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

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Correct answer: 0

Given the pH of the resulting solution is

3.0

, the concentration of

H^+

ions is:

[H^+] = 10^{-3} \text{ M}

For the dissociation of acetic acid (

CH_3COOH \rightleftharpoons CH_3COO^- + H^+

), the acid dissociation constant is given by:

K_a = \dfrac{[CH_3COO^-][H^+]}{[CH_3COOH]}

Assuming

[CH_3COO^-] \approx [H^+]

and the equilibrium concentration of acetic acid is

C

, we can use the approximation:

K_a = \dfrac{[H^+]^2}{C}

1.0 \times 10^{-5} = \dfrac{(10^{-3})^2}{C}

C = \dfrac{10^{-6}}{1.0 \times 10^{-5}} = 0.1 \text{ M}

The number of moles of acetic acid remaining in the

50 \text{ mL}

(

0.05 \text{ L}

) solution at equilibrium is:

n_{eq} = C \times V = 0.1 \text{ mol L}^{-1} \times 0.05 \text{ L} = 0.005 \text{ mol}

The mass of acetic acid remaining in the solution is:

W_{eq} = n_{eq} \times \text{Molar mass} = 0.005 \text{ mol} \times 60 \text{ g mol}^{-1} = 0.30 \text{ g}

The initial mass of acetic acid was

0.45 \text{ g}

. The mass of acetic acid adsorbed (

x

) by the charcoal is:

x = 0.45 \text{ g} - 0.30 \text{ g} = 0.15 \text{ g}

The Freundlich adsorption isotherm is given by:

\dfrac{x}{m} = k C^{1/n}

Taking the logarithm on both sides:

\log_{10}\left(\dfrac{x}{m}\right) = \log_{10}k + \dfrac{1}{n} \log_{10}C

The plot of

\log_{10}(x/m)

against

\log_{10}C

is a straight line with a slope of

1

. Therefore:

\dfrac{1}{n} = 1 \Rightarrow n = 1

Substitute the values into the isotherm equation (

m = 1.0 \text{ g}

):

\dfrac{0.15}{1.0} = k (0.1)^1

k = \dfrac{0.15}{0.1} = 1.5 \text{ L mol}^{-1}

Answer:

1.5

At a given temperature, 0.45 g of acetic acid in 50 mL of water is...

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