Question

.1M HCL is used to titrate 50ml of .115M of a weak base.(C₁₀H₁₄N₂)

Find Initial pH of the base, pH after 20ml HCL,30ml HCl, 60ml HCL,pH at equivalence point. If you can graph this too it would be great.

Answer 1

Kb of C10H14N2 = 1.0 x 10-6

pKb C10H14N2 == -log(1*10^-6) = 6

initial pH = 14 - 1/2(pkb-logC)

            = 14 - 1/2(6-log0.115)

            = 10.53

no of mol of HCl added = 0.1*20 = 2 mmol

no of mol of nicotine = 0.115*50 = 5.75 mmol

pH = 14 - (pkb+log(acid/base))

   = 14 - (6+log(2/(5.75-2))

= 8.27

no of mol of HCl added = 0.1*30 = 3 mmol

no of mol of nicotine = 0.115*50 = 5.75 mmol

pH = 14 - (pkb+log(acid/base))

   = 14 - (6+log(3/(5.75-3))

= 7.96

no of mol of HCl added = 0.1*60 = 6 mmol

no of mol of nicotine = 0.115*50 = 5.75 mmol

concentration of excess HCl= (6-5.75)/(60+50) = 0.002273 M

PH = -log(H+)

    = -log(0.002273)

   = 2.643