Chemical Solution

Solution

"A homogenous mixture of two or more than two substances is called "Solution."

   There are two components of a solution.
   Solute
   Solvent

SOLUTE + SOLVENT= SOLUTION

SOLUTE

   Component of a solution which is in small quantity is called "Solute".

SOLVENT

   Component of solution which is in large quantity is called "Solvent".

Saturated solution

 A solution that can hold no more of the solute at a particular temperature is said to be a saturated solution at that temperature.

Unsaturated solution

An unsaturated solution is a solution, which contains less amount of solute than is required to saturate it    at that temperature.

Super saturated solution

  Solution that is more concentrated than a saturated solution is known as super saturated solution. If a crystal of solute is added to this solution, the excess of solute crystallizes.

Aqueous solution

A solution of any substance in which solvent is water is called "Aqueous solution".

Concentration of solution

Concentration of solution is defined as the amount of solute dissolved in a specific (fixed) amount of solvent.

Difference between Solvent and Solute

Solvent vs. Solute

A solution is a homogeneous mixture of two or more substances. It is called a homogenous mixture, because the composition is uniform throughout the solution. The components of a solution are mainly of two types, solutes and the solvents. Solvent dissolves the solutes and form a uniform solution. So, normally solvent amount is higher than the solute quantity.

What is a Solvent?

Solvent is a substance with dissolving capability, thus can dissolve another substance. Solvents can be in a liquid, gaseous or solid state. Most commonly, liquids are used as solvents. Among liquids, water is considered as a universal solvent, because it can dissolve many substances than any other solvent. Gas, solid or any other liquid solute can be dissolved in liquid solvents. In gas solvents, only gas solutes can be dissolved. There is a limit to the amount of solutes that can be added to a certain amount of solvent. The solution is said to be saturated if the maximum amount of the solute is added to the solvent. Solvents can be organic or inorganic. For example, ether, hexane, and methylene chloride are organic solvents, whereas water is an inorganic solvent. Solvents can be broadly categorized into two as polar solvents and non polar solvents. Polar solvent molecules have a charge separation, therefore, capable of dissolving polar solutes. In the dissolution process, dipole-dipole interactions or dipole-induced dipole interactions may occur. Polar solvents can be further divided as polar protic and polar aprotic solvents. Polar protic solvents are capable of hydrogen bond formation with the solutes. Therefore, they solvate anions by hydrogen bonding. Water and methanol are polar protic solvents. Polar aprotic solvents cannot form hydrogen bonds. However, they have large dipole moments, hence form dipole-dipole interactions with ionic solutes, therefore, solvate them. Acetone is a polar aprotic solvent. Non polar solvents dissolve non polar solutes. Hexane, benzene, and toluene are some common non-polar solvents. Other than the above classified solvents, there are some solvents, which have intermediate polar and non polar properties. According to “like dissolve like” phenomenon, solvents dissolve solutes, which match them.

Properties of solvents are essential to know when we use them in laboratories. For example, knowing the boiling points of solvents helps us to determine how to use distillation methods to separate them. Alternatively, the density of solvents is important in solvent extraction techniques. Volatility, toxicity, and flammability are some of other parameters, which we have to focus, when we are working with different solvents.

What is a Solute?

Solute is a substance that dissolves in a solvent in order to form a solution. Solutes can be in liquid, gaseous or solid phase. Normally, in a solution, solutes are in a lesser amount than the solvents. When a solution has the maximum amount of solutes it can dissolve, then the solution is said to be saturated. The dissolution of a solute in a solvent changes the properties of the solvents.

UNITS OF CONCENTRATION

   Molarity
   Molality
   Normality
   Percentage composition
   Mole fraction

  MOLARITY

Molarity is defined as the number of moles of solute dissolved in one liter (one dm³) of solution.

For example:    If 2 moles of a solute are present in one liter of a solution then the molarity of solution will be (2M).

