Consider the weak organic base n-propylamine. The Kb value for this base is 1.3-10^-5, and its pKa is 10.6. At what pH are the "proton on" and "proton off" forms present in equal concentrations? At what pH values would the "proton on" form have the higher concentration? At what pH values would the "proton off" form have the higher concentration?

Use the Henderson-Hasselbalch equation.

pH = pKa + log (base)/(acid)
(base) is with proton off and (acid) is with proton on.
When (acid) = (base), you see log is 0 and pH = pKa.
Substitute a value for pH and see which ratio (base)/acid or acid/base is higher to get the other parts of the question.

To determine the pH at which the "proton on" and "proton off" forms are present in equal concentrations, we can use the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

In this case, "proton on" corresponds to the protonated form of n-propylamine (HA), and "proton off" corresponds to the deprotonated form (A-). We want to find the pH at which [A-] = [HA], meaning the concentrations of both forms are equal.

Let x represent the concentration of A- and HA. At equilibrium, we can assume that the concentration of HA that deprotonates is x and that x is also the concentration of A-. Thus, the concentration of HA remaining will be [HA] = ([HA]initial - x).

Now, let's plug this information into the Henderson-Hasselbalch equation:

pKa = 10.6
[A-] = x
[HA] = [HA]initial - x --> [HA]initial will be the initial concentration of n-propylamine

pH = 10.6 + log(x/([HA]initial - x))

To find the pH at which [A-] and [HA] are equal, set [A-] = [HA]:

x = [HA]initial - x

2x = [HA]initial

x = [HA]initial/2

Substituting this value of x into the Henderson-Hasselbalch equation:

pH = 10.6 + log(([HA]initial/2)/([HA]initial - [HA]initial/2))

pH = 10.6 + log(1/2)

pH = 10.6 - log(2)

Using the inverse logarithm function (10^x) to find the value of log(2), we get:

log(2) = 0.301

pH = 10.6 - 0.301

pH ≈ 10.299

So, the pH at which the "proton on" and "proton off" forms are present in equal concentrations is approximately 10.299.

To determine at which pH values the "proton on" form has a higher concentration, we need to compare the concentrations of [A-] and [HA]. The "proton on" form (HA) will have a higher concentration when [HA] > [A-]. From the Henderson-Hasselbalch equation, we know that when the pH is above the pKa, HA will be dominant, and when the pH is below the pKa, A- will be dominant.

So, when the pH is higher than the pKa (10.6 in this case), the "proton on" form (HA) will have a higher concentration.

Similarly, when the pH is lower than the pKa, the "proton off" form (A-) will have a higher concentration.

To summarize:

At pH > 10.6, the "proton on" form (HA) will have a higher concentration.
At pH < 10.6, the "proton off" form (A-) will have a higher concentration.