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application of titration

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1. Determine the molar mass of an unknown acid.
Called the unknown acid HX, and it reacts with NaOH.
Equation is:
NaOH(aq)+HX(aq) --> NaX(aq)+H2O(l)
1 mole of NaOH reacts with 1 mole of HX.
Weighed a sample of HX is titrated with standard NaOH.
Weigh the vial containing the acid unknown = 6.9932g
Removed approx 0.5g of the HX sample from the vial to a beaker.
Weighed the vial again = 6.5180g
Now I know the exact weight of the HX sample.
First I will add 10mL of water to the sample to dissolve it completely along with a few drops of an acid base indicator.

Prepared a standard titrant solution. In this example, the titrant concentration is 0.09156mol/L NaOH solution.
31.90mL of the titrant was poured into the solution until the solution was able to change color.
With these results, what is the molar mass of this sample?

2. Analysis of calcium carbonate content of the sample using a back titration.
Equation is:
CaCO3(aq)+2HCl(aq) -->CO2(g)+H2O(l)+CaCl2(aq)
The CaCO3 is dissolved in HCl. Note that the dissolution of the CaCO3 in HCl is a chemical reaction. In order to be surethat all the CaCO3 dissolves an excess of HCl is deliberately added.
Next step involves the titration of the excess HCl using a standard NaOH solution.
HCl(aq)+NaOH(aq) -->NaCl(aq)+H2O(l)
Our sample consists of a fine white powder contained in a vial.
Weigh the vial containing the sample=6.7799g
Removed approx 0.25g of the white powder sample from the vial to a beaker.
Weighed the vial again = 6.5247g
Now I know the exact weight of the whit powder sample.
First I will add 50mL of 0.1007 mol/L HCl delivered by a pipette to the sample to dissolve it completely along with a few drops of an acid base indicator.

Prepared a standard titrant solution. In this example, the titrant concentration is 0.02510mol/L NaOH solution.
41.40mL of the titrant was poured into the solution until the solution was able to change color.
With these results, what is the percentage of calcium carbonate in this sample?

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Solution Summary

It provides examples of applications of titration, such as determining the molar mass of an unknown acid and analyzing the calcium carbonate content in the sample. The solution is detailed and well presented.

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A solution for Allosterity and Cooperativity

When a compound made up of non-polar molecules is mixed with an aqueous solvent such as water, the molecules cluster together into a ball while water tends to form a ring around them. They do so because they are hydrophobic (. Wikibooks 2015).

When more of this solute is added, the water ring is disturbed as more of the hydrophobic molecules join the non-polar core and the displaced water molecules are freed to move around. This causes high disorder in the solution environment referred to as high entropy.
According to the 2nd Law of Thermodynamics, "The total entropy of the system plus its surrounding must always be increasing"(Wikibooks 2015). In this case the release of the water molecules from the cage around nonpolar surfaces is favourable and responsible for phenomenon called the hydrophobic effect.

The degree of freedom i.e. the free movement of drugs in solution or receptor proteins also favours entropy optimization and the binding phenomenon reduces this freedom as well as the binding affinity. Creation of more rigid drug molecules with less interference on the protein degree of freedom results in compensated enthalpy/entropy environments that favour binding affinity.

1. The binding affinity of drug molecules to protein receptor sites is a function of the solubility of the drug molecule and the part of the protein bound to the drug. The less soluble the drug molecule is, the more hydrophobic it is and therefore the higher the entropy as explained in the background above which creates a favourable environment for entropic optimization. Solvation of the protein (hydrophobicity) that is buried during binding will increase binding affinity if it is more hydrophobic. Binding Affinity (Ka) is a function of Gibbs Energy which is a difference between enthalpy and entropy. A more negative enthalpy and more positive entropy are more favourable for binding affinity (Freire 2005).

2. The limitations of this approach to affinity optimization include the following:

2.1. The resulting drug molecules are insoluble in water due to their hydrophobicity

2.2. Drug resistance may result from mutations of the binding site.

3. Enthalpy/Entropy compensation is any gain in enthalpy contributions to binding is opposed by an accompanying loss in entropic contributions. Affinity optimization is accomplished by selecting chemical modifications that carry a low enthalpy/entropy compensation ((Lumry and Rajender, 1970; Eftink et al., 1983). Examples of the practical application of the entropy/enthalpy compensation principle including conformational constraints are all 1st generation HIV Protease Inhibitors Nelfinavir, Saquinavir and Ritonavir (Velasquez et.al 2003)

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