Question

What is the pH of a solution that is prepared by adding 38.34 mL of 1.83 M HCl to 0.772 L of 0.315 M trimethylamine (a weak base)?

Answer 1

$H C l ( a q ) + B ( a q ) → C l ^− ( a q ) + H B ^+ ( a q )$

Molar mass of HCl =36.458g/mol

1.8×36.458= 65.62g

$\frac{38.34 m L H C l}{65.2} = 0.59mol$

Molar mass of trimethylamine = 59.11

0.315×59.11= 18.62

$782/$ 18.62= 42moles

K_a=5.9×10^-6

-log 5.9×10^-6=5.3

Log42/0.59=1.1

pH= 6.4