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Gravimetric Methods Gravimetry includes all analytical methods in which the analytical signal is a measurement of mass or a change in mass. When you step on a scale after exercising you are making, in a sense, a gravimetric determination of your mass. Mass is the most fundamental of all analytical...

Gravimetric Methods Gravimetry includes all analytical methods in which the analytical signal is a measurement of mass or a change in mass. When you step on a scale after exercising you are making, in a sense, a gravimetric determination of your mass. Mass is the most fundamental of all analytical measurements, and gravimetry is unquestionably our oldest quantitative analytical technique. Types of Gravimetric Methods The four examples in the previous section illustrate different ways in which the measurement of mass may serve as an analytical signal. When the signal is the mass of a precipitate, we call the method precipitation gravimetry. The indirect determination of PO33– by precipitating Hg2Cl2 is an example, as is the direct determination of Cl– by precipitating AgCl. In electrogravimetry, we deposit the analyte as a solid film an electrode in an electrochemical cell. The deposition as PbO2 at a Pt anode is one example of electrogravimetry. The reduction of Cu2+ to Cu at a Pt cathode is another example of electrogravimetry. When we use thermal or chemical energy to remove a volatile species, we call the method volatilization gravimetry. In determining the moisture content of bread, for example, we use thermal energy to vaporize the water in the sample. To determine the amount of carbon in an organic compound, we use the chemical energy of combustion to convert it to CO2. Finally, in particulate gravimetry we determine the analyte by separating it from the sample’s matrix using a filtration or an extraction. The determination of total suspended solids is one example of particulate gravimetry. 1 Why Gravimetry is Important Except for particulate gravimetry, which is the most trivial form of gravim- etry, you probably will not use gravimetry after you complete this course. Why, then, is familiarity with gravimetry still important? The answer is that gravimetry is one of only a small number of definitive techniques whose measurements require only base SI units, such as mass or the mole, and de- fined constants, such as Avogadro’s number and the mass of 12C. Ultimately, we must be able to trace the result of an analysis to a definitive technique, such as gravimetry, that we can relate to fundamental physical properties.2 Although most analysts never use gravimetry to validate their results, they often verifying an analytical method by analyzing a standard reference ma- terial whose composition is traceable to a definitive technique.3 1- Precipitation Gravimetry In precipitation gravimetry an insoluble compound forms when we add a precipitating reagent, or precipitant, to a solution containing our analyte. In most methods the precipitate is the product of a simple metathesis reaction between the analyte and the precipitant; however, any reaction generating a precipitate can potentially serve as a gravimetric method. 1.1 Theory and Practice All precipitation gravimetric analysis share two important attributes. First, the precipitate must be of low solubility, of high purity, and of known com- position if its mass is to accurately reflect the analyte’s mass. Second, the precipitate must be easy to separate from the reaction mixture. 2 2. Solubility Considerations To provide accurate results, a precipitate’s solubility must be minimal. The accuracy of a total analysis technique typically is better than ±0.1%, which means that the precipitate must account for at least 99.9% of the analyte. Extending this requirement to 99.99% ensures that the precipitate’s solubility does not limit the accuracy of a gravimetric analysis. We can minimize solubility losses by carefully controlling the conditions under which the precipitate forms. This, in turn, requires that we account for every equilibrium reaction affecting the precipitate’s solubility. For example, we can determine Ag+ gravimetrically by adding NaCl as a precipitant, forming a precipitate of AgCl. Ag+ + CL- ↔AgCL Solubility of AgCl as a function of pCl. The dashed red line shows our prediction for SAgCl if we incorrectly assume that affect silver chloride’s solubility. The solid blue curve is calculated , which accounts for reactions. Because the solubility of AgCl spans several orders of magnitude, S AgCl is displayed on the y-axis in logarithmic form. Avoiding Impurities In addition to having a low solubility, the precipitate must be free from impurities. Because precipitation usually occurs in a solution that is rich in dissolved solids, the initial precipitate is often impure. We must remove these impurities before determining the precipitate’s mass. The greatest source of impurities is the result of chemical and physical interactions occurring at the precipitate’s surface. A precipitate is generally 3 crystalline—even if only on a microscopic scale—with a well-defined lattice of cations and anions. Those cations and anions at the precipitate’s surface carry, respectively, a positive or a negative charge because they have incom- plete coordination spheres. In a precipitate of AgCl, for example, each silver ion in the precipitate’s interior is bound to six chloride ions. A silver ion at the surface, however, is bound to no more than five chloride ions and carries a partial positive charge. The presence of these partial charges makes the precipitate’s surface an active site for the chemical and physical interactions that produce impurities. Occlusions form when interfering ions become trapped within the growing precipitate. Unlike inclusions, which are randomly dispersed within the precipitate, an occlusion is localized, either along flaws within the precipitate’s lattice