N2: Intensified Unit N Reading Notes

UNIT N NOTES

N-1 Reading:

When we first discussed matter, we talked about mixtures. Mixtures consist of 2 or more substances, each of which retains its own properties. Heterogeneous mixtures are ones in which the substances making up the mixture are not evenly distributed, so that samples taken from different locations in the mixture will have differing compositions. A homogeneous mixture is one in which the substances are uniformly distributed, so that samples taken from different locations in the mixture will have the same composition.

Solutions are homogeneous mixtures made up of very small particles (individual molecules, atoms, or ions). They involve the dissolution of a solute (often a solid or a gas) in a solvent (usually a liquid) so that there is only one visible phase present. Typical properties of a solution include the following:

The solution is a homogeneous mixture if it has been well stirred during its formation. This helps to evenly distribute the dissolved particles evenly among the particles of the liquid.
Assuming that the solution is covered and that none of the solvent is permitted to evaporate, the dissolved particles will not come out of solution no matter how long the solution is allowed to stand.
The solution is clear and transparent. The dissolved particles are too small to be seen. A beam of light passing through the solution cannot be seen.
Because of the extremely small size of the dissolved particles, the solution will pass through the finest filters. This means that filtration cannot be used to separate the two substances making up the solution.
A solution is considered to be a single phase even though the constituent substances may have been in different phases prior to mixing.

The substance that has changed its phase upon mixing is the one usually considered to be the solute. If both substances in the solution were in the same phase prior to mixing, then the one that is present in greater quantity is considered to be the solvent.

Solutions in which water is the solvent are called aqueous solutions. Solutions in which alcohol is the solvent are called tinctures.

There are 3 types of solutions:

Gas solutions are those in which two or more gases are dissolved in one another. Air is a perfect example of a gas mixture of mostly nitrogen gas (N2) and oxygen gas (O2). Because gas molecules are usually pretty far apart, gas solutions can be found in any proportion (differing ratios of one gas type to another).
Liquid solutions comprise a solid, gas, or a liquid solute dissolved in a liquid solvent. Examples of a solid dissolved in a liquid would be salt water or sugar water. A carbonated drink (soda) is an example of a gas dissolved in a liquid. Antifreeze is an example of liquid ethylene glycol dissolved in water. When two liquids dissolve in each other in any proportion they are said to be miscible. Two liquids are considered immiscible if they do not dissolve in each other to any appreciable degree.
Solid solutions involve the mixing of two or more solids at the molecular or atomic level. An alloy is a solid solution of two or metals. One example of an alloy is brass (copper and zinc). Amalgams are alloys formed when the solute is mercury (it is considered to be in the solid phase in this case). Solutions of gases in solids are very rare.

You will note in the 3rd characteristic of a solution that solute particles are extremely small (~1 nm diameter). If they are so large (> 1000 nm diameter) that they eventually settle out of the mixture unless the mixture is constantly stirred or shaken, the mixture is called a suspension.Suspension mixtures can be separated using a filter, or settled, and are considered to be heterogeneous mixtures.

When the particle sizes in the mixture are less than 1000nm, but bigger than 1 nm (the order of size for individual ions, atoms, or molecules), the mixture is said to form a colloidal dispersion. The colloid particles make up the dispersed phase, and the liquid in which they are dispersed (often water) is called the dispersing medium. Colloidal particles are too small to be filtered away from their dispersing medium, but light passing through a colloid can be seen in the colloid, so we say that colloids are stable heterogeneous mixtures. Examples of colloidal dispensions in our everyday world are shown in Table 15.3 on page 460.

