Question

Methanol, a common laboratory solvent, poses a threat of blindness or death if consumed in sufficient amounts. Once in the body, the substance is oxidized to produce formaldehyde (embalming fluid) and eventually formic acid. Both of these substances are also toxic in varying levels. The equilibrium between methanol and formaldehyde can be described as follows: $$ \mathrm{CH}_{3} \mathrm{OH}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}(a q)+\mathrm{H}_{2}(a q) $$ Assuming the value of \(K\) for this reaction is \(3.7 \times 10^{-10}\), what are the equilibrium concentrations of each species if you start with a \(1.24 M\) solution of methanol? What will happen to the concentration of methanol as the formaldehyde is further converted to formic acid?

Short answer

The equilibrium concentrations for the reaction are approximately: $[CH_3OH]_{eq} \approx 1.24\:M$, $[H_2CO]_{eq} \approx 2.15 \times 10^{-5}\:M$, and $[H_2]_{eq} \approx 2.15 \times 10^{-5}\:M$. As formaldehyde is further converted into formic acid, the equilibrium shifts to produce more formaldehyde and hydrogen gas, resulting in a gradual decrease in methanol concentration and a corresponding increase in concentrations of the other products.

Step-by-step solution

Write the equilibrium expression

The equilibrium constant (K) can be expressed as the product of the concentrations of the products, divided by the product of the concentrations of the reactants. In this reaction: \(K = \frac{[H_2CO]_{eq}[H_2]_{eq}}{[CH_3OH]_{eq}}\)

Set up an ICE table

We will set up an ICE table to keep track of the initial concentrations, changes in concentrations, and final equilibrium concentrations of all species involved in the reaction. Initial concentrations: \[CH_3OH \: \: 1.24\:M\] \[H_2CO \: \: 0\: M\] \[H_2 \: \: \: \: \:0\: M\] Changes in concentrations: Since we do not know the change in the concentrations, we use variables to represent them. \[CH_3OH \: \: -x\:M\] \[H_2CO \: \: +x\:M\] \[H_2 \: \: \: \: \: +x\:M\] Equilibrium concentrations: \[CH_3OH \: \: (1.24-x)\:M\] \[H_2CO \: \: x\:M\] \[H_2 \: \: \: \: \: x\:M\]

Solve for equilibrium concentrations using the equilibrium expression

Now we substitute the equilibrium concentrations in terms of x into the equilibrium expression: \(3.7 \times 10^{-10} = \frac{x^2}{1.24-x}\) To solve for x, we can assume that the change in the concentration of methanol (x) is negligible compared to its initial concentration (1.24 M), because the K value is very small. Therefore, we can simplify the equation: \(3.7 \times 10^{-10} \approx \frac{x^2}{1.24}\) Now, we can solve for x: \(x \approx \sqrt{(3.7 \times 10^{-10})(1.24)}\) \(x \approx 2.15 \times 10^{-5}\:M\) Now we can find the equilibrium concentrations of all species: \[CH_3OH \: \: 1.24 M - 2.15 \times 10^{-5} M \approx 1.24\:M\] \[H_2CO \: \: 2.15 \times 10^{-5}\:M \] \[H_2 \: \: \: \: \: 2.15 \times 10^{-5}\:M \]

Discuss the further conversion of formaldehyde to formic acid

The concentration of methanol remains approximately 1.24 M at equilibrium, while there is only a slight increase in the concentrations of formaldehyde and hydrogen gas. If formaldehyde is further converted into formic acid, the concentration of formaldehyde will decrease. This will cause a shift in the equilibrium to counteract the change, producing more formaldehyde and hydrogen gas from methanol. As a result, the concentration of methanol will decrease gradually, and the concentrations of the other products will increase until a new equilibrium is established.