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3.3: Solubility of Gas in Water Experiment - Biology

3.3: Solubility of Gas in Water Experiment - Biology


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Scientists hypothesize that approximately 250 million years ago, during the Permian period, the world's oceans became depleted of oxygen. A chain of events then led to the Permian mass extinction, a time during which most living species became extinct.

View the NOVA video and answer the questions

  1. How many years ago did the Permian mass extinction occur?
  2. What % of species became extinct in the Permian mass extinction?
  3. Mammal like reptiles and exotic ocean animals were present during the Permian period. What types of life were NOT on Earth 250 million years ago?
  4. How many major extinctions have occurred on Earth?
  5. What type of gas did the volcanoes in the Siberian Traps release?
  6. Water can hold a type of gas critical to living organisms (fish and other aquatic animals require it to survive). What is this gas?
  7. What type of gas do deadly bacteria in the lower layers of some lakes produce?
  8. What does this gas smell like?
  9. Develop a flowchart of the events that led to the Permian extinction
    a. Anaerobic bacteria thrive in the oceans and produce hydrogen sulfide
    b. Atmospheric carbon dioxide levels increase
    c. Atmosphere warms
    d. Dissolved oxygen levels in the oceans drop
    e. Hydrogen sulfide accumulates in the oceans and atmosphere
    f. Most aquatic life that depends on oxygen dies
    g. 95 percent of Earth's life is killed by hydrogen sulfide
    h. Oceans warm
    i. Volcanoes erupt

Concept question

Explain (briefly) how volcanic eruptions can change the atmospheric and ocean environments.

Experiment

Atmospheric gases, such as oxygen and carbon dioxide, are soluble in water. How much of a particular gas dissolves in water depends on the temperature of the water and on the pressure of that gas above the water.

The gas in carbonated water (seltzer) is carbon dioxide. The high pressure inside the bottle causes more carbon dioxide to dissolve in the water than would dissolve at typical ground-level atmospheric pressures.

Materials

Carbonated water at room temperature
Carbonated water at 4oC
3 glass beakers
40oC water bath
Ice
Thermometer

Hypothesis

State the hypothesis prior to beginning the experiment. The hypothesis indicates at which temperatures you believe CO2 gas will be more or less soluble in water.

Procedure

Read through the entire procedure before beginning the experiment

  1. Obtain 3 - 250 ml beakers. Place one beaker on a bed of ice.
  2. Pour 100 ml of ice-cold carbonated water into the beaker on ice.
  3. Pour 100 ml room temperature carbonated water into the other 2 beakers.
  4. Immediately place a beaker in the 40oC water bath. Bring the beaker on ice and the beaker at room temperature with you so that you can observe them simultaneously.
  5. Record observations in the data table.
  6. After 5 minutes, determine the temperature of each water environment by placing the thermometer in the carbonated water.

Data Table

Questions

Refer to sections 3.1 and 3.2

  1. The aspect that varies between groups in the experiment is called the experimental (independent) variable. Identify the experimental variable in the experiment.
  2. Controlled variables are extraneous factors that are kept constant to minimize their effect on the outcome of the experiment. Identify 3 controlled variables
  3. The control group provides the baseline to which the experimental groups will be compared. Identify the control treatment in the experiment.
  4. The dependent variable changes with respect to the experimental (independent) variable. The dependent variable is what is measured in the experiment. Identify the dependent variable.

COMPOUND PROPERTIES AND DRUG QUALITY

Improving aqueous solubility

Aqueous solubility can be improved by medicinal chemistry despite the blunt SAR feature and our internal record has been quite successful in this respect. However, to improve solubility requires commitment to a combination of computational and experimental interventions and a real effort on the part of chemists to incorporate solubility information into synthesis design. The importance of rapid experimental feedback is particularly important given the current inability to predict computationally poor solubility arising from crystal packing interactions. It is critical not to miss a serendipitous improvement in solubility attendant on a molecular change. Owing to the blunt SAR feature, the easiest way to improve solubility with respect to library design is to try to design the best solubility profile right at the start.


