Question

Calculate the pH at the equivalence point for the following titration: 0.30 M HCL vs. 0.30 M (CH3)2NH. Ka is 1.95 x 10^-11 of (CH3)2NH.

Answer 1

HCl is a strong acid, which reacts with (CH₃)₂NH as follow;

HCl + (CH₃)₂NH ---> Cl⁻ + (CH₃)₂NH₂⁺

At equivalence point; no. of moles of HCl = No. of moles of (CH₃)₂NH

Lets assume volume of HCl = volume of (CH₃)₂NH = a L
Therefore, total volume = 2a L

no. of moles of HCl = 0.3 mol/L x a L = 0.3a mol
no. of moles of (CH₃)₂NH = 0.3 mol/L x a L = 0.3a mol

Ka = 1.95 x 10^-11 ((CH₃)₂NH)

At equivalence point, all the (CH₃)₂NH reacted with HCl to form (CH₃)₂NH₂⁺ ion.
Therefore, [(CH₃)₂NH₂⁺] = 0.3a mol/2a L = 0.15 M
(CH₃)₂NH₂⁺ ion will again in equilibrium with H₂O as follows;

                 (CH₃)₂NH₂⁺ + H₂O ---> H₃O⁺ + (CH₃)₂NH
Initial          0.15M       -          0       0
change       -x       -                 +x       +x
equilibrium      0.15-x   -          x       x

Ka = [H₃O⁺][(CH₃)₂NH]/[(CH₃)₂NH₂⁺]

1.95 x 10-11 = x.x/(0.15 -x)
x² =    1.95 x 10^-11 (0.15-x)
x² = 1.95 x 10^-11 x 0.15 (x is very small compared to 0.15; 0.05 -x ~ 0.15)
x² = 2.925 x 10^-12
x = 1.71 x 10^-6
Therefore, [H3O+] = 1.71 x 10^-6
pH = -log[H₃O⁺]
   = -log(1.71 x 10^-6 )
   = 5.77

Hence, the pH of the solution is: 5.77