What is the pH of the solution in a titration of 50.0 mLs of 0.70 M pyridine wit

Question

What is the pH of the solution in a titration of 50.0 mLs of 0.70 M pyridine with 0.35 M HCl after 25.0 mLs of titrant have been added? please show steps and explain!

Explanation / Answer

let pyridine be POH , POH -----> P+ + OH- , Kb = 10^-8.75 =1.778 x10^ -9 , [POH] = 0.7-x , [OH-] =P+] = x, Kb = [P+][OH-]/[POH] , 1.778 x10^ -9 = x^2/(0.7-x) , x = 3.52 x10^ -5 [OH-] , OH- moles = 3.52 x10^ -5 x (50)/1000 = 1.76 x10^ -6 moles , moles of H+ = 0.35 x25/1000 = 0.00875 , net H+ after neutralisation = 0.00875-1.76 x10^ -6 = 0.00874824, vol = 25+50 = 75 ml , cocn of H+ = ( 0.0087424 x 1000/75) = 0.1166 , pH = -log(0.1166) = 0.933

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