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:

$C{H}_{3}OH\left(aq\right)\iff H_{2}CO\left(aq\right)+H_2\left(aq\right)$

Assuming the value of K for this reaction is 3.7×10^-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 concentration of methanol, formaldehyde, and hydrogen are $\text{1.24} \text{M}$,$2.1\times {10}^{-5}M$and $2.1\times {10}^{-5}M$, respectively.

Step-by-step solution

Equilibrium constant expression and ICE table

The equilibrium constant, K expression in terms of concentrations for the given reaction, is shown below:

$\text{K =}\frac{\left[{\text{H}}_{\text{2}}\text{CO}\right]\left[{\text{H}}_{\text{2}}\right]}{\left[{\text{CH}}_{\text{3}}\text{OH}\right]}$..........(1)

The given equilibrium constant K is $3.7\times {10}^{-7}$.

The initial concentration of methanol is given as 1.24 M.

Let x be the concentration change.

Setup the ICE table: