In this experiment, when you mix methanol and sodium hydroxide together, the rea
ID: 493413 • Letter: I
Question
In this experiment, when you mix methanol and sodium hydroxide together, the reaction below takes place. For this acid/base reaction, the acid on the reactant side is __ and the acid on the product side is __. For CH_3OH, pKa = 15.5 and for H_2O, pKa = 15.7. Which is the stronger acid? Based on the pKa values given above, which is greater at equilibrium, reactant concentrations or product concentrations? If, as is the case in this experiment, a large excess of CH_3OH is used, the equilibrium will shift to (circle one: decrease the concentration of CH_3O^-, increase the concentration of CH_3O^-.)Explanation / Answer
The reaction between methanol and sodium hydroxide will give methoxide ion and water.
Answer for question 3
From the above reaction in reactent side methanol donates the proton and product side water is going to donates the proton. Those which are going to donates the proton they arevacting an acid.
Answer for question 4
In this reaction methanol having pKa is 15.5 and water is having pKa is 15.7 comparing both water and methanol which is having less pKa value that compound is more acidic in this case methanol is stronger acid than water.
Answer for question 5
Based on pka value the above reaction is shifted on product side because water is having more pka value so it is weak acidweak acid and weak base are more stable and lower in ebergy. So for that reaction is shifted in product side means right side of the reaction.
Answer for question 6
The excess amount of methanol used then reaction shifted product side because of higher pka value lower energy compound formation.
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