Question

Given that C4H8COOH has a pKa = 4.8 and C4H8+NH3 has a pKa = 10.7, answer the following:

What pH would you make the water layer in order to cause the acid to dissolve in the water layer and the amine to dissolve in the ether layer?

Express your answer using one decimal place.

Answer 1

Acidic form   Basic Form
C4H8COOH <---> C4H8COO- + H+ (pKa = 4.8)
C4H8-NH3+ <----> C4H8NH2 + H+ (pKa = 10.7)

generally, a compound will exist more in its acidic form when the pH of the solution is less than its pKa.

And the compound exists more in its basic form if the pH of the solution is greater than its pKa.

At pH = pKa; both acidic form and basic form are in equal concentrations.

In our case C4H8COOH; the basic form (ionic form) C4H8COO- is soluble in water. Therefore, at pH = pKa = 4.8 we will
have acidic form and basic form are in equal amounts. If you want to make the compound to dissolve more into the
water, then we should increase the pH of the solution (increasing the ionic component concentration).

Suppose, if you want to make the basic form 10 time more than the acidic form. That means, 10x more compound into
water layer then the pH would be:

Henderson–Hasselbalch equation:

pH = Pka + log[Basic]/[Acidic]
pH = 4.8 + log(10x/x) = 4.8 +1 = 5.8

Therefore, at pH = 5.8 we will have 10X concentration goes into the aqueous layer.

Acid form    Basic form
C4H8-NH3+ <----> C4H8NH2 + H+ (pKa = 10.7)

In this case, the ionic form (acidic form C4H8-NH3+) will go into the water. Therefore, if you want to make more of
acidic form then we should go to the pH less than its pKa value.

Suppose, if you want to make the Acid form 10 time more than the basic form. That means, 10x more compound into
water layer then the pH would be:

Henderson–Hasselbalch equation:

pH = Pka + log[Basic]/[Acidic]
pH = 4.8 + log(x/10x) = 10.7 -1 = 9.7

Therefore, at pH = 9.7 we will have 10X concentration goes into the aqueous layer.