Factors Affecting Solubility

One fun way to start learning about solutions is to open your refrigerator. Do you have any orange juice? Pour yourself a small glass. Do you have any soda or iced tea? Pour another small glass. Look through each liquid. You should notice that you can see clearly thorough both liquids&mdashthey&rsquore transparent. You cannot see through the orange juice, however&mdashit&rsquos opaque. The differences between these liquids are due to the size of particles dissolved in them. Orange juice contains larger particles that are only temporarily suspended in water: if the orange sits for a while, the bigger particles settle to the bottom (that&rsquos why you should always shake a container of orange juice before pouring!). Iced tea and soda, on the other hand, are solutions. The particles within the liquid are small enough remain suspended in the liquid, which allows light to travel through.

A solution is a homogeneous (evenly distributed) mixture of two or more substances. The substance that is present in the largest amount is called the solvent, while the substance that is present in the smaller amount is called the solute. Water is a familiar solvent, as many solutes can be dissolved in it. Check out what happens when sugar is dissolved in water. The three-atom particles are the water molecules, and the bigger white crystals are the sugar molecules. Note how each of the sugar molecules becomes surrounded by water molecules.

Problem

Materials

  • Distilled water (this type of water has absolutely no minerals dissolved in it)
  • White granulated sugar
  • Teaspoon
  • Three straws
  • Thermometer
  • Clear glass

Procedure A: How Sweet It Is!

  1. Add two teaspoons of sugar to one cup of the distilled water.
  2. Stir.
  3. You are now going to see if the concentration of sugar varies in different parts of a sugar solution. Luckily, you have your own sugar detector: your tongue! You are going to draw samples of the sugar solution from the top, middle, and bottom of the cup of the sugar solution. Which location do you think will taste the sweetest?Why?
  4. Dip the straw into the bottom of the cup. When some of the sugar water has entered the straw, put your finger over the top of straw and lift carefully out of cup. Taste the liquid at the bottom of the straw.
  5. Using a fresh straw each time, repeat step 4, sampling at the middle and top of the sugar solution, respectively.
  6. If you are unsure of your results, try again, or have a friend repeat the experiment with you.

Procedure B: How Sweet Can it Get?

  1. Measure one cup of room temperature distilled water into a clear glass.
  2. Add 1 teaspoon of sugar. Stir.
  3. Continue adding sugar to the water, 1 teaspoon at a time, stirring after each addition. Make to keep track of how many teaspoons of sugar you add.
  4. Keep adding sugar to the solution, until the solution reaches a point where sugar no longer dissolves and instead sinks to the bottom of the glass.
  5. At this point, you will have made a saturated solution. A saturated solution is a solution that holds the maximum amount of that particular solute (in this case sugar) for that particular solvent (in this case water) at that particular temperature (room temperature).

Procedure C: What about Hot and Cold?

  1. Heat up one cup of distilled water until hot. Add sugar one teaspoon at a time, stirring after each addition, until you have added the same amount of sugar that you added to the cup filled with room temperature water. Observe.
  2. Fill a second cup with chilled distilled water. Add sugar one teaspoon at a time, stirring after each addition, until you have added the same amount of sugar that you added to the cup filled with room temperature water. Observe.

Results

For Procedure A, the water will be uniformly sweet&mdashit should have made no difference whether the water came from the top, middle, or bottom of the cup. For Procedure B, your results will vary depending on the exact temperature and volume of water in your cup. For Procedure C, the hot water should not have become saturated when the sugar was added to it. The cold water will have become saturated before all the sugar was added.

The reason you did not notice a difference in sweetness when you sampled the solution at different levels is because the solute and solvent were uniformly distributed. You might have thought the solution would taste sweeter near the bottom of the glass&mdashbut that happens with suspensions (like orange juice), not solutions. Of course, if your sugar solution was saturated, it would taste sweeter at the bottom&mdashbut we intentionally had you make an unsaturated solution for this part of the experiment. Speaking of saturated solutions, the reason your solution could only dissolve so much sugar is that eventually there are not enough water molecules around to surround each sugar molecule, so some of the sugar molecules begin to clump together and fall to the bottom. In many solutions in which solids are dissolved in liquids, the warmer the solvent is, the more solution is able to dissolve: temperature is one of the key factors affecting solubility. Cooler liquid solvents are often capable of holding less solute.

