An analytical chemist is titrating 220.3 ml of 0.36 m solution of ethylamine with a 0.26 m hno3. calculate the ph of the base solution after the chemist has added 125 ml of the hno3 solution to it.

The pH of resulting solution is 11

Explanation:

To calculate the number of moles for given molarity, we use the equation:

Molarity of ethylamine solution = 0.36 M

Volume of solution = 220.3 mL

Putting values in above equation, we get:

Molarity of nitric acid solution = 0.26 M

Volume of solution = 125 mL

Putting values in above equation, we get:

The chemical reaction for ethylamine and nitric acid follows the equation:

Initial:          0.079          0.031

Final:             0.048          -                     0.031            

Volume of solution = 220.3 + 125 = 345.3 mL = 0.3453 L    (Conversion factor:  1 L = 1000 mL)

We are given:

= negative logarithm of base dissociation constant of ethylamine = 3.19

pOH = ?

Putting values in above equation, we get:

To calculate pH of the solution, we use the equation:

Hence, the pH of the solution is 11