FORMULA­₌

Molarity = number of moles of solute /volume of solution in liter

OR

Molarity = mass of solute (in grams) /molecular mass x volume of solution in liter

Molality vs. Molarity:
The 'Molarity' of a solution is the number of moles of solute in one liter of solution. To make a 1 N solution of NaOH would be the same as making a 1 Molar solution, (1 eq / mole) X (1 mole / liter). To make one liter of a 1 Molar solution, weigh out one mole of NaOH and slowly, with constant stirring and while monitoring the temperature of the solution (by touching the outside of the beaker), add it to about 750 ml of deionized water in a 1 liter beaker. (If the beaker gets warm to the touch, stop adding the NaOH and continue stirring until all the solid is dissolved and you are sure the solution is not overheating.) When all the NaOH has been added, bring the total volume up to 1.0 liter of solution by adding more deionized water.

To make a 0.1 N solution of NaOH you could follow the above procedure using 1/10 as much NaOH or you can dilute the above solution by a factor of 10. To do this, measure 100 ml of solution in a graduated cylinder. Slowly, and while stirring, add this to about 750 ml of deionized water in a 1 liter beaker. (ALWAYS add the more concentrated solution to the less concentrated solution!) Again, monitor the temperature of the resulting solution. When all the solution from the graduated cylinder has been added, rinse the graduated cylinder several times with 10 or 20 ml of deionized water and finally, bring the total volume of the solution up to 1 liter.


These two words sound similar but they are not synonyms, even though both of them are used for representing solution concentration. By definition, molarity is the number of moles of solute dissolved per liter of solution. We use capital letter “M” to represent molarity and its formula is M= (# mol SOLUTE)/ (Liters of SOLUTION).
Molality is then the number of moles of solute per kilogram of the SOLVENT, NOT solution! We use lower case letter “m” to represent molality and its formula can be represented as: m= (# mole SOLUTE) / (Kilograms of SOLVENT).
Most of the time scientists use either molarity or molality to represent solution concentration, but MOLALITY is preferred when the temperature of the solution varies. That is because MOLALITY does not depend on temperature, (Neither number of moles of solute nor mass of solvent will be affected by changes of temperature.) while MOLARITY changes as temperature changes. (Volume of solution in the formula changes as temperature changes, and that is why.)

Normality

When you need to compare solutions on the basis of concentration of specific ions or the amount of charge that the ions have, a different measure of concentration can be very useful. It is called normality.

The general formula utilized for determining the normality of a solution is:

Normality X Equivalent Weight X Volume = Grams

Equivalents/liter X Grams/equivalent X Liters = Grams needed

The 'Normality' of a solution is the 'Molarity' multiplied by the number of equivalents per mole (the number moles of hydroxide or hydronium ions per mole) for the molecule. For NaOH there is one equivalent per mole (one mole of hydroxide ions release per mole of NaOH dissolved in water) so the 'Normality' is the 'Molarity' times 1 eq / mole.



PERCENTAGE OF COMPOSITION OF SOLUTION

   Concentration of a solution can be expressed in terms of percentage composition. It is based on the weight or volume of the components of solution.

Weight / Weight %

   2% w / w solution means grams of solute is dissolved in 100 grams of solution.

Weight / volume %

   4% w / v solution means 4 grams of solute is dissolved in 100 ml of solution.

Volume / weight %

   3% v/ w solution means 3 ml of solute is dissolved in 100 grams of solution.

Volume / volume %

   5% v / v solution means 5 ml of solute is dissolved in 100 ml of solution.

Percent concentration is calculated using the following formula:

What are Mixtures and Solutions?

A MIXTURE is a combination of two or more substances that are not chemically united and do not exist in fixed proportions to each other. Most natural substances are mixtures. In the graphic on the left there are four substances - water, alcohol, oil, and food color dye.
MIXTURES	PURE COMPOUNDS
A mixture can be physically separated into pure compounds or elements.	A pure compound has a constant composition with fixed ratios of elements.
Just about everything that you can think of is probably a mixture. Even the purest of materials still contain other compounds as impurities.	Although it is almost physically impossible to isolate absolutely pure substances, a substance is said to be pure if no impurities can be detected using the best available analytical techniques.
Mixtures may exhibit a changing set of physical properties. For example, mixture of alcohol and water boils over a range of temperatures.	Physical properties such as boiling point or melting point of pure substances are invariant. For example, pure water boils at 100 degrees C

HOMOGENEOUS MIXTURES AND HETEROGENEOUS MIXTURES

HOMOGENEOUS MIXTURES	HETEROGENEOUS MIXTURES
The prefixes "homo"- indicate sameness	The prefixes: "hetero"- indicate difference.
A homogeneous mixture has the same uniform appearance and composition throughout. Many homogeneous mixtures are commonly referred to as solutions.	A heterogeneous mixture consists of visibly different substances or phases. The three phases or states of matter are gas, liquid, and solid. Graphic on the left of "Dancing Raisins" shows liquid, solid, and gas substances in a heterogeneous mixture.
Particle size distinguishes homogeneous solutions from other heterogeneous mixtures. Solutions have particles which are the size of atoms or molecules - too small to be seen. A colloid is a homogeneous solution with intermediate particle size between a solution and a suspension. Colloid particles may be seen in a beam of light such as dust in air in a "shaft" of sunlight. Milk, fog, and jello are examples of colloids.	In contrast a suspension is a heterogeneous mixture of larger particles. These particles are visible and will settle out on standing. Examples of suspensions are: fine sand or silt in water or tomato juice.
Corn oil is homogeneous, White vinegar is homogeneous. A sugar solution is homogeneous since only a colorless liquid is observed. Air with no clouds is homogeneous.	For example beach sand is heterogeneous since you can see different colored particles. Vinegar and oil salad dressing is heterogeneous since two liquid layers are present, as well as solids. Air with clouds is heterogeneous, as the clouds contain tiny droplets of liquid water.

ppm

While molarity is used for solutions with relatively large concentrations of solute, we use ppm and mg/L to denote much lower concentrations.  This will often be the case when you are measuring solutes in water at a treatment plant, such as the concentration of iron in the water.  We will discuss ppm in this section, then move on to mg/L in the next section. 

Parts per million, or ppm, is just what the name suggests - the number of parts of solute in one million parts of solution.  Concentration in ppm is calculated using the following formula:

Let's consider a simple example:
You add 11 mg of sulfuric acid to 2,000 grams of water.  What is the resulting concentration of sulfuric acid, in ppm?

In order to solve this problem, you first must make the units of the solute the same as the units of the solvent.  So, you will convert from milligrams to grams:

Then you simply plug the numbers into the formula:

So, the concentration of sulfuric acid in the resultant solution is 5.5 ppm. 

Converting Between Types of Concentration

Next, you will be called on to convert between different types of concentration.  For example, if you had a 35.7 ppm solution, what would this be in percent concentration?  If you had a 0.2 M solution, could you convert this to mg/L? 

We've already mentioned a few conversion factors previously.  In aqueous solutions, the following conversion factors are in effect:
1 mg/L = 1 ppm
1,000,000 ppm = 100%
1,000,000 mg/L = 100%

Converting to molarity is a little more complicated.  You must use the following formula to convert from mg/L (or ppm) to molarity in an aqueous solution:

mg/L

Milligrams per liter, or mg/L, can be used to denote concentration in similar circumstances to ppm.  The following formula is used to calculate concentration in mg/L:

As you can see, the primary difference between the two calculations is that ppm is a mass per mass calculation while mg/L is a mass per volume calculation.  Due to the special characteristics of water, the concentration of an aqueous solution is the same when calculated in mg/L as it is when calculated in ppm.  So, the 5.5 ppm sulfuric acid aqueous solution discussed in the last section has a concentration of 5.5 mg/L.

Let's consider the following example problem:

You dissolve 1 mg of salt in water to produce 2 liters of solution.  What is the concentration of salt in the solution?

