Toluene (C_7H_8) is a component of gasoline and it is contaminating a water body

Question

Toluene (C_7H_8) is a component of gasoline and it is contaminating a water body due to a leaky storage tank. The concentration in water of the toluene is 4.25ppm. The starting of the water is 3.0 and this is a completely system lost protons that are liberated remain in solution and protons that are consumed are from solution during the RXNs taking place). Fe(OH)_3 is present and will serve as an electron acceptor for toluene oxidation; it is reduced to Fe^2+. Toluene is oxidized to 3mol acetate (C_2H_3O_2) and 1mol CO_2. How many mol protons are liberated/consumed in this reaction, and assuming no buffer, will the pH increase or decrease)?

Explanation / Answer

Solution:

Toluene + Water ---> Acetate + CO2

3 mol 1 mol

pH of water = 3.0 gives [H+]=0.001 M

[Toluene] = 4.25 ppm = 4.25*10-6 M

as no-buffering takes place the proton of water will be completely liberated i.e., 0.01 M = moles / volume by considering standard volume we get 0.01 moles hence only minor change will be observed in pH increased forming a little acidic taste.

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