This problem is providing the basic dissociation constant of ibuprofen (IB) as 5.20, its pH as 8.20 and is requiring the equilibrium concentration of the aforementioned drug by giving the chemical equation at equilibrium it takes place. The obtained result turned out to be D) 4.0 × 10−7 M, according to the following work:

First of all, we set up an equilibrium expression for the given chemical equation at equilibrium, in which water is omitted for it is liquid and just aqueous species are allowed to be included:

$Kb=\frac{[IBH^+][OH^-]}{[IB]}$

Next, we calculate the concentration of hydroxide ions and the Kb due to the fact that both the pH and pKb were given:

$pOH=14-8.20=5.80$

$[OH^-]=10^{-5.8}=1.585x10^{-6}M$

$Kb=10^{-5.20}=6.31x10^{-6}$

Then, since the concentration of these ions equal that of the conjugated acid of the ibuprofen (IBH⁺), we can plug in these and the Kb to obtain:

$6.31x10^{-6}=\frac{(1.585x10^{-6})(1.585x10^{-6})}{[IB]}$