Sometimes colloids appear to be clear like a solution because the individual particles can’t be seen by the human eye. Colloid particles, however, do reflect light. This means that if we shine a bright beam of light through the colloid, the reflection, or scattering, of that light beam off the colloid particles will allow us to see the path of the light through the mixture. This scattering of light is called the Tyndall Effect. A chart comparing the properties of solutions, colloids, and suspensions appears below:

PROPERTIES OF SOLUTIONS, COLLOIDS AND SUSPENSIONS
SOLUTIONS / COLLOIDS / SUSPENSIONS
Homogeneous / Heterogeneous / Heterogeneous
Particle size: 0.01-1 nm; can be atoms, ions, molecules / Particle size: 1-1000 nm, dispersed; can be aggreagates or large molecules / Particle size: over 1000 nm, suspended; can be large particles or aggregates
Do not separate on standing / Do not separate on standing / Particles settle out
Cannot be separated by filtration / Cannot be separated by filtration / Can be separated by filtration
Do not scatter light / Scatter light (Tyndall Effect) / May scatter light, but are not transparent

Separation of Mixtures

We previously discussed that mixtures can be separated using their physical properties. Differences in density can be utilized to decant a liquid supernatant from a solid precipitate, while centrifugation (high speed spinning of a sample mixture) can be used to separate items of closer densities. Chromatography uses differences in solute attraction to a solid substrate to separate mixtures containing more than one solute. Distillation separates components of liquid solutions according to their boiling point. Constituents with lower boiling points will vaporize before those with higher boiling points. These vapors can subsequently be condensed back to the liquid state in a much purer form (see Figure 2.8, page 47).

N-2 Reading

By definition, the solute is distributed evenly throughout the solvent in a solution. This means that any sample of that solution will have the same ratio of solute particles to solvent paticles. We say that solutions have uniform concentrations. There are quantitative ways to express concentration. One of the ways discussed in the text is called molar concentration, or molarity. We express molarity as the number of moles of solute per dm3 (or Liter) of solution:

Molarity =

To make a 1-molar (abbreviated 1 M) solution, you add enough solvent to 1 mole of solute so that the final volume is 1 L (see Figure 16.8 on page 481 for the making of a 0.5 M solution). A 1 M NaCl solution is made by adding enough water to 1 mole NaCl to make a final volume of 1 L. Here are the calculations you would need to do in order to make this solution:

1 mole Na = 23 g Na1 mol Cl = 35.5 g Cl1 mol NaCl = 23 g + 35.5 g =58.5 g NaCl

This means that 58.5 g NaCl must be present in 1.00 L of solution in order to have a 1 M NaCl solution.

It is important to remember that all solutions of the same molarity and volume will have the same number of solute particles in them before they are dissolved. That is, a 1.0 M solution of sucrose (C12H22O11) will have an identical number of solute particles as a 1.0 M solution of glucose (C6H12O6). The masses of solute added to make the solutions will differ only because of the different molar masses of the solutes.

Molarity is useful because many reactions will only occur in solution. In general it is helpful to remember the following relationship, which is really all factor-label stuff:

Moles of solute in a solution / = / Molar concentration of the solution /  / Volume of solution in dm3 or L
Moles of solute / = / /  /

Once you find the moles of solute in a solution, it is simple to use molar mass to convert it to the mass of solute required to dissolved in a given volume of solution.

Sample Problem:

Some sucrose (common table sugar, C12H22O11) is dissolved in water. How many moles of sugar are dissolved in 202 mL of solution if its concentration is 0.150 M?

Derived quantities (e.g., density and molarity) can always be expressed as equalities, so:

0.150 M sucrose  0.150 mol sucrose = 1.00 L solution

We must also remember that 1000. mL = 1.000 L

Therefore:

202 mL  ? mol sucrose

0.0303 mol sucrose is dissolved in 202 mL of solution to form a 0.150 M sucrose solution.

Sample Problem:

How many grams of sucrose (common table sugar, C12H22O11) is dissolved in 202 mL of a solution with a concentration of 0.150 M?

The first order of business is to find the molar mass of sucrose:

C: 12.0 g /  / 12 mol C / = / 144 g
H: 1.0 g /  / 22 mol H / = / 22 g
O: 16.0 g /  / 11 mol O / = / 176 g
1 mol C12H22O11 = / 342 g C12H22O11

From the results of Sample Problem 5 we can use as our starting point:

0.0303 mol C12H22O11 ? g C12H22O11 so that our calculation is:

10.4 g C12H22O11

Molar Dilutions:

Because we often are supplied with concentrated solutions, but work with lower concentrations in experiments, we must know how much water to add to the concentrated solution. We call this a dilution calculation, and it is based on the principal that the number of moles of solute is unchanged—only the quantity of solvent is increased (see Figure 16.10 on page 483 for a visual example of this process). Because of dimensional analysis, we know that # moles = M  V (in liters). This means that a dilution calculation will be based on the following equation:

MinitialVinitial = MfinalVfinal

Practice Exercise:

How many milliliters of 5.0 M K2Cr2O7 solution must be diluted in order to prepare 250 mL of 0.10 M solution?

