Question

Consider the titration of 50.0 ml of 0.100 M methylamine (Kb = 5.0 × 10- 4)...

Consider the titration of 50.0 ml of 0.100 M methylamine (Kb = 5.0 × 10- 4) with 0.200 M HCl. What is the pH of the solution when 25.00 ml of HCL has been added?

Homework Answers

Answer #1

In this titration we are using 50 ml of methylamine (CH3NH2) with 0.200 M HCl

Moles of H+ = 0.025 L of HCl X 0.200 mol/L= 0.005 moles of H+

Moles of methylamine= 0.05 L of methylamine X 0.100 M = 0.005 moles of methylamine

pH of solution

CH3NH2 + H+ ----------------> CH3NH+3

Initial               0.005   0.005              0

Final              0         0.005                  0.005

Ignoring any hydronium ion are producing in the solution by the ionization of methyamine

H+ = 0.005 / 0.075 (total amount of solution) = 0.067 M

pH = - log 0.067 = 1.17

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