structure or within aggregates of individual precipitate particles. An occlusion usually increases a precipitate’s mass; however, the mass is smaller if the occlusion includes the analyte in a lower molecular weight form than that of the precipitate. We can minimize occlusions by maintaining the precipitate in equilibrium with its supernatant solution for an extended time. This process is called a digestion. During digestion, the dynamic nature of the solubility– precipitation equilibrium, in which the precipitate dissolves and reforms, ensures that the occlusion is exposed to the supernatant solution. Because the rates of dissolution and reprecipitation are slow, there is less opportunity for forming new occlusions. These surface adsorbates comprise a third type of impurity. We can minimize surface adsorption by decreasing the precipitate’s available surface area. One benefit of digesting a precipitate is that it increases the average 4 particle size. Because the probability of a particle completely dissolving is inversely proportional to its size, during digestion larger particles increase in size at the expense of smaller particles. One consequence of forming a smaller number of larger particles is an overall decrease in the precipitate’s surface area. We also can remove surface adsorbates by washing the precipitate, although the potential loss of analyte can not be ignored. Gravimetric Method At the end of this unit , the student is expected to be able to : 1- Understand the fundamentals of gravimetric analysis. 2- Follow the steps of the gravimetric analysis. 3- Choose the appropriate precipitating agent for a certain analyte. 4- Avoid or at least minimize the contamination of the precipitate. 5- Optimize the precipitation conditions in order to obtain a desirable precipitate. 6-Do all sorts of calculations related to gravimetric analysis. Principle of Gravimetric Analysis Gravimetric methods are quantitative methods that are based on measuring the mass of a pure compound to which the analyte is chemically related. Since weight can be measured with greater accuracy than almost any other fundamental property, gravimetric analysis is potentially one of the most accurate classes of analytical methods. However it is lengthy and tedious as a result, only a very few gravimetric methods are currently used. There are 5 three fundamental types of gravimetric analysis. In precipitation gravimetry, which is our subject in this unit , the analyte is separated from a solution of the sample as a precipitate and is converted to a compound of known composition that can be weighed. In volatilization gravimetry, the analyte is separated from other constituents of a sample by conversion to a gas. The weight of this gas then serves as a measure of the analyte concentration. In electrogravimetry, the analyte is separated by deposition on an electrode by an electrical current. The mass of this product then provides a measure of the analyte concentration. What Is Gravimetric Anis GRAVIMETRIC ANALYSIS In precipitation gravimetry, the analyte is converted to a sparingly soluble precipitate. This precipitate is then filtered, washed free of impurities, converted to a product of known composition by suitable heat treatment, and weighed. For example, a precipitation method for determining calcium in natural waters involves the addition of C2O42- as a precipitating agent : Ca2+ (aq) + C2O42- (aq) → CaC2O4 (s) The precipitate CaC2O4 is filtered, then dried and ignited to convert it entirely to calcium oxide: CaC2O4 (s) → CaO (s) + CO (g) + CO2(g) After cooling, the precipitate is weighed, and the calcium content of the sample is then computed. 6 Types of Gravimetric Analysis 1 Precipitation gravimetric: The analytic is separated from a solution of the sample as a precipitate and is converted to a compound of known composition that can be weighed. 2 Volatilization gravimetric: The analytic is separated from other constituents of a sample by converting it to a gas of known chemical composition that can be weighed. 3 Electro-gravimetric: The analytic is separated by deposition on an electrode by an electrical current. Properties of gravimetric analysis: 1- Tradition method 2- Cheap, easily available apparatus, simple to carry 3- Slow, especially when accurate results are required 4- Wide range of concentration (ng-kg) 5- No calibration required except for the balance 6- Accurate The step required in gravimetric analysis: 1- Precipitation 2- Digestion 3- Filtration 4- Washing 7 5- Drying and igniting 6- Weighing 7- Calculation Steps In Gravimetric Analysis 1.Preparation of the Solution: This may involve several steps including adjustment of the pH of the solution in order for the precipitate to occur quantitatively and get a precipitate of desired properties, removing interferences …etc. 2.Precipitation: This requires addition of a precipitating agent solution to the sample solution. Upon addition of the first drops of the precipitating agent, supersaturation occurs, then nucleation starts to occur where every few molecules of precipitate aggregate together forming a nucleus. At this point, addition of extra precipitating agent will either form new nuclei ( precipitate with small particles ) or will build up on existing nuclei to give a precipitate with large particles. Ex: Calcium in water 2NH3 +H2C2O4 →2NH4+ + C2O4-2 Ca +2 (aq)+C2O4-2 (aq) → CaC2O4 (s) filtered, dried, ignited CaC2O4 + Δ →CaO (s)+ CO (g)+CO2 3. This can be predicted by Von Weimarn ratio where, according to this relation the 8 The steps required in gravimetric analysis, after the sample has been dissolved, can be summarized as follows: preparation of the solution , precipitation , digestion, filtration , washing , drying or igniting , weighing and finally calculation. GRAVIMETRIC ANALYSIS particle size is inversely proportional to a quantity called the relative supersaturation where Relative Supersaturation = (Q – S) / S The Q is the