(i) [HCl] = [CH3N2], since :
1) 25cm3 of base required 25cm3 of acid for equivalence
point.
2) HCl is monoprotic and CH3N2 is
monobasic.
(ii) Kb = ( [OH-] x [conjugate acid] ) / [conjugate base]
Since at initial, [OH-] = [conjugate acid], you get
Kb = [OH-]^2 / [conjugate base]
Hence you can find [OH-], hence pOH, hence pH.
(iii) pOH = pKb for maximum buffer capacity.
Notice that the question was much too generous in reminding you
it was “a solution of maximum bufferring capacity”. It was actually
sufficient to simply ask, “Calculate the pH of the solution
obtained when 12.5cm3 of HCl was added during the titration”.
Comments :
This question is too easy for ‘A’ levels exam standard (and your
Promos, and your Prelims, etc). Expect much tougher questions for
the actual ‘A’ levels (especially considering the recent trend that
papers are getting tougher every year – the 2008 exam papers for
Chemistry was much tougher than the 2007 exam papers. Brace
yourself for 2009 and 2010).
