Question

A 25.0 Ml sample of .280 M aqueous trimenthylamine is titrated .350 M HCl. Calculate the pH of the mixture after 10.0, 20.0, and 30.0 mL of acid have been added pKaof (CH3)3N = 4.19 at 25C

Answer 1

No of mol of trimethylamine = 25/1000*0.28 = 0.007 mol

No of mol of HCl added = 10/1000*0.35 = 0.0035 mol

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

    = 14-(4.19+log(0.0035/(0.007-0.0035)))

    = 9.8

2. No of mol of trimethylamine = 25/1000*0.28 = 0.007 mol

No of mol of HCl added = 20/1000*0.35 = 0.007 mol

No of mol of HCl = No of mol of trimethylamine

concentration of salt = 0.007/(20+25)*1000 = 0.161 M

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

   = 7-1/2(4.19+log0.161)

    = 4.95

3. No of mol of HCl excess added = (30-20)/1000*0.35 = 0.0035 mol

concentration of HCl excess added = 0.0035/(55)*1000 = 0.0636 M

Ph = -LOG(H+)

     = -log(0.0636)

     = 1.2