Question

Be sure to answer all parts. A volume of 25.0 mL of 0.150 M HCl is titrated against a 0.150 M CH3NH2 solution added to it from a buret.

(a) Calculate the pH value of the solution after 10.0 mL of CH3NH2 solution has been added.

(b) Calculate the pH value of the solution after 25.0 mL of CH3NH2 solution have been added.

(c) Calculate the pH value of the solution after 35.0 mL of CH3NH2 solution have been added.

Answer 1

a) No of moles of HCl taken = 25*0.15 = 3.75 mmol

   No of moles of CH3NH2 = 10*0.15 = 1.5 mmol

   concentration of excess HCl = (3.75-1.5)/(25+10)

                               = 0.064 M

pH = -log[H3O+]

     = -log0.064

     = 1.194

b) No of moles of CH3NH2 = 25*0.15 = 3.75 mmol

No of moles of HCl taken = 25*0.15 = 3.75 mmol

concentration of CH3NH3cl formed = 3.75/50 = 0.075 M

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

   = 7-1/2(3.38+log0.075)

   = 5.87

c)

pOH = pkb+log(CH3NH3Cl/CH3NH2)

pkb of CH3NH2 = 3.38

No of moles of CH3NH2 = 35*0.15 = 5.25 mmol

No of moles of HCl taken = CH3NH3Cl formed = 25*0.15 = 3.75 mmol

pOH = 3.38+log(3.75/(5.25-3.75))

    = 3.78

pH = 14-3.78 = 10.22