100 mL of 0.3 M acetic acid is shaken with 0.8 g of wood charcoal. The final concentration of acetic acid in the solution after adsorption is 0.125 M. The mass of acetic acid adsorbed per gram of carbon is _____________ × 10⁻⁴ g.
(Given : Molar mass of acetic acid = 60 g mol⁻¹)

Solution

Chemistry ma’am · 3 steps

The acid that left the solution went onto the charcoal, so find the drop in moles first.

1. Moles adsorbed. The concentration falls from 0.3 M to 0.125 M in 100 mL = 0.1 L. The moles lost from solution are (0.3 - 0.125) × 0.1 = 0.0175 mol, and all of them are now on the charcoal.
2. Convert to grams. Mass of acetic acid adsorbed = 0.0175 × 60 = 1.05 g.

   x/m = (1.05 g)/(0.8 g) = 1.3125 = 13125 × 10⁻⁴

3. So. Each gram of charcoal holds 1.3125 g of acetic acid, which is 13125 × 10⁻⁴ g.

The common slip: 10500. Stopping at the total mass adsorbed and not dividing by the 0.8 g of charcoal gives 1.05 g, that is 10500, not 13125.

The idea to keep. In adsorption problems, the amount adsorbed is the amount missing from the solution: (initial minus final concentration) times volume. Then divide by the adsorbent's mass to get x/m.

Answer from the NTA final answer key. Solution by PrepWiser.