What is research? Depending on who you ask, you will likely get very different answers to this seemingly innocuous question. Some people will say that they routinely research different online websites to find the best place to buy goods or services they want. Television news channels supposedly conduct research in the form of viewer polls on topics of public interest such as forthcoming elections or government-funded projects. Undergraduate students research the Internet to find the information they need to complete assigned projects or term papers. Graduate students working on research projects for a professor may see research as collecting or analyzing data related to their project. Businesses and consultants research different potential solutions to remedy organizational problems such as a supply chain bottleneck or to identify customer purchase patterns. However, none of the above can be considered “scientific research” unless: (1) it contributes to a body of science, and (2) it follows the scientific method. This chapter will examine what these terms mean
At the half equivalence point [HA] = [A-] and pH = pKa
<span>if Ka is 5.2e-5 then pKa = pH = 4.28</span>
Henderson–Hasselbalch equation is given as,
pH = pKa + log [A⁻] / [HA]
-------- (1)
Solution:
Convert Ka into pKa,
pKa = -log Ka
pKa = -log 1.37 × 10⁻⁴
pKa = 3.863
Putting value of pKa and pH in eq.1,
4.29 = 3.863 + log [lactate] / [lactic acid]
Or,
log [lactate] / [lactic acid] = 4.29 - 3.863
log [lactate] / [lactic acid] = 0.427
Taking Anti log,
[lactate] / [lactic acid]
= 2.673
Result:
2.673 M
lactate salt when mixed with 1 M Lactic acid produces a buffer of pH = 4.29.
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
Answer:
cinnamic acid - 150 mg
cis-stilbene - 100 μL
trans- stilbene - 100 mg
pyridinium tribromide - 200-385 mg
For this data:
moles of cinnamic acid = 0.150 g/148.16 g/mol = 0.001 mols
Theoretical mass of dibromoproduct formed = 0.001 mol x 307.97 g/mol = 0.312 g
cis-stilbene (100 ul = 0.1 ml)
moles of cis-stilbene = 0.1 ml x 1.01 g/mol/180.25 g/mol = 0.00056 mols
Theoretical mass of dibromoproduct formed = 0.00056 mol x 340.05 g/mol = 0.19 g
trans-stilbene
moles of tran-stilbene = 0.1 g/180.25 g/mol = 0.00055 mols
Theoretical mass of dibromoproduct formed = 0.00055 mol x 340.05 g/mol = 0.19 g
Explanation:
