Блог Анкара

The 5th Wave By Rich Tennant

In this part. . .

Studying chemistry without spending time in a lab is a bit like reading up on the history of skydiving from the confines of an armchair in your basement. Something is missing from the experience. For this reason, AP chemistry courses include extensive labwork as part of the standard curriculum. Although the specific experiments may vary from one high school course to another, they all conform to the same set of guidelines and principles, and most of them are drawn from an official set of recommended experiments put forth by the AP chemistry Powers That Be. In this part, we give you a brief review of common laboratory equipment and techniques, and a complete survey of the Official List of AP chemistry experiments.

In This Chapter

^ Surveying the range of College Board-recommended AP chemistry experiments ^ Highlighting key concepts and calculations from the labs

Oggles are the geeky, timeless emblems of a chemistry lab. But before you laugh off this chapter, remember that the laboratory is where chemistry really takes place. That’s why the College Board has put forth an extensive list of recommended AP chemistry labs, many of them interrelated. Any actual AP chemistry course will not include the full range of these labs — but you’ll probably have done a significant fraction of them. To expose yourself to the pith of any labs your class skipped, and to shore up your understanding of those you did perform, review the following summary of the recommended labs, 22 things you might have done behind goggles. Examples offered for individual experiments may or may not exactly describe what you have done in your own lab classes but reflect the essential principles of the labs, and data like those you might encounter in lab are included in the examples. As you go through the steps of an experiment, keep alert to possible sources of error. Each measurement you make (massing a sample, measuring a temperature, etc.) carries some error with it. That error propagates through every subsequent calculation that you make using those measurements, and results in error (uncertainty) in any final experimental values.

Determination of the Formula of a Compound

By combusting a metal sample of known mass, you form a metal oxide with an unknown formula. Taking the mass of the oxide allows you to calculate the ratio of oxygen to metal within the oxide, and thereby determine the formula.

Example:

3.65g Mg is combusted within a 16.26g crucible. After combustion completes, the mass of the sample-containing crucible is 22.38g. What is the formula of the oxide?

Follow these steps to find the formula of this metal oxide:

1. Mass of crucible + combusted sample = 22.38g

2. Mass of combusted sample = 22.38g – 16.26g = 6.12g

3. Increase in mass of sample upon combustion = 6.12g – 3.65g = 2.47g

Because combustion is essentially a combination reaction with oxygen, any added mass derives from oxygen. The final sample therefore contains the original amount of Mg:

4. Moles Mg = 3.65g Mg x (1mol Mg / 24.3g Mg) = 0.150mol Mg

5. Moles O added = 2.47g O x (1mol O / 16.0g ) = 0.154mol O

6. Moles Mg : moles O ~ 1:1

The formula of the oxide is MgO.

Determination of the Percentage of Water in a Hydrate

By heating a hydrated ionic compound of known mass, you drive off hydrating water molecules. Often, this process can be followed visually by observing the colored hydrate compound lose its color. Taking the mass of the dehydrated compound allows you to calculate the percent of water in the original hydrated compound.

Example:

5.91g of hydrated copper (II) sulfate, CuSO4nH2O, is heated within a 15.73g crucible until its original blue color changes entirely to white. The sample should be cooled and massed, and then heated and cooled again until the mass remains constant indicating all the water has been removed. The mass of the sample-containing crucible after dehydration is 19.48g. What is the apparent percentage of water in the hydrated compound?

Here are the steps to determine the apparent percent water in the original compound:

1. Mass of sample + crucible = 5.91g + 15.73g = 21.64g

2. Mass of dehydrated sample + crucible = 19.48g

3. Mass of dehydrated water = 21.64 – 19.48g = 2.16g

4. Moles of dehydrated water = 2.16g H2O x (1mol H2O / 18.02 g H2O) = 0.120mol H2O

Because the mass that remains in the crucible after dehydration is CuSO4 (not the hydrate), you can calculate the original moles of hydrated compound:

5. Moles CuSO4 • nH2O = moles CuSO4

6. Mass remaining in crucible = 19.48g – 15.73g = 3.75g

7. Moles CuSO4 = 3.75 g CuSO4 x (1mol CuSO4 / 159.6g CuSO4) = 0.0235mol CuSO4

8. Moles dehydrated water : mole CuSO4 = 0.1120 : 0.0235 = 5.1: 1 ~ 5:1

So, given a plausible amount of experimental error, the original hydrate appeared to have the formula CuSO4 • 5H2O.

To calculate the mass percent of water in the original compound, you simply divide the mass of the hydrating waters (5.1 x 18.02g) by the mass of the whole hydrated compound and multiply by 100%:

(91.9g / 249.7g) x 100% = 36.8%

Determination of the Molar Mass by Vapor Density

By evaporating a volatile liquid such that it fills a flask of known volume, and then measuring the mass of recondensed liquid, you determine the vapor density (mass of compound / volume of flask) of the compound. Then, you use the ideal gas law to determine the molar mass of the compound.

Example:

4.00g of an unknown volatile liquid is loaded into a flask of 312mL interior volume. The flask is capped with a single-hole stopper to allow controlled escape of evaporating material. The stoppered flask is immersed in a 373K water bath, causing the volatile liquid to begin to evaporate. Once the last visible trace of liquid evaporates, the flask is plunged into an ice bath, recondensing any volatile liquid that remains as vapor within the flask. By comparing the mass of the flask before and after the heating/cooling procedure, you calculate that 0.854g of recondensed liquid remains. What is the apparent molar mass of the volatile liquid?

Follow these steps to find the apparent molar mass:

1. The experiment takes place in atmospheric conditions, so the pressure ~ 1.0atm.

2. At the moment the flask was immersed in an ice bath to recondense the volatile liquid, the moles of the sample filled the 312mL volume (0.312L). You know that the temperature of the original water bath was 373K and the pressure ~ 1.0atm. Using the ideal gas law, you can calculate the moles of gas in the flask under those conditions:

PV= NRT

N = (PV) / (RT) = (1.0atm x 0.312L) / (0.0821 Latm mol-1 K-1 x 373K) = 0.010mol

3. Because the recondensed liquid had mass 0.854g, we calculate the apparent molar mass as follows:

Molar mass ~ 0.854g / 0.010mol = 85 g mol-1

Determination of the Molar Mass by Freezing Point Depression

You dissolve a known mass of an unknown, nonelectrolyte solute into a known mass of solvent. Then you measure the freezing point of the resulting solution and compare it to the freezing point of the pure solvent. By calculating the change in the freezing point, you determine the molar mass of the unknown solute based on the colligative property of freezing point depression.

Constraining the unknown solute as an nonelectrolyte ensures that each mole of dissolved compound contributes one mole of dissolved particles, as opposed to the more complicated cases of acids, bases, and salts.

Example:

You dissolve 1.32g of an unknown, nonelectrolyte compound into 50.0g of benzene solvent. You measure the freezing point of the solution to be 2.6°C. What is the apparent molar mass of the solute?

