Question

A buffer made with 100.00 mL of 0.95 M lactic acid (Ka=1.4x10-4) and 200.00 mL of 0.50 M lactate has a final volume of 1.0 L. What is the pH after 15.00 mL of 0.75 M HCL is added to the buffer?

Answer 1

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

Answer 2

3.77

Step-by-step explanation:

The pH of a buffer can be calculated by Handerson-Halsebach equation:

pH = pKa+ log [A⁻]/[HA]

Where pKa = -logKa, [A⁻] is the concentration of the conjugate base, and [HA] is the concentration of the acid.

In this case [A⁻] = concentration of lactate. Because the final volume will be the same, and [X] = mol/L, we can use the number of moles instead of the concentration.

nHA = 0.1 L * 0.95 mol/L = 0.095 mol

nA⁻ = 0.2 L * 0.5 mol/L = 0.1 mol

When HCl is added, it will dissociate in H⁺ and Cl⁻. H⁺ will react if A⁻ to form more HA, so the equilibrium will be shift. Because of that, the number of moles of HA will be the initial plus the number of moles of H⁺ added (which is equal to the number of moles of HCl), and the number of moles of A⁻ will be the initial less the number of moles of H⁺.

nH⁺ = nHCl = 0.015 L* 0.75 = 0.01125 mol

nA⁻ = 0.1 - 0.01125 = 0.08875 mol

nHA = 0.095 + 0.01125 = 0.10625 mol

pKa = -log(1.4*10⁻⁴) = 3.85

pH = 3.85 + log(0.08875/0.10625)

pH = 3.77