Answer:
A simple representation of the reaction is below:
Reaction, HC₃H₅O₂ ------ H⁺ + C₃H₅O²⁻
Ka=1.4x10-4 for the 0.95M acid and 0.50M for the lactate
The initial pH can be calculated using the Henderson-Hasselbach equation
pH = pKa + log(base/acid) Eqn 1
pH = -Log(1.4x10-4) + Log (0.5/0.95)
==== 3.854 - 0.279
pH = 3.57
However, we need to know the pH after 15.00 mL of 0.75 M HCL is added to the buffer?
we have the 1st reaction as,
HC₃H₅O₂ ------ H⁺ + C₃H₅O²⁻, we were given 1L of the initial solution and 0.01125 mol HCl (0.015L x 0.75mol/L), we will then analyse the reaction as follows.
Before 0.95 mol +0.01125 0.50 mol
Reactions +0.01125mol -0.01125 -0.01125
After 0.961 mol 0 0.48875 mol
We can then use Eqn 1 to find the pH
pH= -Log (1.4x10-4) + Log (0.48875/0.961)
pH = 3.854 - 0.2937
New pH=3.56