Question

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

.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.

Homework Answers

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

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