The above equation can be algebraically manipulated to obtain:

0.0050 L = 5.0 mL

When diluting concentrated acid or base it is important to remember that the acid or base should be added to water and then further diluted by adding more water until the desired concentration is achieved.

Another type of concentration that is frequently used is called molal concentration, which is defined as . This is used in colligative property calculations, which are discussed later in this unit..

Solution Stoichiometry and Chemical Analysis

As stated above, many reactions will only occur when the reactants are in solution. We therefore need to go through some practice problems with solution stoichiometry

We can use coefficients in neutralization (acids and bases combining completely to form a salt and water) and precipitation reactions to make predictions of amounts of products based on their relative concentration in solution:

Sample Problem:

What volume of 0.500 M HCl(aq) is required to react completely with 0.100 mol of Pb(NO3)2(aq), forming a precipitate of PbCl2(s)?

First, we write the balanced reaction equation and the accompanying recipe:

Pb(NO3)2(aq) + 2 HCl(aq) PbCl2(s) + 2 HNO3(aq)

1 mol 2 mol 1 mol 2 mol

and the appropriate equivalency from the molar concentration of HCl:

0.500 mol HCl = 1.00 L HCl solution

We can now use the recipe and the equivalency from the molarity to convert from 0.100 mol of Pb(NO3)2(aq) to the volume of 0.500 M HCl(aq) required to react with it.

0.400 L HCl solution.

Titrations use standard solutions to determine the concentration of unknown solutions. The equivalence point of a titration is the point in the reaction when stoichiometric equivalent quantities are brought together. Colored indicators are used to show when the equivalence point is reached in acid-base titrations, because their colors change with the concentration of H+ and OH ions.

Sample Problem:

What is the molarity of a NaOH solution if 48.0 mL is needed to neutralize 35.0 mL of 0.144 M H2SO4?

First the equation and recipe:

2 NaOH(aq)+ H2SO4(aq) Na2SO4(aq) + 2 H2O(l)

2 mol NaOH 1 mol H2SO41 molNa2SO4 2 molH2O

0.00908 mol NaOH

M = 0.210 M NaOH.

N-3 Reading

There are 4 main factors that affect the rate of solubility, (the quantity of solute that dissolves per unit of time):

Particle size is an important factor for mainly solid solutes. Dissolution of solid solute occurs at the surface of the solid piece being dissolved. The greater the surface area, the faster the solute will dissolve. Surface area is maximized when a piece of solute is broken into smaller pieces (a powder is the best). Therefore, the smaller the particle size of solute, the faster the dissolution rate.
Stirring of solid and liquid solutes brings fresh portions of the solvent in contact with the solute surface. This also increases the rate of dissolution.
The amount of solute already dissolved can affect the rate of subsequent dissolution. The less solute already dissolved (the less saturated the solvent), the faster solution proceeds.
Higher temperature promotes faster dissolution of solid and liquid solutes. This occurs because solute and solvent particles are both moving faster, thus allowing more frequent collisions so that solute particles can be separated from one another and go into solution. The opposite is true of gases.

A saturated solution is one which has dissolved in it the maximum amount of a solute at the given conditions. If we look at Figure 16.4 on page 474, the slanted red line indicates that sodium nitrate, NaNO3, @ 20C, will dissolve until 83 g of NaNO3 has been added to 100 g of water. If more NaNO3 is added to this solution, it appears to just sit on the bottom of the container .

It is inaccurate to say, however, that the solute added above the saturation point will not dissolve. It will dissolve, but an equal amount of dissolved solute will be crystallized out of solution. When the rate of solute going into solution is equal to the rate of dissolved solute leaving solution, a condition called solution equilibrium or dynamic equilibrium exists. Solution equilibrium is defined as the physical state in which there is a continuous interchange between the dissolved and undissolved portions of the solute. Although this interchange is continuously occurring, there is no net change in the amounts of dissolved and undissolved solute. This is shown in Figure 16.2 on page 472.

