chem stories

Which of following reaction do not produce the NADH as adding product at Krebs Circle?
a. Isocytrate → Oxalosuccinate
b. Alpha ketoglutarate → succinate
c. Succinate → Fumarate
d. Malate → Oxaloacetat

1. Which of the following properties of water explains its ability to dissolve acetic acid?
A. The high surface tension of water, which is due to the formation of hydrogen bonds between adjacent water molecules.
B. The ability to serve as a buffer, absorbing the protons given off by acetic acid.
C. The ability to orient water molecules so that their polarities neutralize the ions formed when the acid dissociates.
D. The ability to form hydrogen bonds with the carbonyl and the hydroxyl groups of acetic acid.
Because acetic acid is a weak acid, its dissociation in water is incomplete. That portion which does ionize, however, is neutralized in solution when the water molecules orient their partially charged atoms around the ions. The un-ionized portion of acetic acid is also soluble due to its polar character as well as via hydrogen bonds. Thus, multiple solvent properties of water are important to solubilize acetic acid.

2. The pH of a solution is equal to:
A. the hydrogen ion concentration, [H +]
B. log [H +]
C. -log [H +]
pH is defined as the negative log of the H + concentration.
D. ln [H +]
E. -ln [H +]

Water and pH
Water can be considered an acid because of its ability to ionize to a proton and a negatively charged hydroxide ion:
H-O-H H + + OH -
The frequency with which this occurs in pure water is very low. In fact, only one water molecule in 500 million will be ionized at any one time in pure water. This means that the molar concentrations of H + and OH - are approximately 10 - 7 each in pure water.
Hydrogen-ion concentrations of aqueous solutions range from greater than 1 M to less than 10 - 14 M. For convenience, H + concentrations are expressed on a logarithmic scale as pH. The pH of a solution is the negative log of its H + concentration, or:
pH = -log [H +]
Therefore, a H + concentration of 10 - 7 M has a pH of 7.0. In pure water, there are always equal numbers of H + and OH - , and the pH is defined to be neutral.(Note that neutral pH is not always exactly 7.00. If you raise the temperature, more H2O dissociates and the pH will be lower). Acids contribute another source of H + causing the pH to decrease from neutrality. Conversely, bases absorb H + from solution and thus make the pH higher than neutrality.

3. Physiological pH is 7.4. What is the hydrogen ion concentration of a solution at physiological pH?
A. -7.4 M
B. 0.6 M
C. 0.6 x 10 - 8 M
D. 1 x 10 - 8 M
E. 4 x 10 - 8

Calculating H+ concentration from pH
pH is the negative log of the hydrogen concentration, i.e. if pH = -log [H +],
then [H +] = antilog (-pH) = 10 - 7.4 M
Thus, the hydrogen ion concentration of a solution at physiological pH (7.4) is
10 - 7.4, which is equal to 3.98 x 10 - 8 M. We can round off to 4 x 10 - 8 M.


4. What is the pH of a 10 - 3 M solution of HCl?
Calculating pH from [H +] Concentration
pH is the negative log of the hydrogen ion concentration, i.e.
pH = -log [H +]
HCl is a strong acid. It completely dissociates in water, i.e.
HCl [H +] + Cl -
A 0.3 M HCl solution has a [H +] of 10 - 3 M.
Therefore, the pH of a 10 - 3 M solution of HCl is -log [10 - 3] = 3

5. What is the pH of a 10 - 10 M solution of HCl?
pH of Solution with Very Dilute Acid
The negative log of 10 - 10 gives a pH of 10 -- which means that the solution is basic. Does it make sense that a very dilute acid would have the pH of a base?
In the case of a very dilute solution of an acid, water may make a greater contribution to [H +] than the dilute acid. In this problem, a 10 - 10 M solution of HCl contributes 10 - 10 M [H +]. The ionization of water contributes 10 - 7 M [H+].
The effective [H +] of this solution is 10 - 7 M and the pH=7.
What is the pH of a 10 - 10 M solution of HCl?
The correct answer is: 7
Although -log [10 - 10] = 10, common sense dictates that a very small amount of acid does not make a solution basic. Rather, such a small concentration of H + ions (10 - 10 M) is far below the H + concentration for water (10 - 7 M), thus the pH remains at 7.

6. If the concentration of H + in a solution is 10 - 3 M, what will the concentration of OH - be in the same solution at 25° C?
A. 10 - 3 M
B. 10 - 11 M
C. 1011 M
D. 2 x 10 - 11 M
E. 10 - 14 M

Relation between H + and OH - concentrations at 25° C
In aqueous solutions at 25° C, the product of the H + and OH - concentrations is always 10 - 14, as expressed in this equation:
[H +] [OH - ] = 10 - 14
This allows calculation of the OH - concentration if the H + concentration is known.
[OH - ] = 10 - 14 / [H +]
[OH - ] = 10 - 14 / 10 - 3 = 10 - 11
This calculation is used when the solution in question is not pure water, but contains some mixture of acids and bases.