concentration of reactants before precipitation at any point , S is the solubility of precipitate in the medium from which it is being precipitated. Therefore, in order to get particle growth instead of further nucleation ( i.e granular precipitate and then low surface area ) we need to make the relative supersaturation ratio as small as possible. In other words conditions need to be adjusted such that Q will be as low as possible and S will be relatively large. The optimum conditions for precipitation which make the supersaturation low are: a. Precipitation using dilute solutions to decrease Q b. Slow addition of precipitating agent to keep Q as low as possible c. Stirring the solution during addition of precipitating agent to avoid concentration sites and keep Q low. d. Increase solubility S by precipitation from hot solution. e. Adjust the pH in order to increase S but not too much increase as we do not want to loose precipitate by dissolution. 9 f. Precipitation from Homogeneous Solution: In order to make Q minimum we can, in some situations, generate the precipitating agent in the precipitation medium rather than adding it. For example, in order to precipitate iron as the hydroxide, we dissolve urea in the sample. Heating of the solution generates hydroxide ions from the hydrolysis of urea. Hydroxide ions are generated at all points in solution and thus there are no sites of concentration. We can also adjust the rate of urea hydrolysis and thus control the hydroxide generation rate. This type of procedure can be very advantageous in case of colloidal precipitates 3- Digestion of the Precipitate: The precipitate is left hot (below boiling) for 30 min to 1 hour in order for the particles to be digested. Digestion involves dissolution of small particles and reprecipitation on larger ones resulting in particle growth and better precipitate characteristics. This process is called Ostwald ripening. An important advantage of digestion is observed for colloidal precipitates where large amounts of adsorbed ions cover the huge area of the precipitate. Digestion forces the small colloidal particles to agglomerate which decreases their surface area and thus adsorption. The precipitate often contains ions that where trapped when the precipitate was formed. This is mostly a problem for crystalline precipitates. If the trapped ions are not volatile, then their presence will corrupt the weighing step. Concentration of interfering species may be reduced by digestion. Unfortunately , postprecipitation as we will see later will increase during digestion. 4-Washing and Filtering Problems with surface adsorption may be reduced by careful washing of the precipitate. With some precipitates, peptization 10 occurs during washing. Each particle of the precipitate has two layers , in primary layer certain ions are adsorbed and in the outer layer other ions of opposite charge are adsorbed. This situation makes the precipitate settle down. If the outer layer ions are removed then all the particles will have the same charge so the particles will be dissonant. This is called peptization. The common ion effect can be used to reduce the solubility of the precipitate. When Ag+ is precipitated out by addition of Cl- Ag+ + Cl- → AgCl (s) The (low) solubility of AgCl is reduced still further by the excess of Ag+ which is added, pushing the equilibrium to the right. It important to know that the excess of the precipitating agent should not exceed 50% of its equivalent amount , otherwise the precipitating agent may form a soluble complex with the precipitate : AgCl + Cl- → AgCl2- ( soluble complex ) The following graph shows that most precipitates follow this pattern, but there are some anomalies such as Hg2I and BaSO4. Example : To precipitate 10 moles of Ag+ as Ag2S , how many moles of the precipitating agent S2- do you need to obtain complete precipitation ? Solution : According to the following precipitation reaction : 2Ag+ + S2- → Ag2S The equivalent amount of S2- = 5 moles. 50% of the equivalent amount = 2.5 moles So the total amount of S 2- needed for complete precipitation of Ag+ = 5 + 2.5 = 7.5 moles 5- Drying and Ignition: The purpose of drying (heating at about 120-150 oC in an oven) is to remove the remaining moisture while the purpose of ignition 11 in a muffle furnace at temperatures ranging from 600-1200 oC is to get a material with exactly known chemical structure so that the amount of analyte can be accurately determined. The precipitate is converted to a more chemically stable form. For instance, calcium ion might be precipitated using oxalate ion, to produce calcium oxalate (CaC2O4) which is hydrophil , therefore it is better to be heated to convert it into CaCO3 or CaO. The CaCO3 formula is preferred to reduce weighing errors as mentioned in previous lectures. It is vital that the empirical formula of the weighed precipitate be known, and that the precipitate be pure; if two forms are present, the results will be inaccurate. 6-Weighing the precipitate : The precipitate can not be weighed with the necessary accuracy in place on the filter paper; nor can the precipitate be completely removed from the filter paper in order to weigh it. Example : A 0.5962 g sample of iron ore is dissolved in perchloric acid (HClO4). All iron present is oxidized to Fe3+. The solution is filtered to remove solid matrix materials and made basic with addition of ammonium hydroxide. The iron precipitates as the Fe(OH)3.xH2O gel. The precipitate is collected in a cistern crucible and ignited to produce Fe2O3. What is the wt. % of iron in the sample if the analysis produced 0.3210 g Fe2O3? Solution: The overall reaction is : 2 Fe3+ + 3 OH- → Fe2O3 + 3/2 H2 From this we derive the gravimetric factor relating weight of final material to the weight of iron analyte : 12 steps including adjustment of the pH of the solution in order for the precipitate to occur quantitatively and desired properties, removing. 