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?

Step by step solution

01

Write the reaction and initial concentrations

First, let's write down the given reaction and the initial concentration of methanol: $$ \mathrm{CH}_{3}\mathrm{OH}_{(aq)} \rightleftharpoons \mathrm{H}_{2}\mathrm{CO}_{(aq)} + \mathrm{H}_{2}_{(aq)}. $$ Initial concentrations: Methanol (\(\mathrm{CH}_{3}\mathrm{OH}\)) = \(1.24 \,\text{M}\), Formaldehyde (\(\mathrm{H}_{2}\mathrm{CO}\)) = \(0 \,\text{M}\), Hydrogen (\(\mathrm{H}_{2}\)) = \(0 \,\text{M}\).

02

Set up an ICE table

Now, let's construct an ICE (Initial, Change, Equilibrium) table to represent the changes in concentration as the reaction proceeds. | | CH₃OH (M) | H₂CO (M) | H₂ (M) | |-------------------|-----------|----------|--------| | Initial (I) | 1.24 | 0 | 0 | | Change (C) |-x |+x |+x | | Equilibrium (E) |1.24 - x | x | x |

03

Write the expression for the equilibrium constant

The equilibrium constant \(K\) is given by the expression: $$ K = \frac{[\mathrm{H}_{2}\mathrm{CO}][\mathrm{H}_{2}]}{[\mathrm{CH}_{3}\mathrm{OH]}}. $$ Substitute the equilibrium concentrations from the ICE table: $$ K = \frac{x^{2}}{1.24 - x}. $$ We are given that \(K\) is equal to \(3.7 \times 10^{-10}\).

04

Solve the equation for x

Now, let's solve the equation for \(x\): $$ 3.7 \times 10^{-10} = \frac{x^{2}}{1.24 - x}. $$ To solve this quadratic equation, you can either use the quadratic formula or factor the equation and solve for x. Alternatively, you can also use numerical methods or software to find the solution. On solving, we will obtain the value for \(x = 2.71 \times 10^{-5}\).

05

Calculate the equilibrium concentrations

Now that we have the value for \(x\), we can find the equilibrium concentrations: Equilibrium concentration of methanol: $$ [\mathrm{CH}_{3}\mathrm{OH}] = 1.24 - x = 1.24 - 2.71 \times 10^{-5} \approx 1.24\,\text{M} $$ Equilibrium concentration of formaldehyde: $$ [\mathrm{H}_{2}\mathrm{CO}] = x = 2.71 \times 10^{-5}\,\text{M} $$ Equilibrium concentration of hydrogen: $$ [\mathrm{H}_{2}] = x = 2.71 \times 10^{-5}\,\text{M} $$

06

Discuss the outcome when formaldehyde is further converted

As the formaldehyde is further converted to formic acid, its concentration decreases. Since the reaction between methanol and formaldehyde is in equilibrium, a decrease in formaldehyde concentration will shift the equilibrium to the right according to Le Chatelier's principle. This will lead to an increase in the concentration of hydrogen gas and a decrease in the concentration of methanol. In conclusion, the equilibrium concentrations of the species are approximately: Methanol: \(1.24\,\text{M}\), Formaldehyde: \(2.71 \times 10^{-5}\,\text{M}\), Hydrogen: \(2.71 \times 10^{-5}\,\text{M}\). As the formaldehyde is further converted to formic acid, the concentration of methanol will decrease.

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