1. The problem statement, all variables and given/known data Ethanol in the body is oxidized to acetaldehyde by liver alcohol dehydrogenase (LADH). Other alcohols are also oxidized by LADH. For example, methanol, which is mildly intoxicating, is oxidized by LADH to the quite toxic product of formaldehyde. The toxic effects of ingesting methanol (a component of many commercial solvents) can be reduced by administering ethanol. The ethanol acts as a competitive inhibitor of methanol by displacing it from LADH. This provides sufficient time for the methanol to be harmlessly excreted by the kidneys. If an individual has ingested 100 mL of methanol (a lethal dose), how much 100 proof whiskey (50% ethanol by volume) must be imbibed to reduce the activity of his LADH towards methanol to 5% of its original value? The adult human body contains ~40L of aqueous fluids throughout which ingested alcohols are rapidly and uniformly mixed. The densities of ethanol and methanol are both 0.79 g/cm3. Assume the KM values of LADH for ethanol and methanol to be 1.0 X 10-3 M and 1.0 X 10-2 M, respectively, and that Ki = KM for ethanol. 2. Relevant equations alpha=1 + ([etOH]/KetOH) (V[meOH]/V[etOH])= (Vmax*[meOH]/KmeOH+[meOH])/(Vmax*[meOH]/alpha*KmeOH+[meOH]) which reduces to (V[meOH]/V[etOH])=(alpha*KmeOH+[meOH])/(KmeOH+[meOH]) 3. The attempt at a solution after determining the concentration of methanol in the body to be 62mM the above equation is filled in with the values 100%/50%= (alpha*1.0e-2+62e-3)/(1.0e-2+62e-3) 20=(alpha*1.0e-2+62e-3)/(1.0e-2+62e-3) alpha=(20*7.2e-2-62e-3)/1.0e-2=137.8 then, using the top equation and the determined and given values we get 137.8=1+[etOH]/1.0e-3 which gives [etOH] as 0.1368M But I can't remember how to convert M to L, and I'm not sure if this is even the right way to solve the problem, so any help on what is done, and where I go from concentration would be great! Thanks!