1.50 g of a weak acid (molar mass 176) is dissolved in 50.0 ml of water, and the resultant solution is titrated with 0.300 m naoh. when 12.5 ml of 0.300 m naoh is added, the ph of the resultant solution is 4.00. calculate ka for this acid.

The dissociation constant of this acid is .

Explanation:

Moles of acid =

Moles of sodium hydroxide:

Moles of sodium hydroxide:

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

Initial:        0.008523    0.00375                0

Final:       (0.008523-0.00375)   0              0.00375

Volume of solution = 50.0 mL + 12.5 mL = 62.5 mL = 0.0625 L    (Conversion factor:  1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

We are given:

= ?

pH = 4.00

Putting values in above equation, we get:

The dissociation constant of this acid is .

ill try to . no promises tho.

explanation: