In This Chapter
^ Seeking certainty about solubility ^ Keeping tabs on concentrations ^ Making dilutions
Compounds can form mixtures. When compounds mix completely, right down to the level of individual molecules, we call the mixture a Solution. Each type of compound in a solution is called a Component. The component of which there is the most is usually called the Solvent. The other components are called Solutes. Although most people think "liquid" when they think of solutions, a solution can be a solid, liquid, or gas. The only criterion is that the components are completely intermixed. We explain what you need to know in this chapter.
Seeing Different Forces at Work in Solubility
For gases, forming a solution is a straightforward process. Gases simply diffuse into a common volume (see Chapter 11 for more about diffusion). Things are a bit more complicated for condensed states like liquids and solids. In liquids and solids, molecules or ions are crammed so closely together that Intermolecular forces Are very important. Examples of these kinds of forces include ion-dipole, dipole-dipole, hydrogen bonding, and London (dispersion) forces. We touch on the physical underpinnings of these forces in Chapter 5.
Introducing a solute into a solvent initiates a tournament of forces. Attractive forces between solute and solvent compete with attractive solute-solute and solvent-solvent forces. A solution forms only to the extent that solute-solvent forces dominate over the others. The process in which solvent molecules compete and win in the tournament of forces is called Solvation Or, in the specific case where water is the solvent, Hydration. Solvated solutes are surrounded by solvent molecules. When solute ions or molecules become separated from one another and surrounded in this way, we say they’re Dissolved.
Imagine that the members of a ridiculously popular "boy band" exit their hotel to be greeted by an assembled throng of fans and the media. The band members attempt to cling to each other, but are soon overwhelmed by the crowd’s ceaseless, repeated attempts to get closer. Soon, each member of the band is surrounded by his own attending shell of reporters and hyperventilating teenage girls. So it is with dissolution.


A.
The tournament of forces plays out differently among different combinations of components. In mixtures where solute and solvent are strongly attracted to one another, more solute can be dissolved. One factor that always tends to favor dissolution is Entropy, A kind of disorder or "randomness" within a system. Dissolved solutes are less ordered than undissolved solutes. Beyond a certain point, however, adding more solute to a solution doesn’t result in a greater amount of solvation. At this point, the solution is in dynamic equilibrium; the rate at which solute becomes solvated equals the rate at which dissolved solute Crystallizes, Or falls out of solution. A solution in this state is Saturated. By contrast, an Unsaturated Solution is one that can accommodate more solute. A Supersaturated Solution is one in which more solute is dissolved than is necessary to make a saturated solution. A supersaturated solution is unstable; solute molecules may crash out of solution given the slightest perturbation. The situation is like that of Wile E. Coyote who runs off a cliff and remains suspended in the air until he looks down — at which point he inevitably falls.
The concentration of solute required to make a saturated solution is the Solubility Of that solute. Solubility varies with the conditions of the solution. The same solute may have different solubility in different solvents, and at different temperatures, and so on.
When one liquid is added to another, the extent to which they intermix is called Miscibility. Typically, liquids that have similar properties mix well — they are Miscible. Liquids with dissimilar properties often don’t mix well — they are Immiscible. This pattern is summarized by the phrase, "like dissolves like." Alternately, you may understand miscibility in terms of the Italian Salad Dressing Principle. Inspect a bottle of Italian salad dressing that has been sitting in your refrigerator. Observe the following: The dressing consists of two distinct layers, an oily layer and a watery layer. Before using, you must shake the bottle to temporarily mix the layers. Eventually, they’ll separate again because water is polar and oil is nonpolar. (See Chapter 5 if the distinction between polar and nonpolar is lost on you.) Polar and nonpolar liquids mix poorly, though occasionally with positive gastronomic consequences.
Similarity or difference in polarity between components is often a good predictor of solubility, regardless of whether those components are liquid, solid, or gas. Why is polarity such a good predictor? Because polarity is central to the tournament of forces that underlies solubility. So, solids held together by ionic bonds (the most polar type of bond) or polar covalent bonds tend to dissolve well in polar solvents, like water. For a refresher on ionic and covalent bonding, visit Chapter 5.
Sodium chloride dissolves more than 25 times better in water than in methanol. Explain this difference, referring to the structure and properties of water, methanol, and sodium chloride.
Sodium chloride (NaCl) is an ionic solid, a lattice composed of sodium cations (atoms with positive charge) alternating with chlorine anions (atoms with negative charge). The lattice has a highly regular, idealized geometry and is held together by ionic bonds, the most polar
Type of bond. To dissolve NaCl, a solvent must be able to engage in very polar interactions with these ions and do so with near-ideal geometry. The structure and properties of water (which is polar) are better suited to this task than are those of methanol (see the following figure). The two O-H bonds of water (on the left) partially sum to produce a strong dipole along the mirror image plane of the molecule that runs between the two hydrogen atoms. Methanol (on the right) is also polar, due largely to its
Own O-H bond, but is less polar than water. In solution, water molecules can orient their dipoles cleanly and in either of two directions to interact favorably with Na+ or Cl – ions. Methanol molecules can engage in favorable interactions with these ions too, but not nearly as well as water.
A)
H H
B)
H
H
H
1. Lattice energy Is a measure of the strength of the interactions between ions in the lattice of an ionic solid. The larger the lattice energy, the stronger the ion-ion interactions. Here is a table of ionic solids and their associated lattice energies. Predict the rank order of solubility in water of these ionic solids.
Sodium Salt Lattice Energy, kJ mol"1
NaBr 747
NaCl
NaF
NaI
787 923 704
Solve It
2. Ethanol, CH3CH2OH, is miscible with water. Octanol, CH3(CH2)7OH, isn’t miscible in water. Is sucrose (as in table sugar) likely to be more soluble in ethanol or octanol? Why?
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Altering Solubility with Temperature

Increasing temperature magnifies the effects of entropy on a system. Because the entropy of a solute is usually increased when it dissolves, increasing temperature usually increases solubility — for solid and liquid solutes, anyway. Another way to understand the effect of temperature on solubility is to think about heat as a reactant in the dissolution reaction:
Solid solute + Water + Heat — Dissolved solute
Heat is usually absorbed when a solute dissolves. Increasing temperature corresponds to added heat. So, by increasing temperature you supply a needed reactant in the dissolution reaction. (In those rare cases where dissolution releases heat, increasing temperature can decrease solubility.)

A.
Gaseous solutes behave differently than do solid or liquid solutes with respect to temperature. Increasing the temperature tends to decrease the solubility of gas in liquid. To understand this pattern, check out the concept of vapor pressure. (If creaking sounds emanate from your skull as you try to remember what vapor pressure is about, take a peek at Chapter 11.) Increasing temperature increases vapor pressure because added heat increases the kinetic energy of the particles in solution. With added energy, these particles stand a greater chance of breaking free from the intermolecular forces that hold them in solution. A classic, real-life example of temperature’s effect on gas solubility is carbonated soda. Which goes flat (loses its dissolved carbon dioxide gas) more quickly: warm soda or cold soda?
The comparison of gas solubility in liquids with the concept of vapor pressure highlights another important pattern: Increasing pressure increases the solubility of a gas in liquid. Just as high pressures make it more difficult for surface-dwelling liquid molecules to escape into vapor phase, high pressures inhibit the escape of gases dissolved in solvent. The relationship between pressure and gas solubility is summarized by Henry’s Law:
Solubility = Constant x Pressure
The "constant" is Henry’s Constant, And its value depends on the gas, solvent, and temperature. A particularly useful form of Henry’s Law relates the change in solubility (S) that accompanies a change in pressure (P) Between two different states:
S1 / P1 = S2 / P2
According to this relationship, tripling the pressure triples the gas solubility, for example.
Henry’s Constant for dinitrogen gas in water at 293K is 0.69 x 10-3 mol L-1 atm-1. The partial pressure of dinitrogen in air at sea level is 0.78 atm. What is the solubility of N2 in a glass of water at 20°C sitting on a coffee table within a beach house?
0.54 X 10"3 mol L"1. This problem requires the direct application of Henry’s Law.
The glass of water is at 20°C, which is equivalent to 293K (just add 273 to any Celsius temperature to get the Kelvin equivalent). Because the glass sits within a beach house, we can assume the glass is at sea level. So, we can use the provided values for Henry’s Constant and the partial pressure of N2.
Solubility = (0.69 x 10-3 mol L-1 atm-1) x (0.78 atm) = 0.54 x 10-3 mol L-1
3. A chemist prepares an aqueous solution of cesium sulfate, Ce2(SO4)3, swirling the beaker in her gloved hand to promote dissolution. She notices something, momentarily furrows her brow, and then smiles knowingly. She nestles the beaker into a bed of crushed ice within a bucket. What did the chemist notice, why was she briefly confused, and why did she place the dissolving cesium sulfate on ice?