Here are the steps to determine the apparent molar mass of this solute:

1. Freezing point depression depends on the molality (m) of the solution and on the solvent according to the freezing point depression equation, ATf = Kf x m.

The parameter Kf is the freezing point depression constant and varies from solvent to solvent. For benzene, Kf = 5.12 °C M-1.

The parameter ATF is the change in freezing point compared to that of the pure solvent, not the freezing point temperature itself. The normal freezing point of pure benzene is 5.5 °C.

Substituting in known values, you get (5.5 °C – 2.6 °C) = (5.12 °C m"1) x M

2. Solving for M Gives you 0.57 molality, or 0.57 moles solute per kg solvent. Because you dissolved the unknown solute into 50g = 0.050kg solvent, you can calculate the apparent moles of solute:

0.57mol kg-1 = N / 0.050 kg

3. Solving gives you 0.029 moles. Those moles had a mass of 1.32g, so the apparent molar mass is 1.32g / 0.029mol = 46 gmol-1.

Determination of the Molar Volume of a Gas

You react a known mass of a known metal with acid within a flask such that the reaction evolves hydrogen gas H2. The volume of gas is measured by its ability to displace water within a connected eudiometer, a graduated column partially filled with water. By measuring the volume of gas evolved, you can use the ideal gas law to calculate the apparent molar volume of hydrogen gas. In so doing, you must account for the contribution of water vapor within the eudiometer to the total pressure.

Example:

You react 0.027g Mg with concentrated HCl within a stoppered flask that is connected by a tube to a eudiometer. As the reaction proceeds, hydrogen gas bubbles from the mixture into the eudiometer. By observing the water displacement, you estimate that the reaction evolved 26.15mL of H2. The temperature of the lab is 22°C, a temperature at which the vapor pressure of water is 0.030atm. What is the apparent molar volume of the hydrogen gas? How do the moles of H2 estimated by this method compare to the number of moles expected from the reaction?

To compare the experimentally estimated number of moles to the expected number, follow these steps:

1. Set yourself on solid ground by writing a balanced equation for the reaction:

Mg(s) + 2HCl(aq) — Mg2+(aq) + 2Cl-(aq) + H2(g)

2. Given the stoichiometry of the balanced reaction equation, you can calculate a theoretical yield of H2 gas from the starting mass of Mg:

0.027g Mg x (1mol Mg / 24.3g Mg) x (1mol H2 / 1 mol Mg) = 1.1 x 10-3 mol H2

Potentially, 1.1 x 10-3 mol H2 gas could coexist with water vapor within the eudiometer tube.

3. You can estimate the moles of water in the displaced volume within the tube (26.15mL = 0.02615L) by using the ideal gas law and the given vapor pressure for water at 22 °C, 0.030atm:

PV = NRT

N = (PV) / (RT) = (0.030atm x 0.02615L) / (0.0821 Latm mol-1 K-1 x 295K) N = 3.2 x 10-5 mol H2O

4. Raoult’s law states that the partial pressure of a substance in the vapor pressure over a sample is proportional to the mole fraction of that substance within the sample. So, you can estimate the fractional pressure exerted by water vapor trapped in the eudiometer tube:

Fractional pressure ~ (3.2 x 10-5 mol H2O) / (3.2 x 10-5 mol H2O + 1.1 x 10-3 mol H2) Fractional pressure ~ 0.028

5. So, the pressure that caused the displacement within the eudiometer derived not only from hydrogen gas but also from about 3 mole percent water vapor. Using this information, you can recalculate the moles of hydrogen within the eudiometer, including a correction for the partial pressure contributed by water vapor:

N = [(1.0atm - 0.028atm) x (0.02615L) / (0.0821 Latm mol-1 K-1 x 295K)

N = 1.0 x 10-3 mol H2

6. The volume occupied by this amount of hydrogen ~ 0.02615L x (1 - 0.028) ~ 0.025L

7. So, the apparent molar volume of the hydrogen ~ 0.025L / 1.0 x 10-3 mol = 25 L mol-1

8. Finally, the percent error between the actual moles H2 produced versus those estimated from the balanced equation calculates as:

[(1.1 x 10-3mol - 1.0 x 10-3mol) / 1.1 x 10-3mol] x 100% = 9%

Standardization of a Solution Using a Primary Standard

To reliably prepare working solutions from stock solutions, it helps to be able to measure the concentration of stock solutions by reacting them with "primary standards" of very certain concentration. This process is called Standardization. To standardize solutions in this way, you must know the balanced reaction equation for the reaction undergone between the primary standard and sample solutions, and you must have some means to assess the extent of the reaction, such as a color change.

Analysis of the data from a standardization titration is essentially identical to the analysis used for acid-base titrations. Both experiments employ titration. Both require you to make use of a balanced reaction equation. Both often use a color change to indicate the endpoint of a titration. A common standard substance is potassium acid phthallate (or KHP for short), which can be prepared very pure, is solid at room temperature, and so easily weighed to obtain an exact mass for reaction. For an example of the types of calculations used in a standardization experiment, see "Determination of the Equilibrium Constant for a Chemical Reaction," later in this chapter.

Determination of Concentration by Acid-Base Titration, Including a Weak Acid or Base

To reliably determine the concentration of acid or base solutions, you can use the solution of unknown concentration in a neutralization reaction against an acid or base of highly certain concentration. To use this method, you must know the balanced reaction equation of the neutralization reaction and you must have a means to assess the extent of the reaction, such as the change in color of an indicator. If you know the number of acid or base equivalents contributed by your sample compound (HCl contributes one acid equivalent, H2SO4 contributes two acid equivalents, for example), you can determine the molar concentration of the compound. If you do not know the number of equivalents per mole of your compounds, you can still determine the normality, N, of your sample solution. (You may or may not have worked with "normality" in your AP class, but the AP exam does not include it.)

Example:

You wish to determine the concentration of a solution of ascorbic acid, C6H8O6, a weak acid. You know that ascorbic acid is monoprotic, contributing one mole of acid equivalent per mole of compound. To determine the concentration of your solution, you titrate 50.0mL of ascorbic acid solution with a 0.10M NaOH standard solution. To follow the progress of the neutralization reaction, you use a colorimetric indicator, phenolph-thalein. The phenolphthalein changes from colorless to pink when you have added 19.6mL of standard to the ascorbic acid sample solution. What is the apparent concentration of the sample solution?

To determine the apparent concentration of the original ascorbic acid solution, follow these steps:

1. To begin the analysis, you must know the stoichiometry of the neutralization reaction, as expressed in a balanced reaction equation:

C6H8O6(aq) + NaOH(aq) — C6H7O6Na(aq) + H2O(l)

Expressed as a net ionic equation,

C6H8O6(aq) + OH-(aq) — C6H7O6- (aq) + H2O(l)

In other words, the soichiometry of the interaction between NaOH and ascorbic acid is 1:1.

2. Phenolphthalein, the colorimetric indicator added to the titration solution, changed from colorless to pink at the equivalence point, indicating that the moles of acid in solution equal the moles of base. The color changed at 19.6mL (0.0196L) added 0.10M NaOH.

3. The moles of base added at the equivalence point is simply the concentration of the standard NaOH solution multiplied by the volume added:

Moles base = 0.10 mol L-1 x 0.0196L = 2.0 x 10-3 mol

4. Because moles base = moles acid at the equivalence point, the original 50.0mL (0.0500L) ascorbic acid solution apparently contained 2.0 x 10-3 mol ascorbic acid. So, the molar concentration of the ascorbic acid solution is

Molarity = (2.0 x 10-3 mol) / (0.0500L) = 4.0 x 10-2 M

Determination of a Concentration

By Oxidation-Reduction Titration

To determine the concentration of a solution of an oxidizing agent or reducing agent, you can use the sample solution in an oxidation-reduction reaction against a reducing agent or oxidizing agent of highly certain concentration. To use this method, you must know the balanced reaction equation of the redox reaction and you must have a means to assess the extent of the reaction, such as a color change.

For example, here is a balanced equation for a redox reaction between hydrogen peroxide and permanganate:

5H2O2(aq) + 6H+(aq) + 2MnO4-(aq) — 5O2(g) + 2Mn2+(aq) + 8H2O(l)

In this reaction, the permanganate ion, MnO4-, itself acts as the indicator because it has an intense purple color. So, as the reaction proceeds and permanganate is reduced, the purple color disappears. When the reaction is complete, the solution entirely loses purple color, becoming colorless.

Analysis of the data from a oxidation-reduction titration is essentially identical to the analysis used for standardization and acid-base titrations. The key elements are the use of a color change to indicate a reaction endpoint and the use of a balanced reaction equation to make correct stoichiometric calculations. The fact that the reaction in question is an oxidation-reduction reaction doesn’t really change the method of analysis. So, to see an example of the types of calculations used in a redox titration experiment, see determination of concentration by acid-base titration, earlier in this chapter.

Determination of Mass and Mole Relationships in a Chemical Reaction

This experiment is not typically performed on its own but rather as a necessary part of other experiments. The basic principle shows up in two main ways. First, you use a measured mass to calculate the apparent number of moles in a sample. Second, you use a measured number of moles to calculate the apparent mass of a sample. The measured quantities may be determined directly or indirectly, depending on the experiment. In either case, the apparent ratio of the mass of a substance to the corresponding number of moles represents the experimental molar mass (g mol-1) of the substance.

Determination of the Equilibrium Constant for a Chemical Reaction

The equilibrium constant, Keq, For a reaction is a ratio of the concentrations of products to reac-tants when the reaction has achieved equilibrium. So, you can determine the equilibrium constant by measuring the concentrations of reactants and products in an equilibrium solution. To use this method, you must know the balanced reaction equation for the reaction in question, and you must have a reliable means to measure the reactant and product concentrations.

Example:

You wish to determine the equilibrium constant for the ionization reaction of ascorbic acid, C6H8O6, a weak acid. You know that ascorbic acid is monoprotic, contributing one mole of acid equivalent per mole of compound. Because one product of the ionization reaction is H+, you choose to measure the concentration of reactants and products by measuring the pH of a standardized, 1.00 x 10-2 M, Solution of ascorbic acid. You observe the pH to be 3.08. What is the apparent Keq of ascorbic acid?

To find the apparent Keq of ascorbic acid by this method, do the following:

1. To calculate the Keq from a set of measured concentrations, you must have access to a balanced reaction equation. For the ionization reaction of ascorbic acid, the equation is

C6H8O6(aq) — C6H7O6-(aq) + H+(aq)

Or to be more rigorously correct,

C6H8O6(aq) + H2O(l) — C6H7O6-(aq) + H3O+(aq)

2. For simplicity, we’ll use the first equation to set up an expression for Keq:

Keq = products / reactants = [C6H7O6-]x[ H+] / [C6H8O6]

So, if you can measure the concentrations of C6H8O6, C6H7O6-, and H+ in a solution of ascorbic acid at equilibrium, you can use those values to solve for Keq.

3. Because the reaction in question is an acid ionization reaction, measuring the pH of an equilibrium solution is a clever way to obtain the needed data. Prior to ionizing, the ascorbic acid in solution exists at 1.00 x 10-2 M Concentration. Each mole of acid that ionizes depletes that initial concentration and adds one mole of ascorbate (the conjugate base) and one mole of H+ to solution. In other words, the equilibrium mixture contains the following concentrations:

[C6H8O6] = (1.00 x 10-2 – X) M

[C6H7O6-]= XM

[H+]= XM

All concentrations are expressed in terms of x, which means that if you can measure x, then you know all concentrations and can calculate the equilibrium constant.

4. The concentration of H+ can be calculated from the measured pH:

PH = -log[H+] therefore 10-pH = [H+] The pH of the solution is measured to be 3.08, so [H+] = 10-3.08 = 8.3 x 10-4M.

5. So, X = 8.3 x 10-4M. Knowing this, you can substitute into the expression for Keq and solve:

Keq = (8.3 x 10-4 x 8.3 x 10-4) / (1.00 x 10-2 – 8.3 x 10-4) = 7.5 x 10-5

Determination of Appropriate Indicators for Various Acid-Base Titrations and Determining pH

This experiment is often performed as part of a larger acid-base titration lab. The core concept is that common colorimetric pH indicators work best within limited pH ranges. Different indicators work best in different pH ranges. The reason for this is that indicator dyes are simply molecules that can exist in protonated or deprotonated form, with each form exhibiting a

Different color. The affinity of a given dye molecule for a proton, H+, is expressed by either of two related quantities: the Ka Or the pKa of the molecule.

The Ka is the acid dissociation constant, which is simply the equilibrium constant for the ionizing dissociation reaction of the molecule, HA H+ + A-. The pKa is simply the negative logarithm of the KA, pKA = – log KA.

The pKA is a convenient quantity to use for thinking about the effective pH range of an indicator dye. The pKA represents the pH at which one half the indicator molecules exist in proto-nated form and one half exist in deprotonated form. Beyond about one pH unit above or below its characteristic pKa, an indicator dye is outside of its useful range — no discernible changes in color can be detected this far from the pKA. So, it is best to use indicator dyes with pKa values as close as possible to the expected pH range of an experiment. If no reasonable pH range can be estimated prior to the experiment, it may be necessary to conduct the experiment multiple times, in the presence of indicators covering a range of pKA values.

Here are the most common indicator dyes alongside their respective pKa values:

Determination of the Rate of a Reaction and Its Order

You determine the rate at which reactant is converted into product and, by varying the initial concentrations of reactants across a series of trials, you gather data that allow you to determine the reaction order. The overall reaction order is the sum of the orders of the individual reactants. Reaction order is a measure of the sensitivity of the reaction rate to changes in concentration.