Solutions don&rsquot always involve a solid dissolved in a liquid. For instance, soda has both solid sugar and carbon dioxide gas dissolved in water. The air we breathe is a solution of several gases, including nitrogen, oxygen, and carbon dioxide. Solids can even be dissolved in solids! Brass is a solid solution (or alloy, because the component materials are metals) of zinc and copper. The combination of the two metals makes brass stronger and more durable than either zinc or copper alone. Bronze is an alloy of copper and tin. Pure gold (24 karat) is too soft and expensive for most jewelry, so it is made it into alloys with stronger, cheaper metals. 18 karat gold earrings are 75% gold and 25% other metals.

Going Further

For a yummy extension, make rock candy by making a supersaturated sugar solution.

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Henry's Law for Gas Solubility Chemistry Tutorial

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Deriving Henry's Law

Consider a cold can of cola on a hot summer day.
Before you open the can, there is carbon dioxide gas dissolved in the cola drink, and, carbon dioxide gas above the drink at a pressure slightly greater than atmospheric pressure.
As soon as you open the can you can hear the "hiss" as gas escapes from the can into the atmosphere in order to equalise the pressure above the cola in the can with atmospheric pressure.
Then you pour the cola into a cold glass.
And you see that bubbles of gas are still rising through the cola to its surface and then escaping into the atmosphere.
Why? Henry's Law!

A gas, such as carbon dioxide, can dissolve in a liquid, such as water, to form a solution.
The gas is the solute and the liquid is the solvent.
In a closed system, solute particles in the gas phase above the solution are in equibrium with solute gas particles in solution as shown by the following equation:

By Le Chatelier's Principle, increasing the pressure on the system will force the equilibrium position to shift to the right, favouring the formation of the solution by dissolving more of the gaseous solute in the solvent.
Similarly, a decrease in pressure will favour the formation of gas and solvent, that is, gas will leave the solution to increase the pressure above the solution.

For the cola in the closed can, the partial pressure of carbon dioxide above the solution is very high.
For the cola in the glass, the partial pressure of carbon dioxide above the solution is very low because the partial pressure of carbon dioxide in the atmosphere is very low.
Therefore, by Le Chatelier's principle, as the cola sits in the glass, carbon dioxide is continually escaping from the solution in order to increase the partial pressure of carbon dioxide in the atmosphere.

solute(g) + solvent(l) ⇋ solution

Now we can see that partial pressure of a gas is proportional to its concentration in solution

partial pressure &prop concentration

So, using a constant of proportionality, K, we can write an expression for Henry's Law:

K is known as the Henry's Law constant.

Consider the system in which oxygen gas dissolved in water is in equilibrium with oxygen gas above the water.

Using the concentration of oxygen gas in water at various partial pressures at 25°C we obtain a graph like the one below:

We can see that K, Henry's Law constant, is the slope (gradient) of the line 4 .
The slope of the line &approx 60 kPa ÷ 0.0008 mol L -1 = 75 000 kPa/mol L -1
Therefore, for an aqueous solution of oxygen gas at 25°C we can write:

The value of K, Henry's Law constant, is dependent on the

The table below gives the value of Henry's Law constant for various aqueous solutions of gases at 25°C given that pressure is measured in kPa and concentration in mol L -1 :

Aqueous Solutions at 25°C
Gas K
( kPa / mol L -1 )
helium 282 700
nitrogen 155 000
hydrogen 121 200
oxygen 74 680
ammonia 5 690
carbon dioxide 2 937

The data in the table above have been plotted on the graph below:

From the graph we we can see that a greater partial pressure is required to dissolve 0.0002 moles of helium gas compared to any of the other gases listed.

Henry's Law only applies when the gas and its solution are essentially ideal, that is,

The curves for He(g), H2(g) and N2(g) are all linear up to about 10,130 kPa (100 atm), that is, these gases are obeying Henry's Law, however, in this range O2(g) starts to deviate.

