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Chem 330 Final Study Guide

by: Leslie Pike

Chem 330 Final Study Guide Chem 330

Marketplace > Western Kentucky University > Chemistry > Chem 330 > Chem 330 Final Study Guide
Leslie Pike
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Everything that we covered in lecture.
Quantitative Analysis Chemistry
Dr. Darwin Dahl
Study Guide
50 ?




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This 11 page Study Guide was uploaded by Leslie Pike on Monday June 27, 2016. The Study Guide belongs to Chem 330 at Western Kentucky University taught by Dr. Darwin Dahl in Fall 2015. Since its upload, it has received 33 views. For similar materials see Quantitative Analysis Chemistry in Chemistry at Western Kentucky University.


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Date Created: 06/27/16
General things from 222: Molarity is in units of moles solute per liter solution, or millimoles of solute per milliliters of solution. Molarity times volume equals moles. You will be making your own solutions in lab now, the TA will not make it for you, so it is important that you know how to prepare a dilute solution from a concentrated stock solution. % by mass: mass solute/mass total * 100%. This is unitless. Parts per million: mass solute/mass total * 1,000,000. Units are mg/kg or mg/L for dilute aqueous solution. (One liter of water weighs one kilogram, so liter and kilogram can be used interchangeably.) To prepare a 20 ppm solution NaCl, you would weigh out 20 mg NaCl and add it to 1 liter of water. Your containers in lab will be marked in % by mass, and specific gravity. In order to perform a dilution, you must convert these figures to moles. Specific gravity is density, and it is in units of grams solution/mL solution. If you were to calculate the density of a solution, you would measure out a certain volume and then weigh it, and divide mass by volume. The mass of the solute is not used in finding the density of a solution; therefore, the units of density are grams solution and mL solution. For example, a solution of HNO tha3 is 70% HNO by mas3 and has a specific gravity of 1.4: 70 g HNO /100 g solution * 1.4 g solution/mL solution * 1000 mL/L * mol HNO /63 g 3 3 HNO =315.6 mol HNO . 3 Use the ideal gas law, PV=nRT, to find the number of moles of a gas from the pressure, volume, and temperature. Pressure must be in atm, volume in L, and temperature in K. Dahl is tricky, he will give you the wrong units on purpose to see if you are paying enough attention. Sample problem: You have a solution that is 0.1 M Mg 2+and 0.1 M Fe . Both of these metals form insoluble hydroxides. You want to separate the two ions by precipitating one out of solution as an insoluble hydroxide. Is this feasible, and if so, by how much do you need to raise the pH in order to accomplish separation? (Dahl made these numbers up) Ksp for magnesium hydroxide is 10^-8, and Ksp for iron(III) hydroxide is 10^-14. Solve for the solubilities of these hydroxides using ICE. (You learned how to do this in 222.) Iron has the lower solubility; therefore, it is the one we will precipitate out. A precipitation is considered complete when only 1% of the original concentration is remaining. The iron is 0.1 M, so 1% of this is 0.001 M. Solve the Ksp expression for the hydroxide concentration needed to achieve this. 10^-14 = (0.001)[OH ] , [OH ]=0.000215 M We don’t want any of the magnesium to precipitate out. Therefore, our hydroxide concentration must not get high enough to cause this. The highest acceptable concentration of hydroxide is: (once again, solve the Ksp expression) 10^-8 = (0.1)[OH ] , [OH] = 0.000316 M Our minimum hydroxide needed to precipitate the iron is smaller than the maximum hydroxide concentration feasible without precipitating the magnesium. Thus, this is doable. Converting the [OH] to pH gives a pH range of 10.3-10.5. The pH must be kept within this range to keep the iron precipitated without allowing the magnesium to precipitate. Chapter 11: Competing equilibria In practice, the solubility of barium carbonate in water far exceeds the value that would have been calculated from the Ksp of barium carbonate. Why is this? Carbonate reacts with water to form carbonic acid. This removes carbonate ion from the water, and more barium carbonate dissolves to compensate. To calculate the observed solubility of barium carbonate, we solve what is known as a competing equilibria problem. This has 7 steps: 1. Find a variable to express solubility. Since the amount number of moles of aqueous barium is equal the number of dissolved moles of barium carbonate (one mole of barium per one mole of barium carbonate), the concentration of barium is equal to the solubility. If we were calculating the solubility of silver carbonate, the solubility would be equal to half of the silver ion concentration (silver carbonate contains two silver atoms, so the silver concentration is double the amount of compound dissolved, so the solubility is half the amount of silver in solution). 2. Generate all K expressions for reactions taking place. (Shown in photo.) 