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
First, we will convert the mass of the gallon to grams:
a gallon of water has a mass of 3.79 * 1000 = 3790 grams of water
number of moles can be calculated using the following rule:
number of moles = mass / molar mass
Therefore,
number of moles = 3790 / 18.02 = 210.32 moles
The empirical formula is K₂O.
The empirical formula is the <em>simplest whole-number ratio</em> of atoms in a compound.
The <em>ratio of atom</em>s is the same as the <em>ratio of moles</em>.
So, our job is to calculate the <em>molar ratio</em> of K to O.
Step 1. Calculate the <em>moles of each element
</em>
Moles of K = 32.1 g K × (1 mol K/(39.10 g K =) = 0.8210 mol K
Moles of O = 6.57 g O × (1 mol O/16.00 g O) = 0.4106 mol 0
Step 2. Calculate the <em>molar ratio of each elemen</em>t
Divide each number by the smallest number of moles and round off to an integer
K:O = 0.8210:0.4106 = 1.999:1 ≈ 2:1
Step 3: Write the <em>empirical formula
</em>
EF = K₂O