13 Quantitative Calculations The stoichiometry of a precipitation reaction provides a mathematical re- lationship between the analyte and the precipitate. Because a precipitation gravimetric method may involve several chemical reactions before the pre- cipitation reaction, knowing the stoichiometry of the precipitation reaction may not be sufficient. Even if you do not have a complete set of balanced chemical reactions, you can deduce the mathematical relationship between the analyte and the precipitate using a conservation of mass. The following example demonstrates the application of this approach to the direct analysis of a single analyte. 14 15 16 17 Introduction to Analytical Chemistry Chemistry is the study of matter, including its composition and structure, its physical properties, and its reactivity. There are many ways to study chemistry, but, we traditionally divide it into five fields: organic chemistry, inorganic chemistry, biochemistry, physical chemistry, and analytical chemistry. Analytical chemistry is often described as the area of chemistry responsible for 1. Characterizing the composition of matter, both qualitatively and quantitatively, 2. Improving established analytical methods, 3. Extending 18 existing analytical methods to new types of samples, and 4. Developing new analytical methods for measuring chemical phenomena. 2. Measurements in Analytical Chemistry Units of Measurement A measurement usually consists of a unit and a number expressing the quantity of that unit. We may express the same physical measurement with different units, which can create confusion. To ensure consistency, and to avoid problems, scientists use a common set of fundamental units. These units are called SI units after the Système International d’Unités. Sometimes it is preferable to express measurements without the exponential term, replacing it with a prefix (Table 1.1). A mass of 1×10–15 g, for example, is the same as 1 fg, or femtogram. What Is a Mole and Why Are Moles Used? Answer: A mole is simply a unit of measurement. Units are invented when existing units are inadequate. Chemical reactions often take place at levels where using grams wouldn't make sense, yet using absolute numbers of atoms/molecules/ions would be confusing, too. Like all units, a mole has to be based on something reproducible ‫ قابل لإلعادة‬. A mole is the quantity of anything that has the same number of particles found in 12.000 grams of carbon-12. That number of particles is Avogadro's Number, which is roughly 6.02 x 1023. A mole of carbon atoms is 6.02x1023 carbon atoms. A mole of chemistry teachers is 6.02x1023 chemistry teachers. It's a lot easier to write the word 'mole' than to write '6.02x1023' anytime you want to refer to a large number of things! Basically, that's why this particular unit was invented. Why don't we simply stick with units like grams (and nanograms and kilograms, 19 etc.)? The answer is that moles give us a consistent method to convert between atoms/molecules and grams. It's simply a convenient unit to use when performing calculations. The number of grams in a mole is different from substance to substance. If you're like most students, it's this that's confusing you. Picture it this way: a dozen elephants have a different weight than a dozen rabbits- but in each case, you have a dozen animals. Similarly, a mole of oxygen gas has a different weight than a mole of water- but in each case, you have 6.02×1023 molecules. Weight g /molecular wt. g/mol.= moles Uncertainty in Measurements A measurement provides information about its magnitude and its uncertainty. Consider, for example, the balance in Figure 1.1, which is recording the mass of a cylinder. When weighing an object on a balance, the measurement fluctuates in the final decimal place. We record this cylinder’s mass as 1.2637 g ± 0.0001 g. mass as 1.2637 g ± 0.0001 g, indicating both its magnitude and its absolute uncertainty. Suppose you weigh a sample on a balance that measures mass to the nearest ±0.1 mg. reporting the sample’s mass as 1.762 g instead of 1.7623 g is incorrect because it does not properly convey the measurement’s uncertainty. Reporting the sample’s mass as 1.76231 g also is incorrect because it falsely suggest an uncertainty of ±0.01 mg or ±0.00001g. Accuracy The closeness of an experimental measurement or result to the true or accepted value. 20 Precision The random or indeterminate error associated with a measurement or result. Sometimes called the variability, it can be represented statistically by the standard deviation or relative standard deviation (coefficient of variation ) Concentration Concentration is a general measurement unit stating the amount of solute present in a known amount of solution: or (The ratio of the amount of solute to the amount of solution.) Although we associate the terms “solute” and “solution” with liquid samples, we can extend their use to gas-phase and solid- phase samples as well. The following table shows examples of solution in liquid, gaseous, and solid states of matter. Solution A homogenous mixture of molecules or ions. Solvent The medium in which the molecules or ions are dissolved. Solute Any substance dissolved in a solvent. Analyte Constituent of the sample which is to be studied by quantitative measurements or identified qualitatively. Molarity Molarity express concentration as moles of solute per liter of solution (number of moles of solute in one litre of solution). Molarity is the concentration of a particular chemical species. Molarity is used so frequently that we use a symbolic notation to simplify its expression in equations and in writing. Square brackets around a species 21 indicate that we are referring to molarity. Thus, [Na+] is read as “the molarity of sodium ions. Normality is a concentration unit that is no longer in common use. Normality defines concentration in terms of an equivalent, in one litre of solution, which is the amount of one chemical species reacting stoichiometrically with another chemical species. Molality Molarity is based on the volume of solution containing the solute. Since density is a temperature dependent property on a solution’s volume, and thus its molar concentration, changes with temperature. By using the solvent’s mass (1 kg) in place of the solution’s volume, the resulting concentration becomes independent of temperature. Molality = no. of moles / weight of solvent (kg) Weight, Volume, and Weight-to-Volume Ratios Weight percent (w/w %), volume percent (% v/v) and weight-to-volume percent 1.