[A level Chem] - Buffer solution

• hoay

  20 Oct 15, 03:55

  A buffer solution is to be made using 1.00 mol dm–3 ethanoic acid, CH3CO2H, and
  1.00 mol dm–3 sodium ethanoate, CH3CO2Na.
  Calculate to the nearest 1 cm3 the volumes of each solution that would be required to
  make 100 cm3 of a buffer solution with pH 5.50.
  Clearly show all steps in your working.

  Ka (CH3CO2H) = 1.79 × 10–5 mol dm–3

  Answer.

  Putting pH and pKa in the Handerson-Hassalbach Equation we will get [A]/[HA] = 0.753..... Now to calulate the volume i want to use c = n / v; c = 1 M but no clue where to find the number of moles ??

• UltimaOnline

  20 Oct 15, 05:11

  Originally posted by hoay:

  A buffer solution is to be made using 1.00 mol dm–3 ethanoic acid, CH3CO2H, and
  1.00 mol dm–3 sodium ethanoate, CH3CO2Na.
  Calculate to the nearest 1 cm3 the volumes of each solution that would be required to
  make 100 cm3 of a buffer solution with pH 5.50.
  Clearly show all steps in your working.

  Ka (CH3CO2H) = 1.79 × 10–5 mol dm–3

  Answer.

  Putting pH and pKa in the Handerson-Hassalbach Equation we will get [A]/[HA] = 0.753..... Now to calulate the volume i want to use c = n / v; c = 1 M but no clue where to find the number of moles ??

  
  Let the volumes (in dm3) of A- and HA be V and (0.1-V) respectively. Then (V dm3 x 1M) / ((0.1-V) dm3 x 1M) = (753 / 1000). Solve for V, then convert from dm3 to cm3, round off to nearest 1cm3.
• hoay

  22 Oct 15, 00:26

  okay i got it. V= 43 cm3 

• UltimaOnline

  22 Oct 15, 06:25

  Originally posted by hoay:

  okay i got it. V= 43 cm3 

  
  :)
  
  You might like to let your students attempt my following acid-base equilibria questions :
  
  
  A couple of BedokFunland JC Buffer Riddles
  
  
  The conjugate acid of a weak base B, has a pKa value of 9.2014 at room temperature. What volumes of 0.1 mol/dm3 of HCl(aq) and 0.05 mol/dm3 of a solution of B, may be mixed to generate 500cm3 of a buffer solution with a pH of 8.90?
  
  Final answer :
  
  125cm3 of 0.1 mol/dm3 of HCl, and 375cm3 of 0.05 mol/dm3 of B, are required to generate 500cm3 of buffer solution with pH 8.9 pH.
  
  --------------------------------------------------------------------------------------------------
  
  

  A buffer solution is titrated with NaOH(aq).  Upon addition of 3 volumes of NaOH(aq), the buffer solution becomes most effective. Equivalence point is attained when 8 volumes of NaOH(aq) is added.

  a) Calculate the ratio of the molarities of both members of the conjugate acid-base pair of the buffer solution, before the NaOH(aq) is added.

  b) Given that the original pH of the buffer solution (ie. before NaOH(aq) is added) is 4.16, calculate the proton dissociation constant of the acidic component of the buffer.

  
  Solution :

  
  a) At maximum buffer capacity :

                             HA    +    OH-    --->    A-    +    H2O

  Initial (mol)         8y            3y              x-8y           n.a.
  Change (mol)   -3y           -3y              +3y            n.a
  Final (mol)         5y             0               x-5y            n.a.

  At equivalence point :

                             HA    +    OH-    --->    A-    +    H2O

  Initial (mol)         8y            8y              x-8y          n.a.
  Change (mol)   -8y           -8y              +8y          n.a.
  Final (mol)          0              0                x              n.a.

  First, complete the ICF table for equivalence point. Next, complete the ICF table for maximum buffer capacity, bearing in mind that the initial moles of HA and A- are the same (for both tables).

  Since at maximum buffer capacity, [HA] = [A-], this implies 5y = x-5y and hence x = 10y.

  Substituting x = 10y into the initial moles of A-, we have moles of A- = 2y.

  Therefore, the ratio of the amounts of both members of the conjugate acid-base pair present in the buffer (before NaOH(aq) is added), is 8y HA : 2y A-, ie. 4 HA : 1 A-

  
  b) Using the Henderson-Hasselbalch equation, we have

  pH = pKa + log ( [base] / [acid] )

  4.16 = pKa + log (1/4)

  pKa =  4.762

  Ka = 1.73 x 10^-5