Question

Calculate the pH during the titration of 20.00 mL of 0.1000 M trimethylamine, (CH3)3N(aq), with 0.1000 M HCl(aq) after 21 mL of the acid have been added. Kb of trimethylamine = 6.5 x 10-5.

Answer 1

20.00 mL of 0.1000 M trimethylamine, (CH3)3N(aq) = volume of solution in L * molarity = 0.020 L * 0.100 mole / L = 0.002 mole.

and

21 mL of 0.100 M HCl = 0.021 L * 0.100 mole / L = 0.0021 mole.

thus

excess mole of HCl = (0.0021 - 0.002) = 0.0001 mole.

total volume of solution = (20 + 21) = 41 ml = 0.041 L

thus

molarity of excess HCl = 0.0001 mole / 0.041 L = 2.439 * 10^-3 M

HCl is strong acid.

so

[HCl] = [H+] = 2.439 * 10^-3 M

PH = - log [H+] = - log (2.439 * 10^-3) = 2.61

therefore,

PH = 2.61