Example:

You measure the initial rate of the following (unbalanced) reaction — the acid-catalyzed iodination (addition of iodine) of acetone — under four different sets of reactant concentrations.

C3H6O + I2 — C3H6IO + H+ + I-

You observe the following results.

What is the reaction order of each species (C3H6O, I2 and H+) and what is the overall reaction order? What are the rate law and the apparent rate constant? To determine the rate law for each reactant, observe the following:

1. To determine rates like those listed in the previous table, you need a means of detecting the progress of the reaction, which typically means having some way to measure the appearance of product or the disappearance of reactant. Often, the simplest and most precise way to measure reaction progress is spectrophotometry, as described in the determination of electrochemical series, later in this chapter.

Because rates are often so dependent on reactant concentrations, it is usually advisable to measure "initial rates," which are fast, linear portions of the early reaction, before depletion of reactants appreciably slows the observed rate.

2. Once you have a set of initial rates at varying concentrations of reactants and/or products, you can seek patterns in the variation of rate with concentration. In the example data, it is clear that doubling the concentration of I2 had no effect on rate when other concentrations remained constant, suggesting that the reaction is zero order in I2.

3. Determining that I2 concentrations do not impact rate allows you to focus exclusively on C3H6O and H+ concentrations. Doubling the concentration of H+ doubles the observed rate suggesting that the reaction is first order in H+.

4. Maintaining the increased concentration of H+ while doubling the concentration of C3H6O causes a further doubling of the rate, suggesting that the reaction is first order in C3H6O. You can convert the individual reaction orders to an overall reaction order and a rate law with a known rate constant, K, As follows:

1. Because the reaction is first order in two reactants, the overall reaction order is 2.

2. The reaction order is reflected in the rate law for the reaction, Rate = K[C3H6O][H+].

3. To determine the value of the rate constant, K, Simply substitute in any set of data from any of the individual rate experiments:

Rate = K[C3H6O][H+]

K = Rate / ([C3H6O][H+]) = (4.8 x 10-7 MS-1) / (0.100Mx 0.100M) = 4.8 x 10-5 M~l S-1

Determination of the Enthalpy Change Associated with a Reaction

This lab involves Calorimetry, The measurement of heat transfer associated with a process (such as a chemical reaction). One common reaction employed in this lab is the dissolution reaction of a solute into water, although many other types of reaction can be used. At constant pressure, the absorption or release of heat that accompanies a chemical reaction is equivalent to the reaction enthalpy, AHRxn . Reactions that release heat are exothermic and have negative reaction enthalpy. Reactions that absorb heat are endothermic and have positive reaction enthalpy. By measuring the change in temperature of a reaction system of known mass and composition, you can calculate the molar reaction enthalpy, the amount of heat released or absorbed per mole of reaction.

Example:

You add 100.mL of water to a well-insulated (probably polystyrene foam) cup. You observe that the water temperature is 22.0°C. You then add 2.70g solid NaOH to the water, cap the cup with an insulated lid, and swirl gently to promote dissolution. While the NaOH dissolves, you monitor the temperature of the solution via a thermometer

Extending through a perforation in the lid. When the temperature levels off for approximately one half minute, indicating that the dissolution reaction has completed, you record the final temperature as 28.5°C. What is the apparent heat of solution, AHsoln, of sodium hydroxide?

Because the reaction performed in this lab is a dissolution reaction, the reaction enthalpy is also known as the "heat of solution." In other words, AHSoln = AHRxn. The dissolution reaction is

NaOH(s) + H2O(l) — Na+(aq) + OH-(aq) AHsoln = ?

The units of the heat of solution are joules per mole, J mol-1. So, if you can measure the heat flow that accompanies the dissolution reaction, you can divide that heat by the moles of NaOH dissolved to find AHSoln.

The key experimental trick is realizing that as NaOH dissolves, heat flows to or from the water in which it is dissolving. We can neglect any heat flow to or through the polystyrene cup as it is such a good insulator. So, qNaOH = – qwater. You can measure the change in the heat content of the water by measuring a change in the temperature of the solution (assuming that the mass of the solute is much smaller than the mass of the solvent — a fair assumption in dilute solutions).

To find the apparent heat of solution of sodium hydroxide from the data in this experiment, do the following:

1. To translate between temperature and heat flow, you need the constant-pressure calorimetry equation,

Q = m x Cp xAT = m x Cp x (Tfinal – Tintitial)

Here, m is the mass of the water, and CP is the specific heat capacity of the water.

2. Because the density of water is 1.00 g mL-1, you can calculate the mass of water in the experiment:

Mass of water = 100mL H2O x (1.00g / mL H2O) = 100.g

3. The specific heat capacity of water is 4.186 J g-1 °C-1. So, you can calculate the heat transfer during dissolution as

Q = (100.g) x (4.186 J g-1 °C-1) x (28.5 °C – 22.0 °C) = 2.72 x 103 J

4. So, as the NaOH dissolved, it released 2.72 x 103 J of heat energy into the water. In other words, QNaOH = -2.72 x 103 J. This amount of heat was released by 2.70g NaOH. You can convert to moles NaOH by using the molar mass:

Moles NaOH = 2.70g NaOH x (1mol NaOH / 40.0g NaOH) = 0.0675mol NaOH

5. So, the apparent heat of solution of NaOH is:

AHsoln = qNaOH / mol NaOH = -2.72 x 103 J / 0.0675mol = -4.03 x 104 J mol-1.

Separation and Qualitative Analysis of Anions and Cations

This lab is essentially about the descriptive chemistry of ionic compounds and their aqueous solutions. Qualitative analysis involves testing for the presence or absence of a compound, as opposed to measuring the precise concentration of that compound. Compounds can often be identified based on the formation of insoluble precipitates or colored compounds. Sometimes it can be difficult to test for the presence of one particular "target" ion within a

Solution of many ions, because the "nontarget" ions can interfere with intended reactions. For example, Cd2+ can be detected in aqueous solution by the formation of a yellow precipitate upon addition of S2-. However, if Pb2+ or Cu2+ happen to be in the sample solution, a black precipitate forms upon addition of S2-, which makes analysis for Cd2+ impossible. In cases like these, it is desirable to be able to selectively separate interfering ions from a solution while preserving other ions targeted for analysis. In other cases, the goal is simply to analyze a mixture of ions for the total ion composition of the solution. Separation schemes typically depend on the differential solubility of different ionic compounds.

Example:

You are given a solution that may contain Cl – anion, I – anion, both, or neither. To determine which of the four possibilities is true, you may use the following reagents: water, nitric acid (HNO3), silver nitrate (AgNO3), and ammonia (NH3). In addition, you have a standard array of glassware and filtration devices.

To separate the anions in the sample solution with the materials provided, do the following:

1. First, acidify the unknown chloride/iodide solution by dropwise addition of HNO3 until the solution is mildly acidic. Because Cl – and I – are weakly basic, addition of acid reduces their solubility and makes them easier to separate.

