pH curves

• hoay

  23 Dec 14, 01:31

  The following is aquestion from CIE A-level (9701) Nov 2010 

  G is NH2(CH2)3NH2 is titrated with HCl(aq).

  A 0.10 mol dm–3 solution of G has a pH of 11.3. When 30 cm3 of 0.10 mol dm–3 HCl is added to 10 cm3 of a 0.10 mol dm–3 solution of G, the final pH is 1.6.

  Using the following axes, sketch the pH changes that occur during this addition of

  HCl(aq). [I cannot show the axes here] 

   

  No other data such as pka values were given. Its was a 2 marks question.

  Solution :  I only knew the starting pH that is 11.3 and the end pH 1.6. There will be two humps as the base is diacidic. The end-point i have no idea. Please help. 

• UltimaOnline

  23 Dec 14, 03:41

  Here is my BedokFunland JC answer to this 2010 Cambridge A level Exam Qn.
  
  Since molarities of both acid and base are the same, the volumes of HCl required for 1st and 2nd equivalent points are 10cm3 and 20cm3.
  
  Show a steeper and longer vertical portion for the 1st equivalence point, with slightly acidic pH, eg pH 6.
  
  Show a less steep and shorter vertical portion for the 2nd equivalence point, with a strongly acidic pH. eg pH 2.
  
  Rationale :
  
  Although technically diprotic, but you can deduce that it in practice it is mostly monoprotic (ie. extremely small Kb2), because of 3 (inter-related) reasons :
  
  1) Considering the unipositive conjugate acid, the unipositive formal charged N atom / unipositive BH+ ion, electrostatically repels the H+ / H3O+ cation. Hence kb2 would be extremely small.
  
  2) Considering the unipositive conjugate acid, the unipositive formal charged N atom is electon-withdrawing by induction, making the lone pair on the other N atom (3 C atoms away) to be less available to accept protons. Hence kb2 would be extremely small.
  
  3) Considering the dipositive conjugate acid, 2 positively formal charged N atoms with only 3 C atoms between them is highly destabilizing, due to inter-nuclei repulsions within the highly unstable high cationic charge density dipositive conjugate acid. The more unstable the conjugate acid, the weaker the conjugate base. Hence kb2 would be extremely small.
  
  As such, regard 1,3-diaminopropane as mostly monoprotic, in which case you can (even without Kb values available) expect the 1st equivalence point to have a acidic pH (due to proton dissociation hydrolysis of the acidic ammonium salt).
  
  Which is why we draw a steeper, longer vertical portion for the 1st equivalence point, and a less steep, shorter vertical portion for the 2nd equivalence point (since Kb2 will be extremely small).
  
  As for the strongly acidic pH at the 2nd equivalence point, as explained earlier, Kb2 is extremely small, meaning Ka1 (ie. proton dissociation for the dipositive conjugate acid) is extremely large, meaning that the dipositive conjugate acid is a strong acid. Hence at 2nd equivalence point, the dipositive diammonium salt is strongly acidic and will have a pH almost as acidic as the final pH of 1.6 (ie. with 10cm3 excess HCl). A reasonable estimate would be approx pH 2.
  
  
• hoay

  23 Dec 14, 03:57

  Thank you for the detailed answer.

  Is there any book you would recommend for these type of pian-staking problems?

   

• UltimaOnline

  23 Dec 14, 03:58

  Originally posted by hoay:

  Thank you for the detailed answer.

  
  Totally welcome, feel free to keep your CIE questions coming! :)
• UltimaOnline

  23 Dec 14, 04:06

  Originally posted by hoay:

  Thank you for the detailed answer.

  Is there any book you would recommend for these type of pian-staking problems?

   

  
  If you're referring specifically to acid-base equilibria problems involving polyprotic / multiprotic acids & bases, there's plenty of free materials available on the internet :
  https://www.google.com.sg/#q=polyprotic+acid+base+equilibrium+problems
• hoay

  25 Dec 14, 01:36

  Excellent site. 

  Can u tell a book with respect to A-level.

• UltimaOnline

  25 Dec 14, 02:49

  Originally posted by hoay:

  Excellent site. 

  Can u tell a book with respect to A-level.

  
  You can check out my List of Recommended Books for Singapore A levels H2 Chemistry.
  
  But there's a problem if you stick to books designed just for A levels, because Cambridge will every now and then, include some problems which go slightly beyond A levels. For instance, polyprotic acid-base equilibria is not strictly within the A level syllabus, but as you pointed out, Cambridge asked it in the recent CIE A level exam.
  
  So I recommend going beyond just A level books, and perusing (at least for topics of your personal interest) a couple of Chemistry books designed for University level, such as those I've included on my website.
  
  Overall, the best books targeted for A level Chemistry are the ones I've indicated as "Super Highly Recommended", which are CS Toh's books, Chan Kim Seng & Jeanne Tan's books, George Chong's books and Jasmine Wong's book. Every Singapore H2 Chemistry student should have these books, and use these books over their school's notes.
• hoay

  25 Dec 14, 04:32

  Thank you for the good advice.

  Two campers are desperately short of calor gas, yet they badly need a hot drink.They estimate that they have the equivalent of 1.12 dm³ of calor gas in their “gas bottle “. Assume that the calor gas is pure butane with a standard enthalpy change of combustion of –3000 kJ mol^-1.

             Given that the process is 77 % efficient, calculate the maximum mass of water at a initial temperature of 20^oC which the campers could boil in order to make some hot coffee at room temperature and pressure? Given that the specific heat capacity of water is 4.2 J g^-1 K^-1 .

  Solution:   The mass will be calculated uaing the relation  q = mcT 

  step I :   For 'q' we need to find moles of butane as enthalpy of combustion is given (3000) 

   n = 1.12 / 24 = 0.0466

  enthalpy of combustion must be converted to J as q gives value in J

  q = 3000,000 x 0.0466 = 139800 J

  step II:   But the process is only 77% efficient hence q is:

  139800 x 0.77 = 107646

  step III : finally we use q = mcT

                                   107646 = m x 4.2x (T2-20)

  What to do next.......T2 is not given in the given or I am missing something?

   

• hoay

  25 Dec 14, 04:36

  Is;nt the T2 is 25-27oC as the campers could boil in order to make hot coffee at room temperature and pressure?? 

• UltimaOnline

  25 Dec 14, 05:52

  Your working is correct.
  
  T2 is 100 deg C. Assume the campers are at 1 atm or 1 bar pressure, ie. water boils at 100 deg C.
  
  Btw Hoay, feel free to start a new thread for a new question each time. Name the threads, eg. "[A level Chem] - Energetics" or "[A level Chem] - pH curves", etc.