7. How many ml of a 0.4 M HCl solution are required to bring the pH of 10 ml of a 0.4 M NaOH solution to 7.0 (neutral pH)?
Note: HCl and NaOH both completely dissociate in water (i.e., no pKa calculation is necessary).
A. 4
B. 40
C. 10
D. 20
E. 2
Neutralizing a basic solution
This question does not require any complicated calculations, but rather can be solved by considering how acids and bases work. In this case, both the acid and the base solutions are at the same concentration. Also, as noted in the question itself, NaOH and HCl completely dissociate.
Thus, a neutral pH can be achieved by adding an equal amount of the acid solution to the base solution.
C. 10 Because both the acid and the base are at the same concentration and both completely dissociate in water, a neutral pH can be achieved by adding an equal amount of the acid solution to the base solution.

8. How many ml of a 0.2 M NaOH solution are required to bring the pH of 20 ml of a 0.4 M HCl solution to 7.0?
A. 4
B. 40
C. 10
D. 20
E. 5
Neutralizing an acidic solution
This question is similar to the previous one. It does not require any complicated calculations, but rather can be solved by considering how acids and bases work.
To begin with, NaOH and HCl both completely dissociate in water, therefore no pKa calculation is necessary. The implications of this are that a given amount of acid solution has the same number of free protons as the same amount of the same concentration of a base solution.
In this case, the concentration of the acid solution (at 0.4 M) is twice that of the base solution (at 0.2 M). Thus, a neutral pH can be achieved by adding twice the amount of base solution to the acid.
B. 40
Because the acid has twice the concentration of the base, a neutral pH can be achieved by adding twice the amount of the base solution to the acid solution.

9.Acids are defined as compounds with pKa values below 7.0.
A. True
B. False
Acids
Acids are defined as compounds which can reversibly lose protons to the solution. The pH at which this occurs (the halfway point) is the pKa. Strong acids such as HCl will give up protons even at very low pH (hence low pKa) and weak acids will only give up protons if the pH is very high (i.e. the free proton concentration is very low). The pH at which this occurs can easily be above 7.
An important example is tyrosine, whose r-group is phenol; an acid with a pKa ~ 10. Conversely, bases can have pKa < 7.0. Can you think of an important biological BASE with a pKa less than 7.0? Most biological acids, however, are weaker acids than HCl. The major class of biological acids is carboxylic acids. Because the difference in electronegativity between oxygen and hydrogen in a carboxylic acid is not as dramatic as it is with HCl, the tendency of a carboxylic acid to give up its proton is much less than that of HCl. However, carboxylic acids dissociate more readily than water due to the presence of two electronegative oxygens. An acid's tendency to dissociate is a function of the strength of the acid and the pH of the solution. Strong acids can still dissociate when the pH is low, whereas weak acids cannot. The convention is to identify the pH at which the acid is half dissociated (e.g. half is protonated and half is deprotonated). This pH value is defined as the pKa of the acid in question. For example: CH3COO (acetic acid) CH3COO- (acetate ion) + H+ ; pKa = 4.8; meaning that, at pH = 4.8, half of the molecules are ionized (acetate) and half are not (acetic acid). The stronger the acid, the lower the pKa. Bases In contrast to acids, bases are able to absorb protons from water and thus they are charged (+1) in the protonated form, and uncharged in the deprotonated form. The most common biological weak base is the amino group (R-NH2). Despite the differences between acids and bases, pKa can also be used to quantify the relative strength of bases. Notice that while the pKa values for acids are generally less than 7, pKa values for bases are usually greater than 7. For instance, ethanolamine has a pKa of 9.5 and is thus half protonated (charge = +1/2) when the pH is 9.5 (Figure 4). An important exception to this rule of thumb is the amino acid histidine, which is a base but has a pKa around 6.0. B. False
Acids are defined as compounds which can reversibly lose protons to the solution.

10.The correct operational relationship between pKa and pH is that:
A. both are log functions.
B. both are always < 7 for acids, and >7 for bases.
C. These two concepts are not operationally related in any way since biological fluids contains mixtures of too many acids and bases.
D. When pH = pKa , the compound in question will have a charge of +0.5.
E. When pH = pKa , the ionizable compound in question (whether acid or base) will be half protonated and half deprotonated.

pKa and pH
The "operational" relationship between pKa and pH is mathematically represented by Henderson-Hasselbach equation:
pH = pKa + log [A-] / [HA]
where [A-] represents the deprotonated form and [HA] represents the protonated form.
One oft-cited solution to this equation is obtained by arbitrarily setting pH = pKa.
In this case, log([A-] / [HA]) = 0, and [A-] / [HA] = 1.
In words, this means that when the pH is equal to the pKa of the acid, there are equal amounts of protonated and deprotonated acid molecules. This same relationship holds for bases as well, with [B] substituting for [A-] as the deprotonated form, and [HB+] substituting for [HA] as the protonated form. It should be emphasized that the Henderson-Hasselbach relationship holds for a specified acid or base even if multiple acids or bases are present.
Other solutions to the Henderson-Hasselbach equation over a range of pH values are displayed in the following figure (for acetic acid).