You could choose to calculate the concentration as either ppm or mg/L, but mg/L is the better choice since you are given the amount of solution in liters.  We would calculate the answer as follows:

You can state the concentration as either 0.5 mg/L or 0.5 ppm.

Dilution

Introduction
In lab, you will often be given a stock solution which you will need to dilute to a given concentration for use in a lab exercise.  Dilution consists of adding more solvent to a solution so that the concentration of the solute becomes lower.  The total number of solutes in the solution remains the same after dilution, but the volume of the solution becomes greater, resulting in a lower molarity, ppm, mg/L, or % concentration. 

In the picture above, I've shown the solute as yellow dots and the solvent as solid blue.  The 1 L beaker on the left shows the initial concentration, which we might represent as 13 dots/L.  The beaker on the right is the result of dilution of the left beaker.  We added more solvent so that the solution's total volume was 3 L.  As a result, the concentration of the diluted beaker is (13 dots)/(3 L), or 4.3 dots/L. 



Calculating Dilution

Dilution calculations are simplified by using the following equation:
M₁V₁ = M₂V₂

Where:

M₁ = concentration of the first solution
V₁ = volume of the first solution
M₂ = concentration of the second solution
V₂ = volume of the second solution

Concentration and volume in the equation above can have any units as long as the units are the same for the two solutions. 

As long as you know three of the four values from the equation above, you can calculate the fourth.  Let's consider a sample problem:
You have 1 L of a 0.125 M aqueous solution of table sugar.  You want to dilute the solution to 0.05 M.  What do you do?

To solve the problem, you simply plug in the three numbers you know:
(0.125 M) (1 L) = (0.05 M) V₂
2.5 L = V₂

Using the equation, you determine that the volume of the diluted solution should be 2.5 L.  So we simply add enough water to the first solution so that the solution's volume becomes 2.5 L. 

More Complicated Calculations

Dilution calculations run the gamut from very simple to very complicate depending on the situation.  Let's consider a more complicated situation so that you'll be prepared for any dilution you may be asked to perform in lab. 
Your plant feeds 50 ppm chlorine using a solution feeder.  You need to fill the 10 gallon tank of the solution feeder, but all you have on hand is a stock solution of 60% sodium hypochlorite.  How much of the sodium hypochlorite do you need to mix with water to fill the tank?

This problem is more complicated because it contains two different units of concentration - ppm and percent.  The first step in the problem is to convert them both to the same measurement of concentration.  We'll convert the 60% sodium hypochlorite into ppm, using one of the equations from the last page:

Now we can plug our known values into the equation:
(600,000 ppm) V₁ = (50 ppm) (10 gal)

V₁ = 0.0008 gal

Finally, we have to convert our answer into a manageable form since it would be impossible to measure out 0.0008 gal.  Let's convert it to mL:

So, in order to achieve a concentration of 50 ppm of chlorine in our solution feeder's tank, we must add 3 mL of the 60% sodium hypochlorite and then fill the rest of the tank with water. 

Titration
We can measure the concentration of a solution by a technique known as a titration. The solution being studied is slowly added to a known quantity of a reagent with which it reacts until we observe something that tells us that exactly equivalent numbers of moles of the reagents are present. Titrations are therefore dependent on the existence of a class of compounds known as indicators.
The endpoint of an acid-base titration is the point at which the indicator turns color. The equivalence point is the point at which exactly enough base has been added to neutralize the acid.
Super saturation

Whenever a solution contains more solute than it can hold, it is said to be super saturated.  We have mentioned one supersaturated solution already in this lesson - the warm air which cooled and thus contained more water vapor than it could hold. 

Solutions can become supersaturated in a variety of ways, but in every case the supersaturated solution is unstable.  If more of the solute is added or if the conditions change in any way, the extra solute will settle out of the supersaturated solution.  In the water vapor and air solution, dust particles in the air provided the slight change which caused water vapor to settle out, forming clouds and rain. 
 
