ω₁ and S₂, Ω₂. If we now   redefine this as  a single system (without actually mixing the two gases), then   the entropy of  the new system will be   S   =   S₁   +   S₂    but the number  of microstates will be the product Ω₁Ω₂   because for each  state of system 1, system 2 can be in any of Ω₂   states. Because  ln(Ω₁Ω₂)   =   ln    Ω₁   +   ln   Ω₂,  the additivity of the entropy   is  preserved. 

  • helpful for the extra lab assignment

• The reason S depends on the logarithm of Ω is easy to  understand. Suppose we have two systems (containers of gas, say) with  S₁, Ω₁ and S₂, Ω₂. If we now  redefine this as a single system (without actually mixing the two gases), then  the entropy of the new system will be  S = S₁ + S₂  but the number of microstates will be the product Ω₁Ω₂  because for each state of system 1, system 2 can be in any of Ω₂  states. Because ln(Ω₁Ω₂) = ln  Ω₁ + ln Ω₂, the additivity of the entropy  is preserved.

• • A step by step explaination of a typical question.
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• The reaction of aldehydes and ketones with ammonia or 1º-amines forms  imine derivatives, also known as Schiff bases, (compounds having a  C=N function). This reaction plays an important role in the synthesis of  2º-amines, as discussed earlier. Water is  eliminated in the reaction, which is acid-catalyzed and reversible in the same  sense as acetal formation.

   R₂C=O   +   R'NH₂      R'NH–(R₂)C–O–H       R₂C=NR'   +   H₂O

  • the ketone or aldehyde reaction with NH2

• • primary alcholol's oxidation form aldehyde
    secondary alcholol's oxidation form ketone
• relationship between the primary alcohol and the aldehyde formed.
• • aldehyde's oxidation forms carboxylic acids
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• Secondary alcohols are oxidised to ketones
• Tertiary alcohols aren't oxidised by acidified sodium or potassium  dichromate(VI) solution. There is no reaction whatsoever.

  • Tertiary alcohol is much harder to oxidize and it doesn't react with potassium dichromate
    this answer the review question
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• • this is our in class experiment

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• Carboxylic Acid
  
  R----C===O
  .........|
  ........OH
• Aldehyde
  
  R----C===O
  .........|
  ........H
  
  Ketone
  
  R----C===O
  .........|
  ........R'
• Ester
  
  .......O
  ........||
  R----C----O----R'
• R----C===O
  .........|
  

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• • I just curious about the alfa helix interaction. I found this rally nice pic since I thought this might on the exam. This relate to our lab question and it's also relate to our review exam. it;s always NH and C=O interat with each other and from hydrogen bond. There is no interaction occurs between R group or C in the mainchain.
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• Secondary structure - this term refers to the arrangement of amino acids that  are close together in a chain. Examples of secondary structures are  helices and pleated sheets. An alpha helix is a tightly  coiled, rodlike structure which has an average of 3.6 amino acids per turn. The  helix is stabilized by hydrogen bonding between the backbone carbonyl of one  amino acid and the backbone NH of the amino acid four residues away. All main  chain amino and carboxyl groups are hydrogen bonded, and the R groups stick out  from the structure in a spiral arrangement. As seen in the table below, there  are several types of alpha helix that arise from the degree of hydrogen bonding  in the helix.
• • beta barral-in our lab, we did a question about this.
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• The interior of a protein molecule contains a preponderance of hydrophobic  amino acids, which tend to cluster and exclude water. The core is stabilized  by Van der Waals forces and hydrophobic bonding.

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• Note that the hydrogen bonds holding together the two peptide chains are not  180 degrees
• • A question relates to this diagram in the practice exam. The side chain reaction is between O H

  • he interaction is between H from -N-H and O from -C=O
    one line is parallel and one line is antiparallel
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• the interaction is between H from -N-H and O from -C=O
  one line is parallel and one line is antiparallel - Ella Li on 2008-10-19