Question

Suppose you titrate 0.310 L of a 0.350 M solution of sodium nicotinate with 6.0 M HCl. K_a for nicotinic acid is 1.5*10^-5. a) What is the pH of the solution before beginning the titration? b) What is the pH of the solution halfway through the titration? c) What is the pH at the stoichiometric point?

Answer 1

before titration,

   pH = 7+1/2(pka+logC)

pka of nicotinic acid = -logka = -log(1.5*10^-5) = 4.824

   pH = 7+1/2(4.824+log0.35) = 9.184

at half equivalencepoint ,

   no of mole of nicotinic acid = sodium nicotinate

so that ,   pH = pka = 4.824

pH at the stoichiometric point

   total salt converts into nicotinic acid

   volume of HCl added = 0.31*0.35/6 = 0.0181 l

pH = 1/2(pka-logC)

c= concentration of acid = n/v = 0.31*0.35/(0.31+0.0181) = 0.33 M

   pH = 1/2(4.824-log0.33) = 2.653

pH chnages from 9.184 to 2.653.so that suitable indicator = Cresol Purple