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Deep-sea divers routinely operate under pressures of multiple atm. One malady these divers must be concerned with is "the bends," a dangerous condition that occurs when divers rise too quickly from the depths, resulting in the over-rapid release of gas from blood and tissues. Why do the bends occur?
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5. Reefus readies himself for a highly productive Sunday afternoon of football watching, arranging bags of cheesy poofs and a six-pack of grape soda around his beanbag chair. At kickoff, Reefus cracks open his first grape soda and settles in for the long haul. Three hours later, covered in cheesy crumbs, Reefus marks the end of the fourth quarter by cracking open the last of the six-pack. The soda fizzes violently all over Reefus and the beanbag chair. What happened?
6. The grape soda preferred by Reefus (the
Gentleman introduced in Question 5) is bottled under 3.5 atm of pressure. Reefus lives on a bayou at sea level (hint: 1 atm). The temperature at which the soda is bottled is the same as the temperature in Reefus’s living room. Assuming that the concentration of carbon dioxide in an unopened grape soda is 0.15 mol L-1, what is the concentration of carbon dioxide in an opened soda that has gone flat while Reefus naps after the game?
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Solve It
Concentrating on Molarity and Percent Solutions
It seems that different solutes dissolve to different extents in different solvents in different conditions. How can anybody keep track of all these differences? Chemists do so by measuring Concentration. Qualitatively, a solution with a large amount of solute is said to be Concentrated. A solution with only a small amount of solute is said to be Dilute. As you may suspect, simply describing a solution as concentrated or dilute is usually about as useful as calling it "pretty" or naming it "Fifi." We need numbers. Two important ways to measure concentration are Molar-.ctABEfl Ity And Percent solution.
Molarity relates the amount of solute to the volume of the solution:
Molarity = (moles of solute) / (liters of solution)
To calculate molarity, you may have to use conversion factors to move between units. For example, if you’re given the mass of a solute in grams, use the molar mass of that solute to convert the given mass into moles. If you’re given the volume of solution in cm3 or some other unit, you need to convert that volume into liters.
The units of molarity are always mol L-1. These units are often abbreviated as M And referred to as "molar." Thus, 0.25M KOH(aq) is described as "Point two-five molar potassium hydroxide" and contains 0.25 moles of KOH per liter of solution. Note that this does Not Mean that there are 0.25 moles KOH per liter of Solvent (water, in this case) — only the final volume of the solution (solute plus solvent) is important in molarity.
Like other units, the unit of molarity can be modified by standard prefixes, as in millimolar (mM, 10-3 mol L-1) and micromolar (uM, 10-6 mol L-1).
Percent solution is another common way to express concentration. The precise units of per-jljjjl ) cent solution typically depend on the phase of each component. For solids dissolved in liquids, mass percent is usually used:
Mass % = 100% x-mass °fs°lute -
Total mass of solution
^ This kind of measurement is sometimes called a mass-mass percent solution because one mass is divided by another. Very dilute concentrations (as in the concentration of a contaminant in drinking water) are sometimes expressed as a special mass percent called Parts per million (ppm) Or Parts per billion (ppb). In these metrics, the mass of the solute is divided by the total mass of the solution, and the resulting fraction is multiplied by 106 (ppm) or by 109 (ppb).
Sometimes, the term Percent solution Is used to describe concentration in terms of the final volume of solution, instead of the final mass. For example:
"5% Mg(OH)2" can mean 5g magnesium hydroxide in 100 mL final volume. This is a mass-volume percent solution.
"2% H2O2" can mean 2 mL hydrogen peroxide in 100 mL final volume. This is a volume-volume percent solution.
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Clearly, it’s important to pay attention to units when working with concentration. Only by observing which units are attached to a measurement can you determine whether you are working with molarity, mass percent, or with mass-mass, mass-volume, or volume-volume percent solution.
Q. Calculate the molarity and the mass-volume percent solution obtained by dissolving 102.9g H3PO4 into 642 mL final volume of solution. Be sure to use proper units. (Hint: 642 mL = 0.642L)
A. First, calculate the molarity:
102.9gH2PO4 0.642L
Mol H 3PO 4 98.0g H3PO4
= 1.64M H3PO4
Next, calculate the mass-volume percent solution: 102.9gH2PO
642 mL
- x 100% = 16.0% mass/volume, or
16.0g H3PO
100 mL
Note that the convention in molarity is to divide moles by Liters, But the convention in mass percent is to divide grams by Milliliters. If you prefer to think only in terms of liters (not mil-liliters), then simply consider mass percent as kilograms divided by liters.