(w/w %)= 𝑾𝒕 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆 𝒈𝑾𝒕 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒈 𝒙 𝟏𝟎𝟎 Example (a): What is the weight percent of 25 g of sodium sulphate dissolved in 200 g of solution? (w/w %)= 𝟐𝟓𝒈𝟐𝟎𝟎 𝒈 𝒙 𝟏𝟎𝟎 = 0.125 x 100 =12.5% Example (b): Calculate the wt% for a solution prepared by dissolving 5 g of AgNO3 in 100 ml water, Water density = 1g/ml 22  100 ml of water=100 g of water = weight of solvent (w/w %)= 𝑾𝒕 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆 𝒈𝑾𝒕 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒈 𝒙 𝟏𝟎𝟎 Weight of solution = solute + solvent = 5+100= 105 g (w/w %)= 𝟓 𝒈𝟏𝟎𝟓 𝒈 𝒙 𝟏𝟎𝟎 =4.76% 2. (v/v %)= 𝒗 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆 𝒎𝒍𝒗 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒎𝒍 𝒙 𝟏𝟎𝟎 Example: Calculate v/v% of a solution that was prepared by adding 50 ml of methanol to 200 ml water. Volume of solution = volume of solute + volume of solvent Vol. of solution= 50+ 200= 250 ml (v/v %)= 𝟓𝟎 𝒎𝒍𝟐𝟓𝟎 𝒎𝒍 𝒙 𝟏𝟎𝟎 = 20% 3. (wt/v %)= 𝒘𝒕 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆 𝒈𝒗 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒎𝒍 𝒙 𝟏𝟎𝟎 Example (a): Calculate wt/v % for 4g of NaOH dissolved in 500 ml solution. wt/v%= 𝟒 𝒈 𝟓𝟎𝟎 𝒎𝒍 𝒙 𝟏𝟎𝟎=0.8% Example: Calculate the molarity of 4 g of NaOH dissolved in 500 ml solution (NaOH Molecular wt.= 40 g/mole No. of NaOH moles= 4 g/ 40 g/mol)= 0.1 mol 𝑴= 𝒏𝒐. 𝒐𝒇 𝒎𝒐𝒍𝒆𝒔 𝒍𝒊𝒕𝒓𝒆 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝑴= 𝟎.𝟏 𝒎𝒐𝒍𝟓𝟎𝟎𝒎𝒍𝟏𝟎𝟎𝟎𝒎𝒍/𝒍= 𝟎.𝟏 𝒎𝒐𝒍𝒙 𝟏𝟎𝟎𝟎 𝒎𝒍 𝒙 𝒍 𝟓𝟎𝟎 𝒎𝒍=0.2 M Parts per Million and Parts per Billion Part per million (ppm) and parts per billion (ppb) are ratios giving the grams of solute to, respectively, one million or one billion grams of sample. For example, 450 ppm Mn in steel that is =steel contains 450 μg of Mn for every gram of steel. If we approximate the density of an aqueous solution as 1.00 g/mL, or 1000 kg/M3 then solution concentrations can be express in ppm or ppb using the following relationships. 23 Ppm = mg/ L = mg/1000 ml = mg/1000g = mg/kg or =1mg/1000000mg = ppm ppb= μg/L = μg/1000ml= μg/1000g = μg/kg= 1 μg /1000000000 μg For gases a part per million usually is a volume ratio. Thus, a helium con- centration of 6.3 ppm means that one liter of air contains 6.3 μL of He.=1μL/L=ppm Constituent A component of a sample; it may be further classified as: Major > 10% Minor 0.01–10% Trace 1–100 ppm (0.0001–0.01%) Ultratrace < 1 ppm 4. PREPARING STOCK SOLUTIONS A stock solution is prepared by weighing out an appropriate portion of a pure solid or by measuring out an appropriate volume of a pure liquid and diluting to a known volume. Exactly how this is done depends on the required concentration unit. For example, to prepare a solution with a desired molarity you weigh out an appropriate mass of the reagent, dissolve it in a portion of solvent, and bring to the desired volume. To prepare a solution where the solute’s concentration is a volume percent, you measure out an appropriate volume of solute and add sufficient solvent to obtain the desired total volume. NOTE: What is the difference between acetic acid and glacial acetic acid? There is technically no difference between the two but the acetic acid you buy in shops (vinegar) is usually 5-6% acetic acid in water and glacial acetic acid is basically 100% and gets its name from its tendency to freeze when it's cold. 24 PREPARING SOLUTIONS BY DILUTION Solutions are often prepared by diluting a more concentrated stock solution. known volume of the stock solution is transferred to a new container and brought toa new volume. The resulting new concentration can be calculated using: C1 V1 = C2V2 Example (dilution) If 25.0 mL of a 2.19 M solution are diluted to 72.8 mL, what is the final concentration? Solution Using the dilution equation, we have (2.19 M) (25.0 mL) = M2 (72.8 mL) M =mol/L= mmol/ml L Solving for the second concentration (noting that the milliliter units cancel), M2 = 0.752 M Example (dilution) If the stock solution is 10.0% KCl and the final volume and concentration need to be 100 mL and 0.50%, respectively, determine how much stock solution to use (10%) V1 = (0.50%)(100 mL) 0.5=.5 V1 = 5 mL Exercises 1. What is the difference between dilution and concentration? Dilution is a decrease in a solution’s concentration, whereas concentration is an increase in a solution’s concentration. 2. What quantity remains constant when you dilute a solution? 25 3. A 1.88 M solution of NaCl has an initial volume of 34.5 mL. What is the final concentration of the solution if it is diluted to 134 mL? Ans. 0.484 M 4. A 0.664 M solution of NaCl has an initial volume of 2.55 L. What is the final concentration of the solution if it is diluted to 3.88 L? 5. If 1.00 mL of a 2.25 M H2SO4 solution needs to be diluted to 1.00 M, what will be its final volume? Ans. 2.25 mL Analytical Chemistry Chapter One Introduction to Fundamentals 13 6. If 12.00 L of a 6.00 M HNO3 solution needs to be diluted to 0.750 M, what will be its final volume? 7. If 665 mL of a 0.875 M KBr solution are boiled gently to concentrate the solute to 1.45 M, what will be its final volume? Ans. 401.0 ml 8. If 1.00 L of an LiOH solution is boiled down to 164 mL and its initial concentration is 0.00555 M, what is its final concentration? 