Answer:
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 )
