Question

Toluene (C7H8) 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 pH of the water is 3.0 and this is a completely un-buffered system (i.e. protons that are liberated remain in solution and protons that are consumed are lost 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 Fe2 . Toluene is oxidized to 3mol acetate (C2H3O2 - ) and 1mol CO2. How many mol protons are liberated/consumed in this reaction, and assuming no buffer, will the pH increase or decrease)

Answer 1

In the given reaction "12 mol" of protons are consumed. The further explanation is given below.

Step-by-step explanation:

Let's compose half of C₇H₈ on such controlled oxidation. For balance, firstly C is controlled, then O would be balanced that used H₂O after this H is balanced that used H⁺and after that change is balanced across each side through using e⁻ on the right.

The equation will be:

8H₂O+C₇H₈ ⇒ 3C₂H₃O₂⁻+CO₂+15H⁺+9e⁻ ...(equation 1)

⇒ e⁻+Fe(OH)₃ ⇒ Fe²⁺+3OH⁻

⇒ 3×(H⁺+OH⁻ ⇒ H₂O)

On adding both the above reactions, we get

⇒ e⁻+Fe(OH)₃+3H⁺ ⇒ Fe²⁺+3H₂O ...(equation 2)

Now,

On multiplying "9" into "equation 2" and adding it on "equation 1", we get

⇒ 8H₂O+C₇H₈+9e⁻+9Fe(OH)₃+27H⁺ ⇒ 9Fe²⁺+27H₂O+CO₂+15H⁺+9e⁻

On canceling the common terms from both sides of the reaction, we get

⇒ C₇H₈+9Fe(OH)₃+12H⁺ ⇒ 9Fe²⁺+19H₂O+CO₂ ...(equation 3)

The above "equation 3" seems to be the overall Redox reaction.

Throughout this reaction, 12 mol of a proton (H⁺) is absorbed. Because H+ is absorbed, pH would also increase.