2. Next, slowly add AgNO3 until any precipitation is complete. Although silver nitrate is soluble, other silver salts, like AgCl and AgI, are insoluble. Filter and wash the precipitate with water. Examine the precipitate for color. AgCl is white, and AgI is yellow.

3. Next, resuspend the precipitate in a small amount of water and add NH3 and AgNO3. Mix well. The added silver nitrate promotes precipitation of silver salts due to the common ion effect. At this point, any iodide anion present in solution should precipitate as yellow silver iodide, AgI. Separate the supernatant from any AgI precipitate.

4. The ammonia that remains in the supernatant forms a cationic complex with silver, Ag(NH3)2+. After dropwise addition of HNO3 to reduce anion solubility, any Cl- in solution precipitates as white AgCl salt.

Synthesis of a Coordination Compound and Its Chemical Analysis

"Coordination compounds" or "coordination complexes" are structures in which a central metal atom is surrounded by a host of nonmetallic groups called Ligands. The ligands are bound to the metal by "coordinate covalent" bonds. These kinds of bonds are particularly common between transition metals and partners that possess lone pairs of electrons. The lone pair containing atom acts as an electron donor (a Lewis base), giving both electrons to a bond with the metal, which acts in turn as an electron acceptor (a Lewis acid). Coordination compounds are often intensely colored and can have properties that are quite different than those of the free metal.

This lab, in which you synthesize a coordination compound, is often combined with the two labs that follow it on this list: gravimetric determination and colorimetric or spectrophoto-metric determination. In keeping with that practice, we’ll stick with one "example" system through these three labs.

Example:

You synthesize a copper-centered coordination compound by the addition of ammonia to a blue, aqueous solution of copper (II) sulfate. When ammonia is added to the copper (II) sulfate, the blue solution turns a deep purple color, indicating that you have successfully

Formed a coordination complex. To separate the complexes from the surrounding solution, you add ethanol to selectively precipitate the complex. This addition results in the formation of crystals of your coordination complex. Using a vacuum filtration apparatus, you repeatedly wash your crystals, rinsing them with an ammonia-ethanol mixture. You allow the crystals to dry overnight. Presumably, you have formed a complex that contains ammonia and/or sulfate ligands surrounding a copper cation center. In the following two labs, you’ll analyze the compound gravimetrically and spectrophotometrically.

Analytical Gravimetric Determination

Gravimetric analysis entails measurements of mass to determine the composition of a compound. Here, we extend the copper coordination complex example from the synthesis of a coordination compound in the previous experiment.

Example:

Based on the chemicals used in the synthesis, the copper coordination complex likely includes ammonia and/or sulfate ligands, although the number of each type of ligand around the copper cation center is unknown. So, at this point, the formula of the complex is

Cua(NH3)b(SO4)c • nH2O

In this formula, a, b and c represent the stoichiometric amounts of copper, ammonia, and sulfate, respectively. The "• NH2O" indicates that some number N Of water molecules may hydrate the complex.

To analyze the composition of your complex, do the following steps:

1. Dissolve 1.00g of your purified crystals in nitric acid, HNO3, which causes the following reaction:

Cua(NH3)b(SO4)c • nH2O + xH+ — aCu2+ + Ј>NH4+ + cSO42- + nH2O

2. To this solution, add lead (II) acetate, Pb(C2H3O2)2, to selectively precipitate sulfate anion as lead sulfate:

Pb2+(aq) + SO42-(aq) — PbSO4(s)

3. Carefully filter the white PbSO4 precipitate and allow it to dry overnight before measuring its mass. From the mass of PbSO4 you can determine the number of moles of sulfate recovered. Within error, this number should be equivalent to the number of moles of sulfate in the original 1.00g of crystals.

4. To determine the amount of ammonia in your compound, you perform an acid-base titration (covered in more detail in previously in the chapter). Briefly, you dissolve 1.00g of your crystals in 50.0mL water, add a pH indicator, and titrate with acid. By doing this, you can indirectly determine the amount of ammonia in your compound. By observing the endpoint of the titration, you can calculate the moles of ammonia that must have been present in the original sample, because the color change of the titra-tion reflects the progress of this reaction:

BNH3(aq) + XH+(aq) — BNH4+(aq)

Within error, the number of moles of ammonia detected by your titration should be equivalent to the number of moles of ammonia in the original 1.00g of crystals.

The analysis for copper content of the coordination complex is done spectrophotomet-rically, as described in determination of electrochemical series, found below.

Colorimetric or Spectrophotometric Analysis

Spectrophotometric analysis measures the amount of a component within a mixture by exploiting the ability of the component to absorb electromagnetic radiation of a specific wavelength. Colorimetric analysis is simply spectrophotometric analysis using wavelength in the range of visible light. In practice, these analyses most often make use of Beer’s law, a mathematical relationship between the concentration of a substance and the absorption of light of a certain wavelength by that substance:

A = Abc

You may have encountered the equivalent equation A = e/c. This equation is absolutely identical to the one given above, but simply uses the variable e for A And the variable / For b.

To properly use Beer’s Law, you must clearly understand the meaning of each variable.

The variable A Is the absorbance of a specific wavelength of light. The wavelength used is typically one that is strongly absorbed by the compound of interest but not well absorbed by other components in the sample.

The variable A Is the molar absorptivity, an experimentally determined constant that depends on the compound of interest and on the wavelength of light used for analysis. Typically, the units of A Are M-1 Cm-1, but other units can be used. The key criterion is that the units used in A Match those used in the variables B And C.

The variable B Is the pathlength, the distance of the path of light as it travels through the sample. Most often, the pathlength is 1cm, but this value can change in the case of strongly or weakly absorbing samples. Whatever the pathlength, it must match the units used in the variable A.

The variable C Is the concentration of the compound of interest. Typically, the units of C Are molar (that is, M, mol L-1), but that can change in special cases. Whatever the concentration units, they must math those used in the variable A.

Example:

To continue the analysis of the copper coordination complex synthesized and gravimet-rically analyzed in the previous two experiments, you subject the compound to spec-trophotometric analysis to determine its copper content.

To subject the coordination compound to spectrophotometric analysis, you must first estimate the molar absorptivity of your copper compound:

1. Calibrate a spectrometer at 645nm wavelength by using a series of Cu2+ standard solutions, diluted as necessary in 1 MHNO3.

2. Dissolve 0.50g of coordination compound crystals in 20.0mL 1M HNO3 and transfer a sample of the solution to a 1cm-pathlength quartz cell.

If the molar absorptivity of your compound, estimated by the Cu2+ calibration, is 5.2 M-1 Cm-1, and the absorbance reading of your dissolved sample is 0.47, what is the apparent copper concentration in the dissolved sample? What is the apparent molar mass of the compound in your crystals?