If you've ever made sweetened iced tea, you will have taken advantage of the characteristics of a supersaturated solution.  By adding sugar to hot tea, you were able to dissolve much more sugar into the water than you would have been able to dissolve into cold tea.  When the tea cooled, the additional sugar remained dissolved in the tea as a supersaturated solution.  If you tried to add more sugar to the cooled tea, however, the excess solute would drop out of solution and the tea would become less sweet. 

Layering

The layering seen in a solution which has passed its saturation point resembles the layering seen when two insoluble substances are mixed.  But, as you can see in the illustration below, these two situations are actually quite different.  One of the layers in the saturated solution contains both solute and solvent while the layers in the insoluble mixture contain only one substance per layer. 

The extra solute added to a saturated solution may settle either to the top or to the bottom of the solution.  The location of the extra solute depends on its density, a concept we will discuss in a later lesson. 

Solubility Product

The concentrations of ions in saturated solutions have a relationship to one another somewhat like the relationship between the concentration of H₃O⁺ and OH^- in water.

Sodium Chloride

Consider saturated sodium chloride solution. Quite a bit of sodium chloride can be dissolved in water, about 6 moles in one liter. That makes the concentration of both the sodium ion and the chloride ion about 6 M.
What happens if we increase the concentration of Cl- by adding some concentrated hydrochloric acid (12M HCl)?. The saturated NaCl is in the test tube and the concentrated HCl is in the dropper. To see the reaction move your mouse over the picture.
If nothing happened, we would still have 6M Na+ and a higher concentration, perhaps, 8M, of Cl-. But something does happen. Crystals of NaCl form from the reaction of some of the extra Cl- with some of the Na+ that was in the solution. The concentration of Na+ goes down to around 5 M as the conc of Cl- increases to somewhere around 7M.	Approximate Concentrations [Na+] [Cl-] Start 6M 6M Potential 6M 8M Actual 5M 7M

As the concentration of one ion increases, the concentration of the other ion decreases. Just as there was an equation that related the concentrations of the dissociated ions of water, there is an equation that relates the concentrations of the dissociated ions of sodium chloride.
Let me draw the parallel.

Water ionizes to form H3O+ and OH-.	2 H2O--> H3O+ + OH-
The reaction is reversible.	2 H2O<---> H3O+ + OH-
The concentrations of H3O+ and OH- are related by this equation:	Kw = [H3O+] [OH-]
When multiplied together, the concentrations of H3O+ and OH- give a fairly constant value called the ionization constant of water, or Kw.

Now sodium chloride.

Sodium chloride dissolves and dissociates in water to Na+ and Cl-.	NaCl(s)---> Na+ + Cl-
The reaction is reversible:	NaCl(s)<---> Na+ + Cl-
The concentrations of Na+ and Cl- are realated by this equation:	Ksp = [Na+][Cl-]
When multiplied together, the concentrations of Na+ and Cl- give a fairly constant value called the solubility product constant, or Ksp. For sodium chloride, the value of Ksp is about 36.

Practice

See if you can figure out what the concentration of Na⁺ would be if we were able to increase the concentration of Cl^- up to 10 M. Take a moment to figure that out.

Answer

You should have calculated about 3.6 M for the sodium ion concentration. I got that value by saying that the concentration of Na+ times the concentration of Cl- is equal to 36 (the Ksp value for sodium chloride). If the concentration of Cl- is going to be 10 M, then the concentration of of Na+ has to be 36 divided by 10. That comes at to 3.6 M.

Ksp = [Na+]x[Cl-] [Na+] = Ksp ÷ [Cl-] [Na+] = 36 ÷ 10 [Na+] = 3.6M

Silver Chloride

The same line of reasoning can be used with any salt that dissolves in water, even if it dissolves only a very small amount.

