Ella Li's List: Chemistry 109H Third and Final Exam preparation

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

• • 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

• 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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• examine acid-base reactions through specific examples
• • don't forget minus the ions created from the total concentration In HW 12 Q 15, BE CAREFUL
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• Weak Acids

   

  Weak acids dissociate only slightly such that if HA in equation [5.14] is a  weak acid, the reaction proceeds only partially to the right and [HA] is still a  significant species in solution. A simple equation can be derived for  determining the pH of a weak acid added to water by making use of equation  [5.14] and equation [5.15], combined with a mass balance (equation [5.18]).

   

  

   

   

   

  Algebraically manipulating equation [5.15] and equation [5.18], while  utilizing the stoichiometry of equation [5.14] results in the following  equation:

   

  

   

  [5.22]

   

  

   

   

   

  [5.23]

   

   

   

  Note that the above simplifying assumption can - and should - be checked once  the value of [H+] is determined. If the final result is not valid, the final  result can be used in the preceding solution (equation[5.22]) of [H+] and  iterated to obtain an improved solution. This iteration of successive  approximation is generally quicker than trying to solve the quadratic  equation.

   

   

   

  Example 5.7 Acetic acid (H3CCOOH) is produced in anaerobic sludge  digesters. Typically, digesters contain 60,000 mg/L or 0.1M acetic acid. What  is the pH of a 0.1 M solution of acetic acid? Acetic acid has a KA of 1.8 x  10^-5. Ignore all species that might be in the digester except acetic acid and  water.

   

  

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• Derivation of Ka x Kb = Kw

   

  For the equation: CH₃COOH CH₃COO^- + H⁺

   

  CH₃COOH is the acid and  CH₃COO^- is its conjugate base.

       

  Ka =	[H+][CH3COO-]
  	[CH3COOH]

   

  And for the conjugate base reaction: CH₃COO^-  + H₂O CH₃COOH + OH^-

       

  Kb =	[OH-][CH3COOH]
  	[CH3COO-][H2O]

   

  As water is in vast excess on both sides of the equilibrium it can be safely  eliminated to give

       

  Kb =	[OH-][CH3COOH]
  	[CH3COO-]

   

  Combining Ka and Kb gives:

   

   

          

  Ka x Kb =	[H+][CH3COO-]	x	[OH-][CH3COOH]
  	[CH3COOH]		[CH3COO-]

   

  Cancelling out terms from top and bottom gives:

          

  Ka x Kb =	[H+]	x	[OH-]

   

  And as: Kw = [H⁺][OH^-]

   

  Then:

   

  Ka x Kb = Kw