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 down the given information

We are given the following information: - The initial concentration of methanol: \([CH_3OH]_0 = 1.24 M\) - The value of the equilibrium constant \(K = 3.7 \times 10^{-10}\) - The chemical reaction: \(\mathrm{CH_{3} OH(aq) \rightleftharpoons H_{2} CO(aq) + H_{2}(aq)}\)

02

Set up ICE table

We will use the ICE (Initial, Change, Equilibrium) table to represent the information and keep track of the changes in concentrations at equilibrium. The table should look like this: | | CH3OH | H2CO | H2 | |---------|-------|------|-----| | Initial | 1.24 | 0 | 0 | | Change | -x | +x | +x | | Equilib.| 1.24-x| x | x | In the table, "x" represents the amount of CH3OH that reacts to produce H2CO and H2. At equilibrium, the concentrations are as follows: [CH3OH] = 1.24 - x, [H2CO] = x, and [H2] = x.

03

Set up equilibrium expression

According to the chemical reaction, the equilibrium constant K can be expressed as follows: \(K = \frac{[H_2CO][H_2]}{[CH_3OH]}\) Plugging the equilibrium concentrations from the ICE table, we get: \(3.7 \times 10^{-10} = \frac{x \cdot x}{1.24-x}\)

04

Solve for x

To find the value of x, we can solve the equation obtained from step 3. Given that K is very small, we can assume that x << 1.24, so the equation can be simplified to: \(3.7 \times 10^{-10} = \frac{x^2}{1.24}\) Now solve for x: \(x^2 = 3.7 \times 10^{-10} \times 1.24\) \(x = \sqrt{3.7 \times 10^{-10} \times 1.24} \approx 2.28 \times 10^{-5}\)

05

Calculate equilibrium concentrations

We can now plug x back into the equilibrium concentrations obtained from the ICE table to get: - \([CH_3OH] = 1.24 - x = 1.24 - 2.28 \times 10^{-5} \approx 1.24 M\) - \([H_2CO] = x = 2.28 \times 10^{-5} M\) - \([H_2] = x = 2.28 \times 10^{-5} M\)

06

Discuss the effect on methanol concentration as formaldehyde is converted into formic acid

As formaldehyde is further converted to formic acid, the concentration of formaldehyde will decrease. This leads to a shift in the equilibrium towards the products according to Le Chatelier's Principle, in an attempt to increase the concentration of formaldehyde. Consequently, more methanol will be converted into formaldehyde and hydrogen, which will decrease the concentration of methanol in the solution.

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