Molar Solubility of Gases

We can rearrange Henry's Law as given above to write an expression for the concentration of a gas:

Henry's Law: P = Kc
divide both sides of equation by K: P
K
= K c
K
Molar Solubility of Gas:
(c in mol L -1 )
c = P
K
= P × 1
K

The table below gives the value of the inverse of Henry's Law constant for various aqueous solutions of gases at 25°C given that pressure is measured in kPa and concentration in mol L -1 :

Aqueous Solutions at 25°C
Gas 1/K
( mol L -1 / kPa )
molar solubility
(101.3 kPa)
helium 3.537 × 10 -6 3.583 × 10 -4
nitrogen 6.452 × 10 -6 6.536 × 10 -4
hydrogen 8.251 × 10 -6 8.358 × 10 -4
oxygen 1.33 × 10 -5 1.356 × 10 -3
ammonia 1.757 × 10 -4 0.01780
carbon dioxide 3.405 × 10 -4 0.03449

We can then use these values to compare the molar solubility of the gases in water at 25°C and the same partial pressure, for example, 101.3 kPa (or 1 atm):

molar solubility of gas at 101.3 kPa

The graph below shows how the solubility of a gas increases as its partial pressure increases according to Henry's Law:

Comparing different solutions of gas dissolved in water with the same partial pressure at the same temperature, we see that oxygen gas is more soluble than hydrogen, nitrogen or helium gases.

Water at sea level can dissolve more oxygen gas than water on top of a mountain.
This is because air pressure at sea level is greater than on top of the mountain, and therefore, the partial pressure of oxygen in air at sea level is greater than on top of the mountain.

Henry's Law Calculations in Air

Air is made up of a mixture of gases.
If a solution is exposed to air, then the concentration of gases in solution will be dependent on the partial pressure of the gases in the atmosphere.
If we use Henry's Law to calculate the concentration of a gas at atmospheric pressure in equilibrium with air, we need to know the partial pressure of the gas in the atmosphere.

Composition of the Atmosphere
(clean dry air at 25°C and 101.3 kPa)
gas concentration
/ ppm by volume
partial pressure
kPa
nitrogen 780 900 79.1
oxygen 209 400 21.2
argon 9 300 0.94
carbon dioxide 315 0.032
neon 18 0.0018
helium 5.2 0.00053
methane 1.1 0.00011
krypton 1.0 0.00010
nitrous oxide 0.5 0.00005
hydrogen 0.5 0.00005
xenon 0.08 0.000008
nitrogen dioxide 0.02 0.000002
ozone 0.02 0.000002
molar concentration of gas = 1/k × partial pressure

c(N2) = 6.452 × 10 -6 × 79.1 = 5.10 × 10 -4 mol L -1

c(O2) = 1.339 × 10 -5 × 21.2 = 2.84 × 10 -4 mol L -1

c(CO2) = 3.405 × 10 -4 × 0.032 = 1.09 × 10 -5 mol L -1

c(He) = 3.537 × 10 -6 × 0.00053 = 1.87 × 10 -9 mol L -1

c(H2) = 0.00005 × 8.251 × 10 -6 = 4.13 × 10 -10 mol L -1

Even though O2(g) is more soluble in water than N2(g), because the partial pressure of nitrogen in the atmosphere is greater, the concentration of nitrogen in water is slightly greater than that of oxygen.

Similarly, H2(g) is more soluble in water than N2(g) or He(g), yet, because the partial pressure of H2(g) in the atmosphere is much less than N2(g) or He(g) its concentration in water is less than that of N2 or He.

Effect of Temperature on Solubility of Gases

As the temperature of an aqueous solution increases, a gas generally becomes less soluble in the water, as shown in the graph below:

Pure water at room temperature is made up of water molecules that are hydrogen-bonded to each other. This creates an arrangement of water molecules containing 'holes'. Molecules of gas solute can occupy these 'holes', but because the intermolecular attraction between the water and gas molecules is less than the intermolecular attraction between water and water molecules, energy is released.

solute(g) + water(l) ⇋ solution(aq) + energy

By Le Chatelier's Principle if the system initially at equilibrium is then heated, the equilibrium position shifts to the left to counteract the effect of the additional heat. Therefore, gas comes out of solution and escapes into the gaseous phase, decreasing the concentration of gas in solution and increasing its partial pressure.

Similarly, if the system is initially at equilibrium and is then cooled, by Le Chatelier's Principle the equilibrium position shifts to the right to provide more energy to offset the reduction in energy. More gaseous solute dissolves in the water which increases the concentration of gas in solution.

