Question

A 22.2 mL sample of 0.367 M dimethylamine, (CH3)2NH, is titrated with 0.277 M hydrobromic acid.

After adding 10.9 mL of hydrobromic acid, the pH is ______

Answer 1

The reaction between (CH3)2NH and HBr is given by

(CH3)2NH + HBr ---> (CH3)2NH2+ + Br-

No of moles of HBr (aq) = 0.277x 10.9/1000 = 3.019 x 10^-3 mol
No of moles of (CH3)2NH = 0.367 x 22.2/1000 = 8.147*10^-3 mole
No of moles of (CH3)2NH after adding HBr = 0.00512 mole
No of moles of (CH3)2NH+ aq = 3.019 x 10^-3 mol
pKb = -log(CH3)2NH = 5.4 x 10-4 = 3.27
so pH = 14 - 3.27 = 10.73
log 8.147*10^-3 - log 3.019 x 10^-3 = 0.431
10.73 + 0.431 = 11.16

The pH of solution is 11.16