Silver chloride will dissolve somewhat in water. However, it reaches saturation very quickly--that is, when the concentrations of silver and chloride ions are about 1.3 x 10-5M.	AgCl(s)<---> Ag+ + Cl-
Still we can write a solubility product equation for it.	Ksp = [Ag+][Cl-]
The value for the Ksp of silver chloride, however, is about 1.8 x 10-10, a very small number.	Ksp = 1.8 x 10-10

Practice

Try using that information to calculate the Ag⁺ concentration if the chloride ion concentration were 3.0 M.

Answer

In this case the answer turns out to be a very small number, which can be calculated using the process shown here.

Ksp = [Ag+]x[Cl-] [Ag+] = Ksp ÷ [Cl-] [Ag+] = (1.8 x 10-10) ÷ 3.0 [Ag+] = 6.0 x 10-11M

Lead(II) Chloride

When the formula for a salt contains more than just one of each ion, the solubility product equation gets a little more complicated.

Let's use PbCl2 as an example. When PbCl2 cissolved in water, two ions of Cl- are released for every one ion of Pb2+.	PbCl2(s)----> Pb2+ + Cl-
When saturation is reached we show that the reaction is reversible.	PbCl2(s) <----> Pb2+ + 2 Cl-
The equation for the solubility product is:	Ksp = [Pb2+][Cl-]2
The concentration of Cl- is squared because the balanced equation for the reaction shows a 2 as the coefficient in front of Cl-.
Looking at the reaction in this way might help you to remember that:	PbCl2(s) Pb2+ + Cl- + Cl- then: Ksp = [Pb2+][Cl-][Cl-] thus: Ksp = [Pb2+] [Cl-]2

Example - Determining concentrations at equilibrium 

Here is an example of how to calculate the equilibrium concentration of one substance given the K_sp and the equilibrium concentrations of the other substance. (Also shown in example 23 in your workbook.)

For the reaction PbCl2(s) Pb+2(aq) + 2 Cl-(aq), what is [Pb+2] at equilibrium if [Cl-] = 2.0 x 10-3 M given that the Ksp = 2.0 x 10-5?	Ksp = [Pb+2] x [Cl]2
	2.0 x 10-5 = [Pb+2] x (2.0 x 10-3)2
	2.0 x 10-5 = [Pb+2] x 4.0 x 10-6
	2.0 x 10-5 4.0 x 10-6 = [Pb+2]
	5.0 M = [Pb+2]

Practice Problems: Equilibrium Concentrations

Try your hand at answering the following questions (also shown in exercise 24 in your workbook.) Check your answers below and then move on to the Wrap-Up.

The Ksp for AgCl is 1.8 x 10-10. If Ag+ and Cl- are both in solution and in equilibrium with AgCl. What is [Ag+] if [Cl-] = .020 M?

If Ag+ and Cl- were both present at 0.0001 M, would a precipitate occur?

What concentration of Ag+ would be necessary to bring the concentration of Cl- to 1.0 x 10-6 M or lower?

Answers: Equilibrium Concentrations

Here are the answers to the questions above.

The Ksp for AgCl is 1.8 x 10-10. If Ag+ and Cl- are both in solution and in equilibrium with AgCl. What is [Ag+] if [Cl-] = .020 M?

[Ag⁺] = 9.0 x 10^-9M

If Ag+ and Cl- were both present at 0.0001 M, would a precipitate occur?

Yes, a precipitate would occur because these concentrations together are higher than what the K_sp allows.

What concentration of Ag+ would be necessary to bring the concentration of Cl- to 1.0 x 10-6 M or lower?

[Ag⁺] = 9.0 x 10^-9M or higher

pH and K_w

When working with acidic or basic solutions, another way of expressing concentration is often used and that is pH. There is more to understanding pH than knowing its definition, so let's set the stage for pH by first looking at the self-ionization of water. After a definition of pH, we will then look at how the hydronium/hydroxide ion balance determines the acidity or basicity of a solution. The concentrations of hydronium and hydroxide ions in an aqueous solution are related to one another by the ionization constant for water, K_w. That constant can be used to calculate hydronium and hydroxide ion concentrations, and these in turn can be used to caculate pH and pOH values. Each of these topics can be found in the pages of this section