The effect of increased temperature on gas solubility is obvious if you leave a cool glass of a carbonated drink (cola for instance) sitting on a table in a warm room. As the solution warms, carbon dioxide gas becomes less soluble and escapes from the solution.

The effect can also be seen if you leave some cool, fresh water in a glass on a table in a warm room, you will see small bubbles of gas appear and float to the surface which will then escape into the atmosphere. As the water in the glass warms, the dissolved atmospheric gases become less soluble and escape from the solution into the atmosphere.

The effect of temperature on gas solubility is extremely important.
Oxygen gas, O2(g), dissolved in water is essential for the survival of living things in water. Oxygen gas is more soluble in cold water than hot water. So, if hot water from a power plant is discharged into a river, the resulting decrease in oxygen gas in the water can lead to the death of fish.

Worked Example of a Henry's Law Question

Question: At 25°C the carbon dioxide gas pressure inside a 1.00 L bottle of cola is 253 kPa.
Determine the amount in moles of carbon dioxide dissolved in the aqueous cola solution.

(K(CO2(g)) = 2,937 kPa/mol L -1 at 25°C)

(Based on the StoPGoPS approach to problem solving.)

Calculate moles of carbon dissolved in solution
n(CO2(g)) = ? mol

Extract the data from the question:

Assume this is an ideal solution.

(A) Henry's Law: P = Kc
P = partial pressure of gas
c = concentration of gas in solution
K = Henry's Law constant in suitable units and at the same temperature

(B) c = n/V
c = concentration of gas in mol L -1
n = amount of gas dissolved in moles
V = volume of solution in litres

Substitute the equation for concentration (B) into Henry's Law (A):

P = K × n/V

Rearrange the equation to find n:

n = PV/K


Solutions of Solids in Liquids

The dependence of solubility on temperature for a number of inorganic solids in water is shown by the solubility curves in Figure 9. Reviewing these data indicate a general trend of increasing solubility with temperature, although there are exceptions, as illustrated by the ionic compound cerium sulfate.

Figure 9. This graph shows how the solubility of several solids changes with temperature.

The temperature dependence of solubility can be exploited to prepare supersaturated solutions of certain compounds. A solution may be saturated with the compound at an elevated temperature (where the solute is more soluble) and subsequently cooled to a lower temperature without precipitating the solute. The resultant solution contains solute at a concentration greater than its equilibrium solubility at the lower temperature (i.e., it is supersaturated) and is relatively stable. Precipitation of the excess solute can be initiated by adding a seed crystal (see the video in the Link to Learning earlier in this module) or by mechanically agitating the solution. Some hand warmers, such as the one pictured in Figure 10, take advantage of this behavior.

Figure 10. This hand warmer produces heat when the sodium acetate in a supersaturated solution precipitates. Precipitation of the solute is initiated by a mechanical shockwave generated when the flexible metal disk within the solution is “clicked.” (credit: modification of work by “Velela”/Wikimedia Commons)


This video shows the crystallization process occurring in a hand warmer.


Medical Biotechnology and Healthcare

5.46.3.3.4 Lipidic Derivatives

Lipid solubility is a key factor in determining the rate at which a drug passively crosses the BBB. Diamorphine, a diacyl, more lipophilic, derivative of morphine, is an excellent example, as this compound crosses the BBB about 100 times more easily than its parent drug. Derivatizing peptides with lipids involves the blocking of polar functional moieties on the peptide backbone with groups that greatly enhance the lipid solubility of the peptide as polar functional groups promote hydrogen bonding and limit membrane permeability. Acetylation of the N-terminus of model peptides, amidation of the C-terminus, and methylation all reduce the hydrogen-bonding potential and increase lipophilicity and BBB transport. Point modifications of a BBB-impermeable polypeptide, horseradish peroxidize (HRP), with lipophilic (stearoyl) or amphiphilic (pluronic block copolymer) moieties, considerably enhances the transport of this polypeptide across the BBB and accumulation of the polypeptide in the brain, while maintaining its enzymatic activity. 22 The modifications of the HRP with amphiphilic block copolymer moieties through degradable disulfide links resulted in the most efficient delivery of HRP to the brain. 22 Stearoyl modifications of HRP improved its penetration by about 60% but also increased clearance from blood while amphiphilic block copolymer modifications increased the passage of HRP across the BBB but had no significant effect on plasma clearance so that uptake by brain was almost doubled. 22