"Net charge on acid" refers to the average of all acid molecules in the solution.
Several points about this graph deserve mention:
• At the pKa , the acid is 50% deprotonated.
• At the 1 pH unit above (below) the pKa , the acid is 90% deprotonated (protonated).
• At 2 pH unit above (below) the pKa , the acid is 99% deprotonated (protonated).
• At 3 pH unit above (below) the pKa , the acid is 99.9% deprotonated (protonated).
• When fully protonated, charge on acetic acid is 0.
• When fully deprotonated, charge on acetate is -1.
E. When pH = pKa , the ionizable compound in question (whether acid or base) will be half protonated and half deprotonated.
This is in fact the very definition of pKa.

11. If equal volumes of 0.05 M NaH2PO4 and 0.05 M H3PO4 are mixed, which of the following best describes the resulting solution? (pKa's for phosphoric acid are 2.0, 6.8 and 12.0)
A. pH 2 and poorly buffered.
B. pH 2 and well buffered.
C. pH 6.8 and well buffered.
D. pH 12 and well buffered.
E. pH 6.8 and poorly buffered.

Recognizing the appropriate protonated and deprotonated forms of an acid with multiple ionizations
Phosphoric acid undergoes three ionizations, hence it has three pKa values (given as 2.0, 6.8 and 12.0 in the question). Phosphoric acid has great biological significance due to its role in DNA/RNA, energy molecules such as ATP, protein phosphorylation, etc; therefore, it is worthwhile to spend a moment examining its ionization reactions.
Imagine the starting form as being completely protonated. Intuitively, complete protonation should occur when the [H+] is very high, ie. at a low pH. Thus, at pH 1 or less, phosphoric acid exists as >90% H3PO4.
Now imagine adding NaOH to a solution of H3PO4 at pH 1. The OH- ion combines with H+ to produce water, raising pH and leaving Na+ in solution. As the pH rises towards 2.0, however, what happens to the H3PO4? Since 2.0 is the first pKa , the first proton will begin coming off. This has two consequences, one concerning the chemical form of the H3PO4 and the other concerning buffering.
The chemical form of the H3PO4
Concerning the chemical form of H3PO4 , after the first H+ has completely come off (which occurs when the pH has risen above 3 or so), we are left with H2PO4 -. But since Na+ is also left by this reaction (the OH- and the H+ having combined to produce water), we can express the ions in solution as being Na+H2PO4 -, or just NaH2PO4 (also called monobasic sodium phosphate).


Buffering
As for the buffering part, one only needs to realize that, during the transition between pH 1.0 and 3.0, a lot of the H+ used to combine with OH- comes from H3PO4. The protons do not come from water, and the relationship [H+] x [OH-] = 10 - 14 still holds; therefore, the pH does not change much when NaOH is added during the 1-3 pH transition. This is not the case between, say pH 3.0 and 5.8, when addition of NaOH does not take H+ from phosphate, because there is no pKa value for phosphate dissociation in this range (the second pKa skips from 2.0 all the way up to 6.8). Thus, any H+ must come from water, and the pH changes rapidly in this range.
The concept of bufferi ng can also be shown graphically, as seen below.

At or near their pKa , both weak acids and weak bases will resist changes in pH, thus acting as buffers. As described above, this is due to their affinity for protons, which are the species that determine pH. Thus, a solution will resist pH change at values near its pKa (see shaded area in figure above).
In this problem, the phosphoric acid solution will continue to resist pH changes as NaOH is added until nearly all of H3PO4 has been converted from to the H2PO4- form. At pH values >> pKa , the affinity of phosphoric acid for protons is not sufficient to bind H+ until the next pKa is reached, and at pH values << pKa , nearly all of the phosphoric acid has already bound H+ and is thus no longer available to bind additional H+. Therefore, phosphoric acid, like any other weak acid or base, is only effective as a buffer at pH values within one pH unit of its pKa . B. pH 2 and well buffered.
Equal amounts of phosphoric acid (H3PO4) and monosodium phosphate (NaH2PO4) will be present at a pH of 2.0, which is the pKa for phosphoric acid. Because both components of the mixture have 50mM phosphate and the solution is poised at the pKa of the first ionization, the solution will be able to absorb as much as 25mM of either acid or base before its buffering capacity is exhausted. The solution is thus judged to be "well buffered."

