Question

Find the volume (mL) and the pH of 0.135 M HCl needed to reach the equivalence point(s) in titrations of the following. (You need to find the pH at the equivalence point, not the initial pH of the solution.

(a) 50.8 mL of 0.272 M NH3

(b) 11.8 mL of 1.76 M CH3NH2

Answer 1

Titrations

(a) moles of NH3 = 0.272 M x 50.8 ml

                            = 13.8176 mmol

volume of acid required to reach equivalence point = 13.8176 mmol/0.135 M

                                                                                 = 102.35 ml

[NH4+] formed = 13.8176 mmol/153.15 ml = 0.09 M

NH4+ + H2O <==> NH3 + H3O+

let x amount has hydrolysed

Ka = 1 x 10^-14/1.8 x 10^-5 = x^2/0.09

x = [H3O+] =7.08 x 10^-6 M

pH = -log[H3O+] = 5.15

(b) moles of CH3NH2 = 1.76 M x 11.8 ml

                                    = 20.768 mmol

volume of acid required to reach equivalence point = 20.768 mmol/0.135 M

                                                                                 = 153.84 ml

[CH3NH3+] formed = 20.768 mmol/165.64 ml = 0.125 M

CH3NH3+ + H2O <==> CH3NH2 + H3O+

et x amount has hydrolyzed

Ka = 1 x 10^-14/4.4 x 10^-4 = x^2/0.125

x = [H3O+] = 1.70 x 10^-6 M

pH = -log[H3O+] = 5.77