9. How much water must be added to 75.0 mL of 0.332 M FeCl3(aq) to reduce its concentration to 0.250 M? Ans 24.6 ml 10. How much water must be added to 1.55 L of 1.65 M Sc(NO3)3(aq) to reduce its concentration to 1.00 M? How to Convert between Percent & Parts per Million (ppm) Suppose we are given a parts per million measurement (ppm of X in Y) and we want to convert that to a percent concentration of X in Y. % Concentration = ppm / 10,000= ppm concentration divided by 10,000 The number 10000 is always used in these conversion. 26 Redox Reaction Q. What is the principle of redox titrations? Answer. The oxidation-reduction titrations work on the principle that the oxidation process involves electron loss while the reduction process involves electron gain. Q. How is the endpoint of a redox titration determined? Answer. When we add a redox indicator to the titrant, the indicator imparts a color based on the potential of the solution. As the potential of the solution changes due to the addition of titrant, the indicator changes oxidation state and color, signaling the endpoint. Q. What is the redox titration curve? Answer. To evaluate a redox titration, first, determine the shape of its titration curve. The titration curve in an acid-base titration or a complexation titration shows how the concentration of H3O+ (as pH) or Mn+ (as pM) changes as the titrant is added. It is more convenient to monitor the potential of the titration reaction rather than the concentration of one species during a redox titration. Q. What are the types of redox titrations? Answer. Redox titrations are named after the titrant used, and they are: A bromine (Br2) titrant is used in bromometry. Cerimetry makes use of cerium(IV) salts. Potassium dichromate is used in dichrometry. Iodine is used in iodometry (I2). 27  Potassium permanganate is used in permanganometry. Q. What is the principle of redox titration? Answer. If the following conditions are met, a redox reaction can be used as the basis for a titration: The redox reaction must be rapid and nearly complete (less than 99% success rate is unacceptable). The endpoint must be quantifiable or identifiable using a colour indicator or potentiometry. Stoichiometric electron exchange means that the redox systems (oxidising and reducing agents) must be adequate and equal. Q. Give the general sequence for calculations in redox titration. Answer. The general sequence for calculating redox titration is as follows: Write down the half equations for the oxidant and reductant. Determine the overall equation. Determine the amount of manganate(VII) or dichromate(VI) used. From the overall redox equation, calculate the ratio of moles of oxidant to moles of reductant. Determine the number of moles in the reductant sample solution. Determine the number of moles of reductant in the original solution. Determine either the concentration of the original solution or the percentage of reductant in a known quantity of the sample. Q. In dilute sulfuric acid, an iron tablet weighing 0.960 g was dissolved. To reach the endpoint, an average titre of 28.50 cm3 of 0.0180 mol dm- 28 3 potassium manganate(VII) solution was required. What is the iron percentage of the tablet in terms of mass? Answer. The equation of the reaction is: MnO4– (aq) + 8H+ (aq) + 5Fe2+ → Mn2+ (aq) + 5Fe3+ (aq) + 4H2O (l) Number of moles of MnO4– = 0.0180 × 0.0285 = 0.000513 = 5.13 × 10 – 4 moles. According to the balanced equation, the molar ratio of permanganate and ferrous ions is 1:5 Therefore, moles of iron (II) = 5 × 5.13 × 10–4 = 2.565 × 10–3 moles Mass of iron (II) = 56 × 2.565 × 10–3 = 0.14364 g % by mass of iron = 90.14364/0.960) × 100 = 15% Q13. A 32.15 mL sample of MoO42–(aq) solution was passed through a Jones reductor (a zinc powder column) to convert all of the MoO 42–(aq) to Mo3+ (aq). For the reaction given, the filtrate required 20.85 mL of 0.09550 M KMnO4–(aq). MnO4¯(aq) + Mo3+(aq) → Mn2+ (aq) + MoO22+ (aq) Balance this equation, and determine the concentration of the original MoO42– (aq) solution. Answer. Half reactions: 5e– + 8H+ +MnO4– → Mn2+ + 4H2O 29  2H2O + Mo3+ → MoO22+ + 4H+ + 3e–. The balanced equation is: 4H+ + 3MnO4– 5Mo3+→ 3Mn2+ + 5MoO22+ + 2H2O Moles of permanganate required: 0.09550 mol L–1 × 0.02085 L = 0.001991175 mol Moles of molybdate required: According to the balanced equation, the molar ratio of permanganate and molybdate is 3:5. Therefore, 0.001991175 × (5/3) = 0.003318625 mol molybdate. Concentration: 0.003318625 mol/0.03215 L = 0.1032 M Q. A flask contains 0.2640 g of sodium oxalate, which requires 30.74 mL of potassium permanganate (from a buret) to titrate and turn pink (the endpoint). This reaction’s equation is: 5Na2C2O4(aq) + 2KMnO4(aq) + 8H2SO4(aq) → 2MnSO4(aq) + K2SO4(aq) + 5Na2SO4(aq) + 10CO2(g) + 8H2O(l) a.) How many moles of sodium oxalate are there in the flask? b.) How many moles of potassium permanganate were titrated into the flask to reach the endpoint? 30 c.) What is the molecular weight of potassium permanganate? Potassium dichromate is used to titrate a sample containing an unknown percentage of iron. To reduce all of the iron to Fe2+ ions, the sample is dissolved in an H3PO4/H2SO4 mixture. The solution is then titrated with 0.01625 M K2Cr2O7, resulting in the formation of Fe3+ and Cr3+ ions in an acidic solution. For 1.2765 g of sample, the titration requires 32.26 mL of K2Cr2O7. (a) Using the half-reaction method, balance the net ionic equation. (b) Determine the iron % in the sample. (c) Does the sample consist of ferrous iodate, ferrous phosphate or ferrous acetate? Answer. a.) 6Fe2+ + Cr2O72– + 14H+ → 6Fe3+ + 2Cr3+ + 7H2O b.) Determination of Fe(II) in solution: (0.01625 mol/L) × (0.03226 L) = 0.000524225 mol dichromate Since the ratio of ferrous ion to dichromate is 6:1 0.000524225 × (6/1) = 0.00314535 mol Fe 0.00314535 mol Fe(II) × 55.845 g/mol = 0.175652 g % of iron in the sample: (0.175652 g/1.2765 g) × 100 = 13.76 % 31 c.) Calculating the % composition of ferrous iodate, ferrous phosphate, or ferrous acetate will help in determining the sample. Ferrous iodate = Fe(IO3)2 %Fe = (55.845 / 405.67) = 13.77 % Ferrous phosphate = Fe3(PO4)2 %Fe = (55.845 / 357.48) = 15.62 % Ferrous acetate Fe(C2H3O2)2 %Fe = (55.845 /173.93) = 32. 