L-Ascorbic Acid, is the bioactive form of Vitamin C, on the other hand D-Ascorbic Acid is biochemically inactive. I wander Why ?
Vitamin C is a product that is prevalent in many foods and produced by an abundance of chemical companies. The chemical side of Vitamin C is what is basically addressed here.Known by most scientists, but virtually unknown by the public and most doctors is the fact that Vitamin C has two totally and distinctly separate sides, as many products do. The two sides consist of "L"-Ascorbic Acid, which is the (-) side , and D-Ascorbic Acid, which is the (+) side. The L side of Vitamin C is the active side, and is the side which is beneficial to mankind . The D-side of Vitamin C is designated as useless and discarded by the body, as most research shows. People ingesting Vitamin C would only benefit from the "L" side of Vitamin C. Vitamin C is not a stable substance, as some people believe, and the L and D sides are subject to change with moisture an if water is added.

Example: In an orange, the Vitamin C is primarily the L side, and remains so in the surroundings of the orange.However, when the j uice has been extracted from the orange for a period of days, the "L" side changes to D and the juice will eventually even off at 50% D and 50% L Vitamin C content. Frozen concentrate orange juice, when thawed and water added, will produce exactly the same action.The exact same action takes place again when Vitamin C is added by manufacturers to juices or drinks.

The assumption the public is under is that the Vitamin C added as a supplement during the manufacture of any liquid is all beneficial, which is a myth. The facts are that adding Vitamin C to drinks may constitute only 50-60% "L" , with the remainder being the D side. Although it is believed that the body discards the D side of Vitamin C, further research should be undertaken to confirm this theory, as it may reveal that in some individuals, an abundance of the D side of Vitamin C could conceivably be detrimental to a person's health.

http://www.xpressnet.com/bhealthy/vitaminc.html

Definition reducing sugar
If a sugar contains aldehyde groups that are oxidised to carboxylic acids, then that sugar is classified as a reducing sugar. It is called a reducing sugar because it reduces the number of chemicals present in its structure through oxidation.

Examples
Reducing sugars include glucose, fructose, glyceraldehyde, lactose, arabinose and maltose.

Reducing Capabilities
Reducing capability is defined by the presence of potential aldehyde or ketone group, anomeric carbons found in sugars, which determine a substance's ability to lose or gain electrons to form new or more stable solutions or its reaction to other substances.

Testing
Benedict and Fehling's reagent are two solutions used to determine the reducing capability of a sugar. These solutions are used to determine the presence of free aldehyde or ketone group in sugars.

Non-Reducing Sugar
The reason that sucrose is a non-reducing sugar is that it has no free aldehyde or keto group. Additionally, its anomeric carbon is not free and can't easily open up its structure to react with other molecules.


SUCROSE
The common name for sucrose is table sugar. Just as the name implies, sucrose is used throughout the world as a sweetener for cooking, baking and sweetening edibles. Sucrose is commonly made by refining plant matter, such as sugarcane or sugar beets. In its most commonly used form, it is a white, orderless powder that is sweet to the taste. Although sucrose is a disaccharide, it is not a reducing sugar.

Sucrose is the technical term for table sugar such as cane sugar or white sugar. It is composed of the combination of one glucose molecule and one fructose molecule.

Formation
The process starts with a condensation reaction, a process involving the release of water. This process is followed by the formation of a glycosidic bond between two available and appropriate monosaccharide molecules, creating disaccharides like sucrose.

Properties
Sucrose is soluble in water but its molecules are too big to pass through the cell membrane during diffusion. It can only be broken down through a hydrolysis reaction, a reverse condensation reaction.

Disaccharide
Sucrose is a complex carbohydrate known as a disaccharide, meaning made up of two simple carbohydrates or monosaccharides. The monosaccharides that form sucrose are glucose and fructose. Disaccharide is the most common form of sugar that is found in nature. It results from the combination or reaction of two simple sugars (monosaccharides). It has two types, the reducing and non-reducing sugar. Sucrose is a classic example of a non-reducing sugar.


Significance
Although both glucose and fructose are reducing sugars, sucrose is not because it does not contain anomeric hydroxyl groups, and does not reduce the chemicals present in its structure through oxidation.

Identification
It is possible to test a substance such as sucrose for the qualities that make it a reducing sugar using Fehling's solution, a mixture of copper sulfate, distilled water, Rochelle salt and sodium hydroxide. In the presence of reducing sugars, the copper sulfate in the solution will oxidize and turn red.






Read more: Why Is Sucrose a Non-Reducing Sugar? | eHow.com http://www.ehow.com/facts_5882980_sucrose-non_reducing-sugar_.html#ixzz15o22o2m6

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