Question

9. Explain as much as you can! I alreadyknow the answer, but need to know why! Use written words as well.

sp (9) Calculate the pH of a buffer solution that contains 0.820 grams of sodium acetate and 0.010 moles of acetic acid in 100 ml of water. The Ka of acetic acid is 1.77x10

Answer 1

Acidic buffer is a solution of a weak acid and its salt. The Henderson-Hasselbalch approximation can be used to calculate the pH of a buffer solution. The balanced equation for an acidic buffer is:

HA⇌H⁺+A⁻

The strength of a weak acid is usually represented as an equilibrium constant. The acid-dissociation equilibrium constant

K_a = [H+][A-]/[HA]

or, -logK_a = -log[H+] -log[A-]/[HA]

or, pK_a = pH - log[A-]/[HA]

or, pH = pK_a + log [A-]/[HA]

HA is a weak acid. By definition, HA does not dissociate completely and we can write,

[HA] = [HA]_initial

[A-] = [A-]_initial

So, pH = pK_a + log [A-]/[HA] = pH = pK_a + log [A-]_initial/[HA]_initial

= pK_a + log[salt]/[acid]

This equation is known as Henderson-Hasselbalch equation.

According to the queation,

pK_a of acidic acid = -logK_a = -log (1.77*10^-5) = 4.75

We need to calculate the concentrations of the salt(sodium acetate) and the acid first.

Molar mass of sodium acetate = 82 g/mol

So, 0.82 g sodium acetate = 0.82/82 mole = 0.01 mol

100 mL solution contains 0.01 moles of sodium acetate. So [sodium acetate]= 0.01*1000/100 = 0.1 M

[acid] = 0.010 * 1000/100 = 0.1 M

So, pH = pK_a + log[salt]/[acid]

= 4.75 + log 0.1/0.1

= 4.75