When discussing solution saturation, temperature conditions must be stated because solubility will change with temperature. For solutions of gases in liquids, both temperature and pressure must be stated. Any solution that has less solute dissolved in it than it can hold for a given set of conditions is unsaturated. Using the NaNO3 example from above, this means that an aqueous solution containing less than 83 g of NaNO3 at 20C will be unsaturated.

If the temperature of a saturated solution is increased, the solubility of many solid and liquid solutes will also increase. The solution equilibrium is disturbed or perturbed, and the higher temperature solution can hold more solute because the rate of solute dissolution will be larger than the rate at which it crystallizes out. This continues until a new equilibrium at the higher temperature is reached.

Under certain conditions, there are some solutions that can be ‘tricked’ into holding more solute than a normal saturated solution for those conditions. These are called supersaturated solutions, and they are prepared differently from saturated solutions: instead of adding solute at a constant temperature until no more dissolves, the solute is added until the saturation point at a higher temperature. The solution is then allowed to cool as it sits undisturbed. While most solutions will crystallize out some solute as they cool, a solution capable of supersaturation will retain all of the dissolved solute so that it has a final solute concentration that is higher than the usual saturation point for that solute-solvent system.

Supersaturated solutions, however, are unstable. They will crystallize out all of the excess solute if even a single small crystal of solute is added to the system, or the system is physically stressed. The system returns to the correct saturation point when it is perturbed in this manner. The text gives an example of a solute-solvent system that forms a supersaturated state (see the pictures of this in Figure 16.6 on page 475).

Continuous solute solubility of various compounds in water vs temperature are shown in Figure 16.4 and solubilities of other substances in water at specific temperatures, are summarized in Table 16.1 on page 475.

Solubility is expressed as grams of solute per mass of solvent: Solubility = , and we can use information from these graphs and tables to predict the solubility of various solutes in larger or smaller quantities of solvent:

The rate at which a solute dissolves is unrelated to how much solute will actually dissolve in a given quantitiy of solvent. There are 3 main factors that affect solubility:

The nature of the solute and solvent is very important. The type of material being dissolved can determine how much of that solute will dissolve in a given quantity of solvent.
Temperature affects solubility. In general, solubility of solid solute increases with increasing temperature. Solubility of gaseous solutes in liquids decreases with increasing temperature (think of a cold glass of soda left out on a counter in a room temperature environment. Raising the temperature of a gas-liquid system can be used to separate the gaseous solute from the liquid solvent.
The solubility of a solute in a given solvent will vary with temperature. We use solubility curves like Figure 16.4 on page 475 to illustrate the mass of solute that will dissolve in 100 g of water.

Sample Problem

Based on the information given in Figure 16.4, what is the solubility at 50C of the solid ionic substance potassium nitrate, KNO3?

You will want to use a ruler or straight-edge to do this. Line your ruler up with the vertical grid line that runs through 50C. Find the curve that represents the solubility of KNO3 and note where the ruler intersects that curve. Now turn your ruler at a 90 angle to the temperature grid line. Make sure it is still intersecting the KNO3 curve at the same point. Note where the ruler intersects the vertical axis labeled “Grams of solute/100g H2O.” This is the solubility of KNO3 in 100 g of water. You should see that the solubility of KNO3 @ 50C should be approximately 75 g KNO3/100 g H2O.

When the amount of solute dissolved is small compared to the amount of solvent, a dilute solution is formed. Conversely, a concentrated solution is one in which the amount of dissolved solute is relatively large compared to the amount of solvent. The terms concentrated and dilute do not indicate whether or not a solution is saturated. Let me give an example of this from Table 16.1 on page 475: At 20C, 166 g LiBr will dissolve in 100 g of water, while only 0.99 g PbCl2 will dissolve in the same amount of water at the same temperature. Both solutions are saturated, but the LiBr solution is concentrated while the PbI2 solution is dilute.