Question

The equilibrium constant for the reaction shown below Nat NaH B. The equilibrium constant for the reaction shown below H₂0 CH₃MgBr HOMEBr + CH₄ pka = pka = + + H₂ pka = pka = K= K= C. Complete the acid-base reaction below by providing an inexpensive base and the corresponding conjugate acid such that the reaction has K > 103. Calculate the equilibrium constant for your specific acid-base reaction

Answer 1

Choose an alkoxide, like ethoxide, as a base which has a stronger base and weaker conjugate acid than ammonia to create an equilibrium constant (K) much greater than 10^3. Use Kw and the known Kb for ethoxide to calculate the equilibrium constant for the given reaction.

Step-by-step explanation:

To complete the acid-base reaction with a K > 103, we need to choose a base that is stronger than NH3 but weaker than OH−. Given the information, we can use the ammonia system as a reference. The reaction of NH4+ with water forms NH3 and H3O+, with Ka being Kw/Kb of ammonia.

For our base, we could choose an alkoxide, such as ethoxide (C2H5O−), because alkoxides are known to be strong bases in aqueous solution. The conjugate acid of ethoxide is ethanol, which has a notably higher pKa than ammonia, indicating a stronger base with a weaker conjugate acid.

The expected reaction would be as follows:

C2H5O− (aq) + H2O(l) ⇒ C2H5OH(aq) + OH−(aq)
By using the values for Kb for ethoxide (which is much larger than 103), we can infer that the equilibrium constant for this reaction, K, would also be much greater than 103, fitting the criteria.

To approximate an actual value for K, you would use the known Kb for ethoxide and Kw to calculate Ka for ethanol, then use Ka = Kw/Kb to find the equilibrium constant K.