7. Calculate the molarity of these solutions:
A. 2.0 mol NaCl in 0.872L solution
B. 93g CuSO4 in 390 mL of solution
C. 22g NaNO3 in 777 mL of solution
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8. How many grams of solute are in each of these solutions?
A. 671 mL of 2.0MNaOH
B. 299 mL of 0.85MHCl
C. 2.74L of 258 mMCa(NO3)2
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9. A 15.0MSolution of ammonia, NH3, has density 0.90g mL-1. What is the mass percent of this solution?
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10. A chemist dissolves 2.5g of glucose, C6H12O6, into 375 mL of water. What is the mass percent of this solution? Assuming negligible change in volume upon addition of glucose, what is the molarity of the solution?
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Changing Concentrations by Making Dilutions
Real-life chemists in real-life labs don’t make every solution from scratch. Instead, they make concentrated Stock solutions And then make Dilutions Of those stocks as necessary for a given experiment.
To make a dilution, you simply add a small quantity of a concentrated stock solution to an amount of pure solvent. The resulting solution contains the amount of solute originally taken from the stock solution, but disperses that solute throughout a greater volume. So, the final concentration is lower; the final solution is less concentrated and more dilute.

But how do you know how much of the stock solution to use and how much of the pure solvent to use? It depends on the concentration of the stock and on the concentration and volume of the final solution you want. You can answer these kinds of pressing questions by using the dilution equation, which relates concentration (C) And volume (V) Between initial and final states:
C1 X V1 = C2 X V2

This equation can be used with any units of concentration, provided the same units are used throughout the calculation. Because molarity is such a common way to express concentration, the dilution equation is sometimes expressed in the following way, where M1 and M2 refer to the initial and final molarity, respectively:
M1 X V1 = M2 X V2

Q. How would you prepare 500 mL of
200 mM NaOH(aq), given a stock solution of 1.5M NaOH?
A. Add 67 mL 1.5M NaOH stock solution to 433 mL water.
Use the dilution equation: M1 x V1 = M2 x V2
The initial molarity, M1, derives from the stock solution, and so is 1.5M. The final molarity is the one you want in your final solution, 200 mM, Which is equivalent to 0.200M. The final volume is the one you want for your final solution, 500 mL,
Which is equivalent to 0.500L. Using these known values, you can calculate for the initial volume, V1:
V1 = (0.200MX 0.500L) / 1.5M= 6.7 x 10-2L
The calculated volume is equivalent to 67 mL. The final volume of the aqueous solution is to be 500 mL. 67 mL of this volume derives from the stock solution. The remainder, 500 mL – 67 mL = 433 mL, derives from pure solvent (water, in this case). So, to prepare the solution, add 67 mL 1.5MStock solution to 433 mL water. Mix and enjoy.
11. What is the final concentration of a solution prepared by diluting 2.50 mL of 3.00M KCl(aq) up to 0.175L final volume?
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12. A certain mass of ammonium sulfate,
(NH4)2SO4, is dissolved in water to produce 1.65L of solution. 80.0 mL of this solution is diluted with 120 mL of water to produce 200 mL of 200 mM(NH4)2SO4. What mass of ammonium sulfate was originally dissolved?
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Answers to Questions on Solutions
By this point in the chapter, your brain may feel as if it has itself dissolved. Check your answers, boiling away that confusion to reveal crystalline bits of hard-earned knowledge. In other words, make sure you know what you’re doing. Solutions are critically important. Really.
D The rank order from most to least soluble is: Nal, NaBr, NaCl, NaF. As the question indicates, the larger the lattice energy is, the stronger the forces holding together the ions. Dissolving those ions means outcompeting those forces; a solution forms when attractive solute-solvent forces dominate over others (such as solute-solute bonds). So, salts with lower lattice energy are typically more soluble than those with higher lattice energy.
CM Sugar should be more soluble in ethanol than in octanol. Like dissolves like. Chemists know from experience that sugar dissolves well in water. Therefore, we expect sugar to dissolve best in solvents that are most similar to water. Because ethanol is more miscible with water than is octanol, we expect that ethanol has solvent properties (especially polarity) more like water than does octanol.