To determine the apparent copper concentration of your sample, do the following:

1. The apparent concentration of Cu2+ in the sample can be directly determined by using Beer’s Law:

A = Abc Therefore C = A / (ab)

C = 0.47 / (5.2 M~l Cm-1 x 1 cm) = 9.0 x 10-2M Cu2+

2. Assume for the moment that each coordination complex centers on a single Cu2+ ion

(a reasonable working assumption), then the concentration of coordination complex in your sample is also 9.0 x 10-2M.

You can proceed from these results to estimate the molar mass of your copper-centered coordination complex as follows:

1. You prepared the spectrophotometer sample by dissolving 0.50g compound into 20.0mL (0.0200L) 1 MHNO3. Using this information, you can estimate the moles of coordination compound in the sample:

9.0 x 10-2mol L-1 x 0.0200L = 1.8 x 10-3mol

2. So, 1.8 x 10-3 moles of compound appear to have been present in the 0.50g sample you dissolved. Therefore, the apparent molar mass of the compound is

0.50g / 1.8 x 10-3mol ~ 280 g mol-1

Separation by Chromatography

Chromatography is a useful method for separating the components of a mixture. Although many variations of chromatography allow for quantitative analysis of the components of a mixture, the most frequent uses of chromatography in AP chemistry are qualitative, revealing how many components are in a mixture or determining whether a certain component is in the mixture or not.

In a chromatographic analysis, the sample to be analyzed is included in a "mobile phase," a fluid mixture that passes over a "stationary phase," a surface or porous medium that interacts with the mobile phase as it passes. Separation of components within the sample depends on those components interacting differently with the stationary phase. Some components of the sample may be more strongly attracted to the stationary phase than others, and these strongly attracted components take longer to be swept along by the flowing mobile phase. Separations can be based on charge, size, polarity, or other properties.

One common way to characterize the separated components of a mixture is the retention factor, Rf. To calculate an Rf, You must measure two quantities: the distance covered by the fastest-moving part of the mobile phase (often called the "solvent front") and the distance covered by the component in question. The retention factor is simply the ratio of the two quantities.

Rf = (distance of solvent front) / (distance of sample)

Preparation and Properties of Buffer Solutions

Buffer solutions are mixtures of conjugate acid-conjugate base pairs that resist changes in pH. Typically weak acids and their conjugate bases are used as buffers. Essentially, as extra acid or base is added to a buffer solution, the proportion of conjugate acid to conjugate base shifts in response, thereby minimizing the effect of the acid or base addition. The ability of a buffer solution to resist pH change depends on the concentration of the buffer compounds in solution and on the proportion of acid and base conjugate forms (that is, the ratio of HA to A-). A buffer is at maximum buffering capacity when [HA] = [A-]. Under this optimal-buffering condition, the pH of a buffer solution equals the PKa Of the weak conjugate acid. Buffers tend to be useful within about 1 pH unit of their characteristic pKa.

The Henderson-Hasselbach equation describes the relationship between pH, pKa and the relative concentrations of conjugates acid and base in a buffer solution:

PH = pKa + log ([A-]/[HA])

Example:

You are preparing an acetic acid/acetate buffer solution. At room temperature, acetic acid has pKa = 4.76. You have 500.mL of 0.25M acetic acid. You want the final pH of your solution to be 5.00. What mass of sodium acetate salt must you add to achieve the desired pH?

To determine the mass of sodium acetate you must add to achieve a solution with pH 5.00, do the following:

1. Substitute into the Henderson-Hasselbach equation your known values for pH and pKa, and then solve for the desired ratio of conjugate base to conjugate acid:

5.00 = 4.76 + log ([A-]/[HA])

0.24 = log ([A-]/[HA])

100.24 = [A-]/[HA] = 1.74

2. So, for every mole of acetic acid (the conjugate acid, HA) in solution you want to add 1.74 moles acetate (the conjugate base, A-).

Moles acetic acid = (0.25mol L-1) x (0.500L) = 0.13mol

Moles acetate = 1.74 x 0.13mol = 0.23mol

3. So, you want to add 0.23mol sodium acetate, C2H3O2Na. The molar mass of anhydrous (that is, not hydrated) sodium acetate is 82.0 g mol-1.

0.23 mol C2H3O2Na x 82.0 g mol-1 = 19 g C2H3O2Na

Determination of Electrochemical Series

In this lab, you use your knowledge of electrochemical cells to explain and predict observed voltage potentials in cells of differing composition. Key tasks are to identify the anode and cathode of a cell and to use a table of standard reduction potentials to predict the voltage potential of a cell given its composition.

By measuring the voltage potentials of a series of half-cells, all against the same counterpart half-cell, you can construct an electrochemical series, a ranked list of redox reactions in

Order of reduction potential. Tables of standard reduction potentials represent just such a series, one in which the constant counterpart half cell is the standard hydrogen electrode:

2H+(a<7) + 2e – — H2(g) E =0 (by definition)

Example:

A series of half-cell solutions is tested against a constant half-cell counterpart of 1.0M Copper (II) nitrate, Cu(NO3)2, with a solid copper electrode. In all trials, an ammeter indicates that electrons flow toward the copper electrode from the opposing electrode. The following voltage potentials are observed:

1.0M lead (II) nitrate, Pb(NO3)2 0.47 V

1.0M zinc (II) nitrate, Zn(NO3)2 1.10 V

1.0M iron (II) nitrate, Fe(NO3)2 0.78 V

Given that E = +0.34 V for the reduction of copper (II) cation, which divalent cation is the strongest reducing agent: lead (II), zinc (II) or iron (II)?

To determine the strongest reducing agent from the given data, observe the following:

1. The observed voltage potentials are standard cell potentials:

E°cell = E°red(cathode) – E°red(anode)

2. Because electron flow was toward the copper electrode, you know that the copper electrode was the cathode in each case.

3. Because each observed cell potential is positive and larger than +0.34 V, you know that each metal tested has a negative standard reduction potential.

The most negative of the standard reduction potentials must belong to zinc (II) cation. Based on the data given, zinc (II) cation, Zn2+, is the strongest reducing agent.

To calculate the standard reduction potential of the zinc (II) cation, do the following:

1. Substitute into the equation for standard cell potential:

E°cell = E°red(Cu2+) – E°red(Zn2+) 1.10 V = +0.34 V – E°red(Zn2+) E°red(Zn2+) = -0.76 V

2. Because zinc (II) has the most negative standard reduction potential of the three metal ions tested, it is the most difficult to reduce and is therefore the strongest reducing agent.

Measurements Using Electrochemical Cells and Electroplating

This lab requires you to combine your knowledge of electrochemical cells with basic stoi-chiometry and gravimetric analysis. Electroplating occurs at a cathode when electrons reduce metal cations in solution, causing them to deposit onto the cathode. By measuring the mass of a cathode before and after an electrochemical reaction in which electroplating occurs, you can directly determine the mass of metal deposited onto the cathode during the reaction and compare that mass with one expected from calculations.

