A buffer contains 0.020 mol of lactic acid (pKa = 3.86) and 0.100 mol sodium lactate per liter of aqueous solution.

a. Calculate the pH of this buffer.

b. Calculate the pH after 8.0 mL of 1.00 M NaOH is added to 1 liter of the buffer (assume the total volume will be 1008 mL).

pH = 4.8

Explanation:

We will use the Henderson-Hasselbach equation to calculate the pH of the buffer:

pH = pKₐ + log [A⁻]/[HA]

From the information given:

pKₐ = 3.86

[A⁻] = 0.100 M

[HA] = 0.020 M

Plugging our values:

pH = 3.86 + log (0.100/0.020) = 4.6

For part b the same equation is utilized.

However we have to realize that the concentrations of the acid and its conjugate base have changed according to the neutralization reaction:

NaOH + lactic acid ⇒ sodium lactate + H₂O

# mol NaOH reacted = (8.0 mL x 1 L / 1000 mL) x 1.00 M

= 8.0 x 10⁻³ mol

mol sodium lactate produced = 8.0 x 10⁻³ mol (1:1)

number of moles mol lactic acid originally = 1 L x 0.020 mol/L = 0.020 mol

new mol lactic acid after reaction = 0.020 - 8.0 x 10⁻³ = 0.012 mol

new mol sodium lactate after reaction = 0.100 mol/L x 1 L + 8.0 x 10⁻³ = 0.108

Here we do not need to calculate the new concentrations since molarity is mol/V, and the volumes cancel each other in the Henderson-Hasselbach equation because they are in a ratio.

Now we are in position to determine the pH.

pH = 3.86 + log (0.108/0.012) = 4.8

This the usefulness of buffers, we are adding a 1.00 M strong base NaOH, and the pH did not change that much (a long as they are small additions within reason)