Another lipophilic prodrug of D-Ala 2 , D-Leu 5 -Enkephalin (DADLE) created by esterification of the free C-terminal with cholesterol and by amidation of the free N-terminus with the 1,4-dihydrotrigonellinate gains access to the brain via passive nonsaturable processes due to its increased lipophilicity. 23 Following localization in the brain to the 1,4-dihydrotrigonellinate moiety undergoes an enzyme-mediated oxidation that converts the dihydrotrigonellinate to the hydrophilic, membrane-impermeable trigonellinate ion, trapping the prodrug beyond the BBB. 23 Increased antinociception was shown for this prodrug approach after IV administration of a formulation of the prodrug in a vehicle (mixture of dimethylsulfoxide, ethanol, and 50% aqueous 2-hydroxypropyl-β-cyclodextrin solution).

Halogenation of peptides has been shown to significantly enhance lipophilicity and BBB permeability, although this is highly dependent on the halogen used. Addition of chlorine on the phenylalanine 4 residue of DPDPE led to significant increase in BBB permeability, which was further increased by addition of two chlorine atoms.

Increasing drug compound lipophilicity still creates certain limitations, especially as highly lipid-soluble drugs may be extensively plasma protein bound. Plasma protein binding reduces the amount of free or exchangeable drug in the plasma, therefore compromising brain uptake. Furthermore, an increased rate of sequestration by cytochrome P450 and other enzymes tends to be increased with increase in lipophilicity as shown for barbiturates. Additionally, the site of modification must also be carefully considered, as receptor-binding affinity may be diminished if alterations are within the pharmacophore region, thus reducing biological activity. To complicate things further, factors such as the increase in molecular size following lipid derivatization, derivative stability, intracellular sequestration, nontarget organ uptake, volume of distribution, and susceptibility to P-gp efflux activity (which increases with increased lipophilicity) could all contribute to a reduced level of brain delivery.


Gas Solubility In Organic Solvents

The trend that gas solubility decreases with increasing temperature does not hold in all cases. While it is in general true for gases dissolved in water, gases dissolved in organic solvents tend to become more soluble with increasing temperature.

There are several molecular reasons for the change in solubility of gases with increasing temperature, which is why there is no one trend independent of gas and solvent for whether gases will become more or less soluble with increasing temperature.

Boundless vets and curates high-quality, openly licensed content from around the Internet. This particular resource used the following sources:


  1. Pour 10–20 cm 3 of soda water into the beaker and add a few drops of methyl red indicator to give a red solution.
  2. Remove the nail from the syringe and insert the plunger completely. Draw about 5 cm 3 of the soda water and indicator solution into the syringe. Place a syringe cap over the end of the syringe (or use a finger), pull the plunger out to the 50 cm 3 mark and lock it with the nail. Bubbles of carbon dioxide will be seen out-gassing from the water and the indicator will begin to turn orange. Shake the syringe to speed up the out-gassing.
  3. Hold the syringe vertically with the nozzle pointing upwards, remove the syringe cap and the nail, and push in the plunger to expel the gas but not the solution. Seal the syringe again and repeat the out-gassing cycle in step 2. More bubbles will be seen and the indicator will turn further towards a yellow colour. Several more such cycles can be repeated until the indicator becomes completely yellow.

A white background helps. Place the syringe next to the original red solution to emphasise the colour change.

An additional demonstration that shows the effect of temperature on the solubility of a gas, and the associated indicator colour changes, involves boiling some soda water containing a little methyl red indicator in a boiling tube. This will expel the carbon dioxide, which is less soluble at high temperatures, and shows the colour change of the indicator from red to yellow.

Soda water contains carbon dioxide that has been dissolved in it under pressure. The equilibria involved in this experiment are:

  1. CO2(g) ⇌ CO2(aq)
  2. CO2(aq) + H2O(l) ⇌ H2CO3(aq) (carbonic acid)
  3. H2CO3(aq) ⇌ H + (aq) + HCO3 – (aq) (hydrogencarbonate ions)
  4. HCO3 – (aq) ⇌ H + (aq) + CO3 2– (aq) (carbonate ions)

(For simplicity, teachers may prefer not to discuss this last equilibrium).

