A student performs a titration of 25.0 mL of 0.100 M lactic acid (HC3H5O3), using 0.050 M sodium hydroxide (NaOH). The Ka for lactic acid is 1.4 x 10-4. a) (10 points) What is the pH of the solution after the addition of 23.5 mL of sodium hydroxide solution

The pH of resulting solution is 3.815

Explanation:

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

Molarity of lactic acid solution = 0.100 M

Volume of solution = 25.0 mL

Putting values in above equation, we get:

Molarity of NaOH solution = 0.050 M

Volume of solution = 23.5 mL

Putting values in above equation, we get:

The chemical reaction for NaOH and lactic acid follows the equation:

Initial:        0.0025    0.0012      

Final:         0.0013          -               0.0012          

Volume of solution = 25 + 23.5 = 48.5 mL = 0.0485 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:

= negative logarithm of acid dissociation constant of lactic acid = 3.85

pH = ?

Putting values in above equation, we get:

Hence, the pH of resulting solution is 3.815

it takes on an electronical energy