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]}](https://tex.z-dn.net/?f=Kb%3D%5Cfrac%7B%5BIBH%5E%2B%5D%5BOH%5E-%5D%7D%7B%5BIB%5D%7D)
Next, we calculate the concentration of hydroxide ions and the Kb due to the fact that both the pH and pKb were given:

![[OH^-]=10^{-5.8}=1.585x10^{-6}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E%7B-5.8%7D%3D1.585x10%5E%7B-6%7DM)

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]}](https://tex.z-dn.net/?f=6.31x10%5E%7B-6%7D%3D%5Cfrac%7B%281.585x10%5E%7B-6%7D%29%281.585x10%5E%7B-6%7D%29%7D%7B%5BIB%5D%7D)
Finally, we solve for the equilibrium concentration of ibuprofen:
![[IB]=\frac{(1.585x10^{-6})(1.585x10^{-6})}{6.31x10^{-6}}=4.0x10^{-7}](https://tex.z-dn.net/?f=%5BIB%5D%3D%5Cfrac%7B%281.585x10%5E%7B-6%7D%29%281.585x10%5E%7B-6%7D%29%7D%7B6.31x10%5E%7B-6%7D%7D%3D4.0x10%5E%7B-7%7D)
Learn more:
(Weak base equilibrium calculation) brainly.com/question/9426156
The charge of Br changed from –1 to 0, therefore it is the
element which is oxidized. Since it is oxidized then Br is also the reducing
agent.
The charge of Mn changed from +4 to +2 therefore it is the
element which is reduced. Since Mn is reduced, then MnO2 is the oxidizing
agent.
The half –reactions are:
Br: 2Br --> Br2 + 2e-
Mn: MnO2 --> Mn2+
First balance oxygen by adding H2O:
MnO2 --> Mn2+ + 2H2O
Then balance hydrogen by adding H+ ions:
4H+ + MnO2 --> Mn2 + 2H2O
Then the appropriate electrons:
4e- + 4H+ + MnO2 --> Mn2 + 2H2O
Multiply the half-reaction of Br by 2 because the half-reaction
of Mn has 4 electrons.
4Br --> 2Br2 + 4e-
Combine the two half reactions and cancel common factors:
4Br- + 4H+ + MnO2 --> 2Br2 + Mn2 + 2H2O
Sublimation. It's basically, in simple terms, when a solid changes to a gas without going into liquid form.