Question

The pH of a buffer prepared by mixing 0.05 mol of acetic acid (pKa=3.86) with 0.03 mol of sodium lactate per liter of aqueous solution is

Answer 1

Answer 2

To determine the pH of the buffer solution prepared by mixing acetic acid and sodium lactate, we need to consider the dissociation of acetic acid and the formation of the acetate ion.

The equilibrium equation for the dissociation of acetic acid (CH₃COOH) can be written as follows:

CH₃COOH ⇌ CH₃COO⁻ + H⁺

Given that the pKa of acetic acid is 3.86, we can use the Henderson-Hasselbalch equation to calculate the pH of the buffer:

pH = pKa + log ([A⁻] / [HA])

Where: pH = The pH of the buffer solution pKa = The dissociation constant of the acid (acetic acid) [A⁻] = The concentration of the conjugate base (acetate ion) [HA] = The concentration of the acid (acetic acid)

In the given problem, we have: [A⁻] = 0.03 mol/L (concentration of sodium lactate) [HA] = 0.05 mol/L (concentration of acetic acid)

Now, let's substitute these values into the Henderson-Hasselbalch equation:

pH = 3.86 + log (0.03 / 0.05)

Calculating the ratio: (0.03 / 0.05) = 0.6

Taking the logarithm of the ratio: log (0.6) ≈ -0.2218

Now, substituting the value into the equation: pH ≈ 3.86 - 0.2218

Calculating the subtraction: pH ≈ 3.6382

Therefore, the pH of the buffer solution prepared by mixing 0.05 mol of acetic acid with 0.03 mol of sodium lactate per liter of aqueous solution is approximately 3.6382.