CM The chemist noticed as she swirled the beaker of dissolving cesium sulfate that the beaker
Was becoming noticeably warmer. This observation momentarily confused her, because it suggested that the dissolution of cesium sulfate released heat, a state of affairs opposite to that usually observed with dissolving salts. Having diagnosed the situation, she cleverly turned it to her advantage. With typical salts, increasing temperature increases solubility in water, so heating a dissolving mixture can promote dissolution. In the case of cesium sulfate, however, the reverse is true: By cooling the dissolving mixture, the chemist promoted solubility of the cesium sulfate.
MM At the high pressures to which deep-sea divers are exposed during their dives, gases become more soluble in the blood and tissue fluids due to Henry’s Law (Solubility = Constant X Pressure). So, when the divers do their thing at great depth, high concentrations of these gases dissolve into the blood. If the divers rise to the surface too quickly at the end of a dive, the solubility of these dissolved gases changes too quickly in response to the diminished pressure. This situation can lead to the formation of tiny gas bubbles in the blood and tissues. These bubbles can be deadly.
MM Nothing dramatically fizzy happened when Reefus opened the first soda because that soda was still cold from the refrigerator. As the game progressed, however, the remaining sodas warmed to room temperature as they sat beside Reefus’s beanbag chair. Gases (like carbon dioxide) are less soluble in warmer liquids. So, when Reefus opened the warm, fourth-quarter soda, a reservoir of undissolved gas burst forth from the can.
CM 4.3 X 10"2 mol L"1. To solve this problem, use the two-state form of Henry’s Law:
51 / P1 = S2 / P2
The initial solubility and pressure are 0.15 mol L-1 and 3.5 atm, respectively. The final pressure is 1.0 atm. Using these known values, solve for the final solubility:
52 = (0.15 mol L-1 / 3.5 atm) X 1.0 atm = 4.3 X 10-2 mol L-1
MM Solve these kinds of problems by using the definition of molarity and conversion factors: A. 2.3MNaCl
2.0Am0o’NaCl = 2.3M NaCl 0.872L
B. 1.5M CuSO4
93gCuSO4 103 mL molCuSO
390 mL
^ = 1-5M CuSO4 160gCuSO4 4
C. 0.33M NaNO3
22iNaN, O3 X L°^mL X J??^ = 0.33M NaNO3 777 mL L 85.0gNaNO3
Efl Again, conversion factors are the way to approach these kinds of problems. Each problem features a certain volume of solution that contains a certain solute at a certain concentration. Begin each calculation with the given volume. Then convert to moles by multiplying the volume by the concentration. Finally, convert from moles to grams by multiplying by the molar mass of the solute.
A. 54g NaOH
671 mL x L x 2.0 mol NaOH x 40.0gNaOH = 54gNaOH 1 10 3 mL L mol NaOH
B. 9.3g HCl
299 mL x L x 0.85 mol HCl x 36.5gHCl = 9 3gHCl 1 103 mL L mol HCl
C. 116g Ca(NO3)2
2 74L 258 mmol Ca (NO3 )2
Mol x 164gCa (NO3 )2
10
Mmol mol Ca(NO3 )2
= 116gCa (NO3 )2
H 28%. To calculate mass percent, you must know the mass of solute and the mass of solution. The molarity of the solution tells you the moles of solute per volume of solution. Starting with this information, you can convert to mass of solute by means of the gram formula mass (see Chapter 7 for details on calculating the gram formula mass):
15.0 mol L-1 X 17.0g mol-1 = 255g L-1
So, each liter of 15.0M NH3 contains 255g NH3 solute. But how much mass does each liter of solution possess? Calculate the mass of the solution by using the density. Note that the problem lists the density in units of milliliters, so be sure to convert to the proper units:
1.0L solution X (0.90g / 1.0 X 10-3L) = 9.0 X 102g solution
So, 255g NH3 occur in every 900g of 15.0MNH3. Now you can calculate the mass percent:
Mass percent = 100% X (255g / 900g) = 28%
1
L
IfiJ The mass percent is 0.66%; the molarity is 3.7 X 10-2M To calculate the mass percent, you
Must use the estimate that 1.0 mL of water has 1.0g mass. This is a very good approximation at room temperature, and one with which you should be familiar. So, 375 mL water has 375g mass. Adding 2.5g glucose increases that mass to 377.5g for the final solution. Calculate the mass percent as follows:
Mass % = 100% (2.5g / 377.5g) = 0.66%
To calculate the molarity, you must know the final volume of the solution. Although adding 2.5g to 375 mL water increases the volume from 375 mL, the increase is very small compared to the volume of the water. So, 375 mL is a good approximation of the final volume of the solution. Next, convert from grams of glucose to moles of glucose by means of the gram formula mass:
Moles glucose = 2.5g glucose (1 mol / 180.2g) = 1.4 10-2 mol glucose
Now that you know the moles of glucose and the final volume of solution, calculating molarity is easy:
Molarity = (1.4 10-2 mol glucose) / (0.375L solution) = 3.7 10-2M
IF| 4.29 X 10-2M Use the dilution equation: M1 X V1= M2 X V2
In this problem, the initial molarity is 3.00M, the initial volume is 2.50 mL (or 2.50 X 103L), and the final volume is 0.175L. Use these known values to calculate the final molarity, M2:
M2 = (3.00 mol L-1 2.50 10-3L) / 0.175L = 4.29 10-2M
U 109g (NH4)2SO4. First, use the dilution equation to find the concentration of the original solution:
M1 = (200 X 10-3 mol L-1 X 200 X 10-3L) / (80.0 X 10-3L) = 0.500M
This calculation means that the original solution contained 0.500 mol (NH4)2SO4 per liter of solution. The question indicates that 1.65L of this original solution were prepared, so:
I65i 0.500 mmol (NH4) SO4 132g(NH4) SO4 , .
