Calculate the pH during the titration of 20.00 mL of 0.1000 M

Question

Calculate the pH during the titration of 20.00 mL of 0.1000 M dimethylamine, (CH₃)₂NH(aq), with 0.1000 M HCl(aq) after 21 mL of the acid have been added. K_b of dimethylamine = 5.4 x 10^-4.

The answer is 2.61

Answer 1

(CH3)₂NH + HCl ===> (CH3)₂NH₂⁺ + Cl-

0.002 mol.....0.0021 mol.........0.....................initial

-0.002............-0.002.............+0.002.............change

0....................0.0001............0.002...............equilibrium

Volume = 20 ml + 21 ml = 41 ml = 0.041 L

(CH3)₂NH₂⁺ ==> (CH3)₂NH + H⁺ Ka = 1x10^-14/Kb = 1x10^-14/5.4x10^-4 = 1.85x10^-11

Final [(CH3)₂NH₂⁺] = 0.002 mol/0.041 L = 0.049 M

Ka = [(CH3)₂NH][H⁺]/[(CH3)₂NH₂⁺]

1.85x10^-11 = (x)(x)/0.049

x² = 9.1x10^-13

x = [H⁺] = 9.5x10^-7 M

[H⁺] from HCl = 0.0001 mol/0.041 L = 0.0024 M

Since the [H⁺] from the excess HCl is so much greater than that contributed by the (CH3)₂NH₂⁺, the pH will be the negative log of the [H⁺] from the HCl

pH = -log 0.0024

pH = 2.61