Example:

Identical carbon electrodes, each with a mass of 5.50g are immersed into a 1.0MSolution of nickel (II) chloride, NiCl2. The poles of the electrodes are connected to a battery, and current flows through the cell for 30. minutes at 1.5 amperes. What is the expected mass of the cathode after 30 minutes?

To determine the expected mass of the cathode, make the following observations and calculations:

1. Deposition of Ni(s) happens at the cathode, where the following reaction takes place:

Ni2+ (aq) + 2e – — Ni(s) So, it takes 2 moles of electrons to electroplate 1 mole of nickel.

2. Calculate the total moles of electrons that flow through the cathode during the experiment. This calculation requires you to use the unit of the Faraday, F. One Faraday is charge associated with one mole of electrons. Each electron has 1.602 x 10-19 Coulombs of charge, so:

F = (6.022 x 1023mol-1)(1.602 x 10-19 C) = 9.649 x 104 C mol-1

3. The experiment lasted for 30 minutes (1.8 x 103 seconds) at 1.5 amperes. One ampere is one Coulomb per second (C s-1), so the total moles of electrons are

Moles e- = (1.8 x 103 s) x (1.5 C s-1) x (1 mol e – / 9.649 x 104 C) = 2.8 x 10-2 mol e-

4. Calculate the mass of nickel that you expect to electroplate as a result of transferring 2.8 x 10-2 moles of electrons into the NiCl2 solution:

Mass Ni(s) = (2.8 x 10-2 mol e-) x (1mol Ni / 2mol e-) x (58.7g / 1mol Ni) = 0.82g

5. Because the initial mass of the carbon electrode was 5.50g, the expected final mass (after electroplating) is 5.50g + 0.82g = 6.32g.

Synthesis, Purification, and Analysis of an Organic Compound

Typically, this lab involves a relatively straightforward synthesis — one with few steps. The synthesis is followed by a purification of the synthesized compound. In organic synthesis, you typically calculate a percent yield following each major step to track the progress of your reactions. Whatever the chemical details of the compound synthesized, the principles of "yield" are the same:

Percent yield = 100% x (actual yield) / (theoretical yield)

Actual yield is simply the amount of synthesized compound actually obtained after purification. Theoretical yield is the (never-actually-obtained) amount of compound that would result if each step of synthesis occurs with perfect efficiency and if there is no loss of compound between synthetic steps or during the purification.

In This Chapter

^ Getting familiar with common laboratory equipment ^ Reviewing common laboratory techniques

Chemistry is a laboratory science. Although there are such things as purely "theoretical" chemists who do all their reactions on computers, other chemists tend to make fun of these types behind their backs. Most chemistry still involves things like beakers, tubes, and interesting odors (beyond those possibly emanating from the chemists themselves). This state of affairs is not lost on the creators of the AP exam, who expect you to know your basic laboratory equipment and techniques. As you’ll know if you’re taking an AP chemistry course, lab is an important component. This chapter will help you to review the equipment and procedures that you’ve employed to create interesting odors of your own.

Gearing Up: An OverView of Common Lab Equipment

Figure 27-1 shows all of the lab equipment that an AP chemistry test-taker will need to be familiar with. You will not be directly tested on your knowledge of lab equipment, however the AP chemistry exam will often describe and/or diagram experimental setups so you will need to be able to recognize and understand the purpose of each of the pieces of equipment below.

Figure 27-1:

Common chemistry lab equipment.

Platform balance (triple beam)

Ring stand

Burner

Figure 27-1 shows you what the equipment looks like, and the list below tells you how each piece of lab equipment functions:

Balance: Used for obtaining the masses of solid and liquid samples

Beaker: A flat-bottomed, cylindrical piece of glassware used for mixing and heating compounds

Bunsen burner: Attached to a gas line and lit to provide heat for your experiments

Buret: An extremely accurate device with a stopcock at the bottom used to measure volumes of reagents

Ceramic square: Used to avoid burning the surface of your lab bench and incurring your chemistry teacher’s wrath

Clamps: Used to hold a variety of things in place, particularly test tubes Clay triangle: Used to hold a crucible while it is being heated

Condenser: Used to collect vapors by condensing them into liquid as they contact the liquid-cooled inner surface of the condenser

Crucible: A cup-shaped container capable of sustaining high temperatures. It is used to heat chemicals.

Crucible tongs: Used to handle the hot crucible

Erlenmeyer flask: Used to hold liquids. The small upper opening slows evaporation, so for some volatile liquids, a flask is a better choice than a beaker. The shape also makes it suitable for mixing and swirling liquids during a titration.

Florence flask: A type of flask, generally round-bottomed, usually suspended and heated from below. Its shape makes it easy to swirl and mix liquids inside of it.

Funnel: Used together with filter paper to filter precipitates out of solutions

Graduated cylinder: Used to precisely measure volumes

Metal spatula: Used to measure out solid substances

Mortar and pestle: Used to grind sesame seeds for cooking and chemical compounds for chemistry experiments, though we recommend using a different set for each

Pipette bulb: Used to transfer accurately measured amounts of liquid from one container to another

Rubber stoppers: Used to close flasks or test tubes to prevent evaporation of liquids or

Escape of gases

Scoopula: Another instrument used to transfer solids from one place to another

Test tube: Cylindrical open-topped piece of glassware that comes in varying sizes

Thermometer: Used to measure temperatures. Thermometers generally contain liquid mercury.

Watch glass: A piece of glassware in the shape of a large contact lens used for evaporating liquids

Wire gauze: Generally used as a surface for a beaker or flask to rest when being heated by a Bunsen burner

Following Protocol

What makes an effective laboratory "hunchback" is an intimate knowledge of two things. First, the hunchback knows the right tool for the job. Second, the hunchback knows how to use that tool. If you don’t know the first thing, you may find yourself attempting to measure the pH of a solution by placing it dropwise onto the tip of your tongue (don’t ever do this). If you know the first thing but not the second thing, you may find yourself staring in admiration at a stack of litmus paper that you have no idea how to use. To prevent both these kinds of embarrassing episodes, and to shore up your lab knowledge for the AP exam, make sure you thoroughly review this section.

Measuring mass, volume, and density

Mass and volume are by far the most frequently measured quanities in your basic chemistry lab. From mass and volume, you can calculate density. Alternately, you can use density to calculate mass or volume.

Mass: You measure mass in the laboratory using a balance. Most laboratory balances are extremely accurate, capable of measuring masses with a precision of hundredths of grams.

Volume: Techniques for measuring volume vary depending on what type of substance you are measuring, with liquids being the easiest to measure. Graduated cylinders are generally used to make approximate volume measurements for liquids. Pipettes and burets are used to measure precise volumes of liquids. Obtaining the volume of a solid can be somewhat tricky and, in some cases, impossible to do without altering the substance. If the substance does not react with water, you can submerge it in water and measure the volume of water displaced by it. Luckily, in most experiments it is mass of solid used that matters and not volume. You cannot measure the volume of a gas because gases have no definite volume. Recall that it is a fundamental property of gases that they expand to fill the entire volume available to them. However, the amount of gas can be obtained by measuring its pressure when held in a known volume.

