At a given temperature, 0.45 g of acetic acid in 50 mL of water is shaken with 1.0 g of charcoal and the pH of the resulting solution is 3.0. Assume, the adsorption of acetic acid from the aqueous solution by charcoal follows Freundlich isotherm, xm=kC1/n If the plot of 10(x/m) against 10C gives a straight line with slope 1, the value of k in L mol-1 is . Given: The molar mass of acetic acid is 60 g mol-1. The acid dissociation constant of acetic acid is 1.0 10-5 at the given temperature. x is the mass (in grams) of acetic acid adsorbed. m is the mass (in grams) of charcoal. C is the equilibrium concentration of acetic acid in the solution after the adsorption is complete. k and n are constants for acetic acid–charcoal system at the given temperature.
Answer
1.45 to 1.52
Solution
Slope 1n=1, so xm=kC.
After adsorption: [H+]= KaC 10-6=10-5C C=0.1 mol L-1
Initial: 0.4560=7.5 10-3 mol in 0.05 L, i.e. 0.15 mol L-1
Adsorbed: (0.15-0.1) 0.05=2.5 10-3 mol =0.15 g, so xm= 0.151.0=0.15
k= 0.150.1=1.5
Answer: 1.5