The solution of carbon dioxide is thus acidic because of the increase in concentration of H + (aq) ions resulting from these reactions. Reducing the pressure causes CO2 to come out of solution, ie equilibrium 1 moves to the left. The result is that the other three equilibria also move to the left, removing H + (aq) ions from the solution and making the solution less acid.

Additional information

This is a resource from the Practical Chemistry project, developed by the Nuffield Foundation and the Royal Society of Chemistry. This collection of over 200 practical activities demonstrates a wide range of chemical concepts and processes. Each activity contains comprehensive information for teachers and technicians, including full technical notes and step-by-step procedures. Practical Chemistry activities accompany Practical Physics and Practical Biology.


Abstract

By means of the saturation shake-flask technique, the saturation solubility data of 3,3′-diaminodiphenyl sulfone in 14 monosolvents (n-propanol, N,N-dimethylformamide, methanol, ethanol, ethylene glycol, cyclohexane, acetonitrile, isopropanol, water, n-butanol, ethyl acetate, 1,4-dioxane, isobutanol, and 1-heptanol) was achieved at temperatures from 283.15 to 328.15 K and ambient pressure (p = 101.2 kPa). No phenomenon of crystalline form transformation or solvation occurred after dissolution of 3,3′-diaminodiphenyl sulfone in different solvents. The solubility values (mole fraction) of 3,3′-diaminodiphenyl sulfone in above 14 solvents increased as the temperature elevated and obeyed the decreasing tendency in different monosolvents as follows: N,N-dimethylformamide > ethyl acetate > acetonitrile > ethylene glycol > 1,4-dioxane > methanol > ethanol > isobutanol > n-propanol > n-butanol > isopropanol > 1-heptanol > water > cyclohexane. The mutual miscibility of solvent and 3,3′-diaminodiphenyl sulfone was explained through the three-dimensional Hansen solubility parameter. Examination was carried out for molecular interactions between the solute–solvent and solvent–solvent species by means of the linear solvation energy relationships. The solubility data obtained through experiments was correlated by the use of four models/equations, namely, the NRTL model, Apelblat equation, λh equation, and Wilson model. Correlation resulted in the maximum root-mean-square and relative average deviation values of, respectively, 397.1 × 10 –5 and 7.57 × 10 –2 . The Apelblat equation gave lower relative average deviations than the other models/equations for a certain neat solvent. Also, the mixing thermodynamic properties, infinite-dilution activity coefficient, and reduced excess enthalpy were obtained in terms of the Wilson model.


Solubility

Solubility is a property referring to the ability for a given substance, the solute, to dissolve in a solvent.

It is measured in terms of the maximum amount of solute dissolved in a solvent at equilibrium.

The resulting solution is called a saturated solution.

Certain substances are soluble in all proportions with a given solvent, such as ethanol in water.

This property is known as miscibility.

Under various conditions, the equilibrium solubility can be exceeded to give a so-called supersaturated solution, which is metastable.

The solvent is often a solid, which can be a pure substance or a mixture.

The species that dissolves, the solute, can be a gas, another liquid, or a solid.

Solubilities range widely, from infinitely soluble such as ethanol in water, to poorly soluble, such as silver chloride in water.

The term insoluble is often applied to poorly soluble compounds, though strictly speaking there are very few cases where there is absolutely no material dissolved.

The process of dissolving, called dissolution, is relatively straightforward for covalent substances such as ethanol.

When ethanol dissolves in water, the ethanol molecules remain intact but form new hydrogen bonds with the water.

When, however, an ionic compound such as sodium chloride (NaCl) dissolves in water, the sodium chloride lattice dissociates into separate ions which are solvated (wrapped) with a coating of water molecules.

Nonetheless, NaCl is said to dissolve in water, because evaporation of the solvent returns crystalline NaCl.


Watch the video: 5 κόλπα με νερό που θα σας κάνουν να τρελαθείτε!5 tricks with water that will make you go crazy (October 2022).