11% Hence, the sample consists of ferrous iodate. Practice Questions on Redox Titration Q. For redox titrations, KMnO4 is acidified by: a.) HCl b.) HNO3 c.) H2SO4 d.) All of the above Q. Which of the following is the colour change when acidified potassium dichromate (VI) is used as an oxidising agent? a.) Orange to red b.) Orange to green 32 c.) Yellow to red d.) Yellow to green Q. What are redox indicators? Q. 0.500 g of zinc powder reduced a 25.50 mL acidified solution of 0.200 M AO2+. What is metal A’s final oxidation number? Q. Titration with a solution containing a known concentration of S2O32– (aq) can be used to determine the amount of I3– (aq) in a solution (thiosulfate ion). The balanced equation is used to make the determination: I3¯(aq) + 2S2O32– (aq) → 3I– (aq) + S4O62¯(aq) Calculate the molarity of I3¯ (aq) in the solution, given that it takes 36.40 mL of 0.3300 M Na2S2O3 (aq) to titrate the I3– (aq) in a 15.00 mL sample. Oxidation reduction reaction in electrochemical cell Oxidation reduction reaction: In an oxidation/reduction reaction electrons are transferred from one reactant to another. An example is the oxidation of iron (II) ions by cerium(IV) ions. The reaction is described by the equation In this reaction, an electron is transferred from Fe2+ to Ce4+ to form Ce3+ and Fe3+ ions. A substance that has a strong affinity for electrons, such as Ce4+, is called an oxidizing agent, or an oxidant. A reducing agent, or reductant, is a species, such as Fe2+, that donates electrons to another species. To describe the chemical behavior represented by, we say that Fe 2+ is oxidized by Ce4+ ; 33 similarly, Ce4+ is reduced by Fe2+. We can split any oxidation/reduction equation into two half-reactions that show which species gains electrons and which loses them Balancing oxidation reduction equation 1- Electronically balanced Fe+3 +e- →Fe+2 2- Charge balance Fe+3 +e- →Fe+2 3- Mass balance MnO4- + 5e- → Mn+2 MnO4- + 5e- +8H+ → Mn+2 MnO4- + 5e- +8H+ → Mn+2+4 H2O (Fe+3 +e- →Fe+2)*5 MnO4- +5 Fe+2+ 5e- +8H+ → Mn+2+4 H2O+ 5 Fe+3 H.W: 1- I2+ H2S → 2 I- + S 2- MnO4- + H2SO3 → Mn2+ + SO4 2- 3- Cu2S + HNO3 → Cu2+ + NO + SO4 2- 34 Half reaction : it is often helpful to separate an oxidation – reduction equation two half – reactions one describes the oxidation process and the other the reduction process. Ex : MnO4- + 5e- +8H+ → Mn+2+4 H2O (Fe+3 +e- →Fe+2)*5 MnO4- +5 Fe+2+ 5e- +8H+ → Mn+2+4 H2O+ 5 Fe+3 Oxidation –reduction reactions in electrochemical cell: Oxidation –reduction reactions may be the result of a direct transfer of electrons from the donor to acceptor Ex : Cu 2+ + 2 e - → Cu(s) reduction Zn (s) → Zn 2+ + 2 e - oxidation The most red-ox reaction can be carried out in to two ways. 1- Oxidant and reductant are brought together in the medium.(direct transfer of electrons can take place) 2- The reactants are made a pant of an electrochemical cell in which the two half reactions are physically separated, the electron transfer then takes place via a metallic conductor of electricity. 35 Electrochemical cell : An electrochemical cell consists of two conductors called electrodes, each of which is immersed in an electrolyte solution. In most of the cells that will be of interest to us, the solutions surrounding the two electrodes are different and must be separated to avoid direct reaction between the reactants. The most common way of avoiding mixing is to insert a salt bridge, between the solutions. Conduction of electricity from one electrolyte solution to the other then occurs by migration of potassium ions in the bridge in one direction and chloride ions in the other. However, direct contact between copper metal and silver ions is prevented. The Component of cell 1-Cathodes and Anodes The cathode in an electrochemical cell is the electrode at which reduction occurs. The anode is the electrode at which an oxidation occurs. 2 Salt bridge and liquid junctions Salt bridge : device consists of a U shaped tube that contains a concentrated solution of salt such as KCl. Liquid junction: the porous disk prevents direct reaction between the components of the half-cell if mixing were allowed Cu would deposit directly on the Zn electrode. The voltmeter measures the potential difference, or voltage between two electrodes at any instant. The voltage is a measure of the tendency of the cell reaction to proceed toward equilibrium. 36 Types of electrochemical cell 1- Galvanic cell They act as a source of electrical power, they provides electrical energy. 2- Electrolytic cell They requires an external source of electricity, thy consume electrical energy. The denial cell is galvanic. it could operate as electrolytic cell if using battery having potential grating than 1 v in to circuit in such way as force electrons to flow in opposite direction. Ex: Cu (s) + Zn2+→ Cu 2++ Zn (s) Irreversible cell : in which changing the direction of the current causes entirely different reactions at one both electrodes. Reversible cell : in which reversing the current direction simply results a reversal of the chemical processes that take place I α t the two electrodes. I = Q/ t = coulomb/ second =A Ex: if we have a current 50 A passing through an NaCl (l) electrolysis cell in 1 hour. How mach sodium and chlorine will we produced? Q = 50 * 3600 = 180 c 37  N= Q/{ F ( 96.485) cons.} = 1.87 mol * avagdros no.