Density: Mass and volume measurements can also be used to determine the density of a substance using the formula density=mass^volume. Density is a characteristic physical property for solids and liquids and can often be looked up in a table.

Determining melting point

Melting Is the phase transition between the solid and liquid states and is equivalent to the freezing point. Because melting point is a characteristic physical property of a solid, you can use melting point to help to determine the identity of an unknown solid or to verify that an experiment has been successful and you have isolated the desired product.

You can use several different types of apparatus for measuring melting points, but they all have the same general setup. A sample of the solid in question is placed in a thin glass tube and heated slowly from the bottom. The experimenter watches the solid through a magnifying glass until it reaches the melting point and records it. Pure crystalline solids have very distinct melting points and the phase transition between solid and liquid occurs in a few seconds, giving a very accurate melting point determination. Impurities lower the melting point of a pure solid, by an amount depending on their concentration. So it is vital to have a pure

Sample if you are using melting point to confirm an identity. Amorphous solids such as glass and plastic do not have precise melting points and so chemists make special types of melting measurements for them.

Determining boiling point and doing distillation

The boiling point of a substance is slightly more difficult to measure than the melting point. This is because it is the phase transition between a liquid and a gas and gases are difficult to see so you need a way of verifying that your liquid is at its boiling point and not merely very, very hot. One convenient phenomenon to help you in measuring a boiling point is the tendency for a liquid to maintain a constant temperature during a phase transition. All of the energy being added to the solution goes into changing its phase rather than into heating it.

Distillation is a process that can aid you not only in determining a precise boiling point for a liquid but also in purifying that liquid. Say you have a mixture of two different liquids in a flask, one with a lower boiling point than the other. If you could somehow trap the gas being released as the first liquid boils you would have isolated the pure first substance. This is the basic purpose of a distillation apparatus such as the one shown in Figure 27-2.

The flask on the left is being heated, while the one on the right is cool. Connecting the two is a piece of lab equipment called a Condenser. The inner tube in a condenser connects the first flask to the second, while cold water is run through the outer tube (this water is kept isolated and does not come into direct contact with the chemicals in the experiment). This causes the gas that is escaping from the boiling liquid in the first flask to condense in the inner tube of the condenser, which is angled downward so that it will collect in the second flask.

As the experimenter, you watch a thermometer connected to the first flask. When the first liquid begins to boil and condense in the condenser, the temperature of the liquid will remain constant until all of the lower boiling point liquid has escaped. Once it has all escaped, the condenser will collect no more liquid and the temperature in the first flask will again begin to rise. You then turn off the heat and disconnect your apparatus, having isolated a pure substance in your second flask. If your mixture contains more than two liquids or you wish to isolate a liquid with a higher boiling point, you can continue this process, switching out the second flask and isolating one substance at a time.

Dealing with solids in precipitation and filtration

In chemistry, you will often create solid precipitates in aqueous solutions. You separate the solution from the precipitate using a process called Filtration, Which involves pouring your solution/precipitate mixture through a funnel lined with filter paper. The filter paper allows the solution to pass through, but catches the precipitate. The precipitate should be washed to remove impurities.

Positive spin: Centrifugation

Centrifugation is another nifty way of separating compounds, this time according to their density. Centrifuges are quite expensive and most high school chemistry labs are not equipped with them, but you will often find reference to centrifugation in your chemistry text or on the AP exam. A centrifuge is essentially a big, extremely fast-spinning trough. A mixture in the centrifuge will separate according to density as the denser components migrate away from the center of the apparatus and the less dense migrate toward it.

Thinning out with dilutions

Rather than storing lots of different concentrations of the same type of solution, most chemistry labs will store concentrated stock solutions of each chemical. However, you don’t always want to be dealing with extremely concentrated solutions, particularly when they are strong acids and bases that have a propensity to react with skin and are generally unpleasant. Often, a dilution is required before an experiment is begun. In a dilution, a concentrated stock solution is diluted by adding water to a desired molarity. Dilutions utilize the equation Mi x Vi = Mf x Vf, where Mi And Mf are the initial and final molarities and Vi And Vf are the initial and final volumes. The amount of water added in these cases is equal to Vf—Vi.

Drying off

Because so many compounds react with water, it is often desirable to eliminate water from certain substances. Since so many substances are soluble in water, it is often very difficult to get rid of. In many cases, you can simply heat the substance you wish to rid of water to higher than 100 degrees C and wait for the water to evaporate. However, high temperatures can also cause decomposition of the substance you wish to isolate so sometimes another method is needed.

This alternative method takes advantage of substances called Desiccants That draw in water. If you leave your sample in a sealed container also containing a desiccant, over time the desiccant will absorb all of the water from your sample. Desiccants are slow acting, and your sample will generally have to be left in the container overnight to complete the drying process. Desiccants are capable of absorbing only a certain amount of water, so it is important that your sample have as much water as possible removed from it before being sealed with the desiccant. Silica gel-based desiccants are common in chemistry labs because they can be easily regenerated by heating. Most commercially available dessicants are prepared with additives so that they change color as they become more hydrated.

Making pH measurements

When doing acid/base reactions in the laboratory, you will often wish to know the pH of the solution you are dealing with. There are several ways to measure pH. Often, a material known as litmus paper is used. Litmus paper changes color as it reacts with solutions of different pH. Proper technique is to place a drop of solution onto the paper with a stir rod. When the color of the paper develops, you match its color to the pH scale that comes with the box of litmus paper. Many labs now use "pH paper" or "pH strips," that change to different colors over a range of pH values.

Many chemists prefer to use a pH indicator such as phenolphthalein instead of litmus paper, however, probably because it makes the solution itself turn pink when a pH above 9 is reached. Universal indicators can provide a range of color changes against a scale. pH meters have now become quite cheap and small, so a probe dipped in a solution can give you a direct digital readout of the pH in the solution.

Using spectroscopy/spectrometry to measure concentration

Spectroscopy Is a technique for separating light into its component colors. Spectrometry Means making measurements of the separated light. Sophisticated chemistry labs use spec-trophotometers to create a graph showing the absorption and/or emission at each wavelength of the visible spectrum. However, as is the case with centrifuges, not all high school chemistry labs are equipped with spectrophotometers. If you are asked about spectroscopy on the AP chemistry exam, you will be required to know that it relies on the separation of light into its component colors. Recall from Chapter 3 that the colors absorbed or emitted by a pure element are unique to that element and are related to its atomic structure.

In addition, you will be expected to know that a spectrometer can be used to measure the concentration of a colored substance by using Beer’s Law, and measuring the amount of light transmitted by the solution. The solution is poured into a clean cell and placed in the spectrometer. By comparing this to a cell containing a standard amount of the substance, the spectrometer can determine an unknown concentration by a simple ratio. It is important that no other light scattering (such as by a fingerprint on the cell) happens, as it will affect the result.