What is the pH of a 0.127 M solution of (CH3)2NH when it is titrated with 0.274 M of HCl to the half-equivalence point?

Question

Note: (CH₃)₂NH  has a K_b = 5.42 x 10^-4 .

Answer 1

What is the pH of a 0.127 M solution of (CH3)2NH when it is titrated with 0.274 M of HCl to the half-equivalence point?

Note: (CH3)2NH  has a Kb = 5.42 x 10-4 .

Hello Elva,

1) Balanced neutralization reaction:

(CH₃)₂NH + HCl ---> (CH₃)₂NH₂⁺ + Cl^-

2) A very important concept here is that the acid HCl, was used to titrate the base dimethylamine (CH₃)₂NH, only to the half equivalence point. Knowing, that at the half equivalence point, pH = pKa of the weak acid. This tells us that the dimethylammonium cation (CH₃)₂NH₂⁺ acts as a weak acid, after being protonated. Thus, the pH can be determined from the Kb given for (CH₃)₂NH, by the fact that Kw = Ka * Kb

Kw = 1 x 10^-14

1 x 10^-14 = Ka * 5.42 x 10^-4 .....*Solve for Ka

1 x 10^-14 / 5.42 x 10^-4 = Ka

Ka = 1.84 x 10^-11

Use -log[Ka] to get pKa:

pKa = -log[1.84 x 10^-11]

pKa = 10.73

Since pH = pKa

The pH of the solution is 10.7 at the half equivalence point.