( 6.02 *10 23) To give the no. of cations Danial cell The Daniell gravity cell was one of the earliest galvanic cells to find widespread practical application. It was used in the mid-1800s to power telegraphic communication systems. the cathode was a piece of copper immersed in a saturated solution of copper sulfate. A much less dense solution of dilute zinc sulfate was layered on top of the copper sulfate, and a massive zinc electrode was located in this solution. The electrode reactions were Currents in Electrochemical Cells 1. Electrons carry the charge within the electrodes as well as the external conductor. Notice that by convention, current, which is normally indicated by the symbol I, is opposite in direction to electron flow. 2. Anions and cations are the charge carriers within the cell. At the left-hand electrode, copper is oxidized to copper ions, giving up electrons to the electrode. As shown in Figure above, the copper ions formed move away from the copper electrode into the bulk of solution, while anions, such as sulfate and hydrogen sulfate ions, migrate toward the copper anode. Within the salt bridge, chloride ions migrate toward and into the copper compartment, and potassium ions move in the opposite 38 direction. In the right-hand compartment, silver ions move toward the silver electrode where they are reduced to silver metal, and the nitrate ions move away from the electrode into the bulk of solution. 3. The ionic conduction of the solution is coupled to the electronic conduction in the electrodes by the reduction reaction at the cathode and the oxidation reaction at the anode. The cell potential : is the difference between two half-cell or two electrode potentials. Electrochemical potential The potential difference between the electrodes of the cell is a measure of the tendency for the reaction reference electrode 1- The Standard Hydrogen Reference Electrode 2-Saturated calomel electrode (SCE) 3- silver chloride electrode Electrode potential : An electrode potential is defined as the potential of a cell in which the electrode in question is the right-hand electrode and the standard hydrogen electrode is the left-hand electrode. Standard electrode potential The standard electrode potential, E0, of a half-reaction is defined as its electrode potential when the activities of the reactants and products are all unity 39 Titrations Based on Redox reactions Analytical titrations using redox reactions were introduced shortly after the development of acid–base titrimetry. The earliest Redox titration took advantage of the oxidizing power of chlorine. Redox Titration Curves To evaluate a redox titration we need to know the shape of its titration curve. In an acid–base titration or a complexation titration, the titration curve shows how the concentration of H3O+ (as pH) or Mn+ (as pM) changes as we add titrant. For a redox titration it is convenient to monitor the titration reaction’s potential instead of the concentration of one species. A red + Box ⇌ Bred + A ox where Aox is the titrand’s oxidized form, and Bred is the titrant’s reduced form. The reaction’s potential, Erxn, is the difference between the reduction potentials for each half-reaction. Erxn = E Box / Bred – E Aox /Ared After each addition of titrant the reaction between the titrand and the titrant reaches a state of equilibrium. Because the potential at equilibrium is zero, the titrand’s and the titrant’s reduction potentials are identical. EBox/Bred = EAox/Ared This is an important observation because we can use either half-reaction to monitor the titration’s progress. 40 Before the equivalence point the titration mixture consists of appreciable quantities of the titrand’s oxidized and reduced forms. The concentration of unreacted titrant, however, is very small. The potential, therefore, is easier to calculate if we use the Nernst equation for the titrand’s half-reaction Note : Although the Nernst equation is written in terms of the half-reaction’s standard state potential, a matrix-dependent formal potential often is used in its place. After the equivalence point it is easier to calculate the potential using the Nernst equation for the titrant’s half-reaction. Calculating the Titration Curve Let’s calculate the titration curve for the titration of 50.0 mL of 0.100 M Fe2+ with 0.100 M Ce4+ in a matrix of 1 M HClO4. The reaction in this case is Fe2+(aq) + Ce4+(aq) ⇌ Ce3+(aq) + Fe3+(aq) 12.1 Note In 1 M HClO4, the formal potential for the reduction of Fe3+ to Fe2+ is +0.767 V, and the formal potential for the reduction of Ce4+ to Ce3+is +1.70 V. Because the equilibrium constant for reaction 12.1 is very large—it is approximately 6 × 1015—we may assume that the analyte and titrant react completely. Note Step 1: Calculate the volume of titrant needed to reach the equivalence The first task is to calculate the volume of Ce4+ needed to reach the titration’s equivalence point. From the reaction’s stoichiometry we know that 41 moles Fe2+ = moles Ce4+ MFe × VFe = MCe × VCe Solving for the volume of Ce4+ gives the equivalence point volume as Note Step 2: Calculate the potential before the equivalence point by determining the concentrations of the titrand’s oxidized and reduced forms, and using the Nernst equation for the titrand’s reduction half-reaction. Before the equivalence point, the concentration of unreacted Fe2+ and the concentration of Fe3+ are easy to calculate. For this reason we find the potential using the Nernst equation for the Fe3+/Fe2+ half-reaction. Note Step 4: Calculate the potential at the equivalence point. Adding the equations together to gives Because [Fe2+] = [Ce4+] and [Ce3+] = [Fe3+] at the equivalence point Volume of Ce4+ E (V) Volume Ce4+ E (V) (mL) (mL) 10.0 0.731 60.0 1.66 20.0 0.757 70.0 1.68 30.0 0.777 80.0 1.69 40.0 0.803 90.0 1.69 50.0 1.23 100.0 1.70 42 The pH Scale The concentration of H+ or OH− in aqueous solution can vary over extremely wide p Scales are used to compress and more conveniently express range of numbers that span several decades in magnitude. ranges, from1 M or greaterto10−14 M or less. To construct a plot of H+ concentration against somevariablewouldbeverydifficultiftheconcentrationchangedfrom,say, 10−1 M to 10−13 M. This range is common in a titration. It is more convenient to 43 Q 44 Salts of Weak Acids and Bases 45 46 Buffers—Keeping the pH Constant 47 48 Salt Protic acids 49