I^k X–i-^–x-^— = 109g (NH4) SO4
1 L mol (NH4)2 SO4 2
In This Chapter
^ Knowing what to expect in a typical consultation


Chiropractic adjustments open up the spaces between the spinal bones (vertebrae) and loosen up other joints, thereby increasing flexibility and range of motion. Adjustments also improve circulation through muscle tissue and help to reduce inflammation.
Research on other conditions has also been inconclusive. Some have shown good results for infant colic and headaches whereas others for these conditions, and also asthma, carpal tunnel syndrome (nerve compression in the wrist), and painful menstruation, have not. However, many of these studies have had design flaws and more good research is needed.
Neck adjustments should only be performed with great care in people with a history of stroke or high blood pressure. Particular care should also be taken with the elderly and young children or infants.
Chiropractic has a good safety record and the risk of serious complications from it is very low, although a very slight risk of complications with neck manipulations does exist.


Chiropractors may also give nutritional, ergonomic and general lifestyle advice.
At the end of your treatment, you may receive the following additional tips and advice:
To find a qualified chiropractor in your area, contact the General Chiropractic Council (GCC) (Tel: 020 7713 5155; 


U Insurance: To be a registered chiropractor, your practitioner must have appropriate indemnity insurance.
Some GP practices offer chiropractic treatment on the NHS. Ask your doctor for details. Some private health insurances cover chiropractic treatment. Check with your provider for advice.
1. Lie flat on the floor in a relaxed position with legs outstretched together.
In This Chapter-5.jpg)
$0
October $40 $760 $800
What drops you into the doughnut hole-135.jpg)
Ii$jAB£W You get out of the gap as soon as your Own Out-of-pocket spending on drugs since the beginning of the year reaches the limit set by law ($4,050 in 2008, $4,350 in 2009). In this case, what your Plan Has paid doesn’t count. And other rules (no surprise here) outline what counts as "out-of-pocket" spending and what doesn’t.
I Payments made by your plan for your drugs during the initial coverage period or in the doughnut hole. (Some plans, as you see later in this chapter, cover some drugs in the gap.)
Rules for buying your drugs in the doughnut hole-77.jpg)
Making sure your gap purchases count-157.jpg)
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-152.jpg)
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If you go into the gap, the statements also show how much more you have to spend to get out of it before you can start receiving catastrophic coverage. Statements are normally sent monthly, unless you haven’t filled any prescriptions in a particular month. You can also request a statement at any time by calling your plan. Some plans offer personalized access to their Web sites so enrollees can keep track of their status online.-61.jpg)
Every chart varies according to the drugs you take and the plan’s overall design. Figures 15-1 and 15-2 are two examples of what the charts look like.
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11th
Figure 15-2:-139.jpg)
7th
8th
However, you may be able to get other benefits that help fill in the doughnut hole, either completely or partially. In the following sections, I consider help given through employer or union health plans, State Pharmacy Assistance Programs, and the few Medicare drug plans that offer some coverage in the gap. (I outline other ways of reducing your drug costs, which may also help you avoid the doughnut hole or lessen its impact, in Chapter 16.)
Consumer advocates vigorously opposed this clause in the law, arguing that buying additional