Question

A 25.0 mL sample of a 0.290 M solution of aqueous trimethylamine is titrated with a 0.363 M solution of HCl. Calculate the pH of the solution after 10.0, 20.0, and 30.0 mL of acid have been added; pKb of (CH3)3N=4.19 at 25 degrees C

Answer 1

No of mol of trimethylamine = 25/1000*0.29 = 0.00725 mol

No of mol of HCl added = 10/1000*0.363 = 0.00363 mol

pH = 14 - PKB+LOG(salt/base)

    = 14-(4.19+log(0.00363/(0.00725-0.00363)))

    = 9.8

2. No of mol of trimethylamine = 25/1000*0.29 = 0.00725 mol

No of mol of HCl added = 20/1000*0.363 = 0.00726 mol

No of mol of HCl = No of mol of trimethylamine

concentration of salt = 0.00726/(20+25)*1000 = 0.161 M

pH = 7-1/2(pkb+logC)

   = 7-1/2(4.19+log0.161)

    = 5.3

3. No of mol of HCl excess added = (30-20)/1000*0.363 = 0.00363 mol

concentration of HCl excess added = 0.00363/(55)*1000 = 0.066 M

Ph = -LOG(H+)

     = -log(0.066)

     = 1.18