3. Generate a mass balance expression for the reaction. (Shown in photo.) Barium carbonate contains equal amounts of barium and carbonate. However, some of the carbonate is present as bicarbonate ion or carbonic acid. 4. Generate a charge balance expression for the reaction. (Shown in photo.) Cations on the left, anions on the right. Anything with a charge of +2 or +3 is multiplied by the corresponding coefficient. (An ion with a +2 charge contains twice the charge of an ion with a +1 charge; therefore, its concentration is multiplied by 2 to compensate for this.) 5. Make sure you have an equation for every unknown. 6. Simplify everything that can be simplified. In this case, since barium carbonate produces a basic solution, we know automatically that this solution will contain very few (if any) acidic molecules. Thus, we remove carbonic acid from the mass balance equation, and hydronium from the charge balance equation. 7. Solve. If you get a nasty expression, it is not necessary to solve it numerically for the exam, but you must at least come up with an expression for solubility. Since the barium concentration is equal to the solubility, we devise an equation in which the only unknown is barium concentration. (All K will be given to you for the exam.) Mass balancing an equation for ammonium sulfate, (NH ) S: 4 2 + 2- - [NH ]3+ [NH ] =42([S ] + [HS ] + [H S]) 2 There is a 2 in front of all the sulfur things because ammonium sulfate contains two ammonium ions and one sulfur ion, meaning that the ammonium concentration would be double the sulfur concentration. To make the ammonium concentration and the sulfur concentration equal, sulfur must be multiplied by 2. Charge balancing barium phosphate: 2+ + 3- 2- - - 2[Ba ] + [H ] = 3[PO ] + 2[4PO ] + [H PO ]4+ [OH ] 2 4 Competing Equilibria with a buffer involved Solution of calcium oxalate is buffered at a pH of 4. What is the solubility of calcium oxalate? 1. Oxalate will react with water to form bioxalate and oxalic acid. Set solubility equal to the calcium ion concentration. 2+ 2- - - 2- 2. Write all K expressions. Ksp=[Ca ][C O ], Kb1=[2C 4 ][OH ]/[C O ],2 4 2 4 Kb2=[H C O ][OH ]/[HC O ], Kw=[H ][OH ] + - 2 2 4 2+ 2 4 - 2- 3. Mass balance. [Ca ]=[H C O ]+[H2 O2]+4C O ], 2H+4=10^-42 [4H-]=10^- 10 (we can get these because we are given the pH at the beginning of the problem) 4. Cannot charge balance because we do not know what the buffer is. 5. Number of equations equals number of unknowns, despite that we do not have a charge balance equation. 6. No simplifications can be made, but that is ok because pH problems are simpler anyway. 7. Using the mass balance equation: 2+ 2+ 2+ 2+ [Ca ]= Ksp/[Ca ] + Kb1*Ksp/([Ca ]*10^-10) + Kb2*Kb1*Ksp/([Ca ]*10^- 20) Chapter 10: Effects of electrolytes on equilibria An electrolyte’s interaction with water molecules affects its solubility. Ksp for an electrolyte is actually the product of activities, not the product of concentrations. Activity is equal to the activity constant times the concentration of the ion, Activity constant is calculated with this equation: Log(γ)= -0.51*Z *sqrt(μ)/(1+α*sqrt(μ)/305) Z is the charge on the ion. Alpha, the effective hydrated radius, is a constant that would be looked up in a table. A hydrated ion is surrounded by water molecules that travel with it (they are attracted to its charge), making its effective radius larger. Dahl will give you different alphas than the textbook. To use the above equation, you need to multiply the textbook alpha by 1000. Mu symbolizes ionic strength. This is calculated as .5 of the summation of the concentration of each ion times its charge squared. Ionic strength for 0.1 M magnesium chloride, MgCl , i2 calculated this way (magnesium is the first ion and chloride is the second ion, note that the chloride concentration is double the magnesium concentration by default and that the negative charge on chloride disappears because the term is squared): μ=0.5(0.1*2 + 0.2*(-1) )=0.3 Statistics stuff Standard deviation, when you have fewer than 20 data samples, is calculated as sqrt((sum of square of differences between each data point and the average value of all data points)/(number of data points minus one)) When you have multiple subsets of data, calculate the sum of square of differences for every subset. Add all of these up, and divide by the total number of data points minus the number of subsets. If you have nine data points as three sets of three, you divide by 9-3=6. Lastly, you take the square root of this number. Round your standard deviation to the nearest nonzero digit. If you get .0004283 as your standard deviation value, report it as .0004. If you get 46,792.56, report it as 50,000. Round your calculated number to however many decimal places your standard deviation went to. If your standard deviation is .002, round your calculated number to three decimal places. The reason we do it this way is because the standard deviation marks the point at which we cannot measure any more precisely. If you calculate an average temperature to be 21.5839204 C, but your thermometer has increments of 0.1 C, all digits after the tenths’ place are meaningless because your instrument cannot measure that precisely to begin with. This limit on precision will be reflected in your standard deviation calculations (say you get 0.3265 after calculating standard deviation, you report your value as 21.6 +/- 0.3). When you add two numbers and both have a standard deviation, take the square root of the sum of the squares of your standard deviation values. When you multiply (or divide, it makes no difference) numbers with standard deviations, use this equation: 2 2 If (a +/- x)(b +/- y)=(c +/- z), z is calculated as: sqrt((x/a) + (y/b) )*c Acids and Bases, advanced You can’t take quantitative analysis chemistry without having taken intro chemistry, so I’m assuming that you’ve been introduced to the concept of acids and bases and pH. You were probably taught to calculate the pH of an acidic solution as –log(acid concentration). This works just fine, until we make up a solution of 10^-8 M HCl and you are asked to calculate the pH. –log(10^-8) = 8, but a pH of 8 means that you have a basic solution. This is obviously not the case, seeing as your solution consists of water and hydrochloric acid! Where did we go wrong? Water can auto-ionize, K=10^-14. As you can see from the very small K value, the auto-ionization of water has a very slight effect on pH (especially if there are other, stronger, acids and bases in solution), and in freshman chemistry you were taught to ignore it. However, in very dilute acidic and basic solutions, the auto-ionization of water makes an appreciable impact on pH. Therefore, we must take it into account in our pH calculation. First, we write our mass balance equation. In this case, hydrogen ion comes from two sources, dissociation of hydrochloric acid and auto-ionization of water. [H+]=[Cl-]+[OH-] The Cl- concentration is known, because it is equal to the HCl concentration. Using the Kw expression, we can remove the variable OH- from the equation: [H+]=10^-8 + Kw/[H+] This is then solved with the quadratic equation. We find that the pH is 6.98. This is a reasonable answer, seeing that the pH of pure water is 7, and this solution is known to be very slightly acidic. Preparing a Gran plot from a titration The convenient thing about a Gran plot is that it is linear (unlike the typical titration curve plot), enabling us to easily fit an equation to the data gathered in lab. When you are titrating an acid with a base, you use this equation: Vb[H+] = -Ka*Vb + Ka*Veq [H+], dependent on Vb (the volume of basic titrant you have added, a variable), is the “y” of your y=mx+b equation. Vb is your “x”. All other values are constants. Your linear fit to your data will be a line with a negative slope; the point at which the line crosses the x-axis is your Veq, or equivalence point volume. Ka can be determined from a gran plot as: -1*slope of fit line. Alpha fractions Alpha fractions represent how much of a weak acid is present in the protonated form, and how much is present in the deprotonated form. α0is the amount of fully protonated acid (0 protons have been lost) divided by the total amount of acid, α r1presents the amount of acid that has lost 1 proton (also the Ka 1cid) divided by the total amount of acid. If the acid is polyprotic, α is 2he amount of acid that has lost 2 protons (also the Ka aci2) divided by the total amount of acid, and so on. It is easy to get this numbering backwards, so remember that the subscript number represents the number of protons that have been lost. The sum of all the alpha fractions for any given acid is 1. We need know only the dissociation constants and the pH to calculate alpha fractions. The denominator is calculated as follows: (# of protons acid has)(# of protons -1) (# of protons-2) [H+] + [H+] Ka 1 [H+] Ka 1a +2… Continue in this fashion until you have accounted for all the protons your acid has. Your final term will consist of no [H+] and will contain all of the Ka’s. The numerator will be one of the terms in the denominator equation. For α , it is the 0 first term. For α ,1it is the second term, etc. Sample problem: we have 0.1 M phosphoric acid, H PO . This is a weak acid. The 3 4 solution pH is 4. What fraction of the acid is present as HPO ? What is4the 2- concentration of HPO ? 4 We want the alpha expression for the acid that has lost two protons: 3 2 α 2 [H+]Ka Ka 1([H2] + [H+] Ka + [H+]Ka 1a + Ka Ka Ka1) 2 1 2 3 We are given a pH of 4, so [H+]=10^-4. The Ka’s we would be given, or we could look up in a table. We could find the concentration of the acid missing two protons by multiplying its alpha fraction by the total acid concentration. Titration of a Dilute Polyprotic Weak Acid You titrate 50 mL 0.05 M H A wi2h NaOH. K1=0.021, K2=6*10^-6. What does the resulting titration curve look like? Calculate the pH when 0, 10, 25, 40, 50, and 60 mL of NaOH have been added. If K1/K2 > 1000, you will see two distinct breaks in the titration curve, instead of just one break like you see for a monoprotic acid. This is the case in this situation. For this problem, our weak acid is actually quite strong, and it is also dilute. We CANNOT use simplifications because of this; we have to work everything the long way. We will use K1 for our Ka until we have used up all of the H A, then we2will use K2. (Makes sense because K1 is the dissociation constant for H A and K2 is2the dissociation constant for HA-.) 2 At 0 mL NaOH added, we have no base in solution. Ka=x /([H A]-x), solvin2 for x and taking the negative log of this gives pH 1.63. At 10 mL NaOH added, we use ICE and find that we have 1.5 mmol H A and 1 mmol 2 HA in solution. We DO NOT use Henderson-Hasselbalch equation, because our Ka is too high and our concentration is too low. We use the unsimplified version: Ka=[H+]([HA-]+[H+])/([H A]-[H+]) 2 Plugging in numbers and solving gives [H+]=0.0108 and a pH of 1.97. At 25 mL, the OH- has converted all H A to HA2. We only have HA- in solution, but HA- is an amphiprotic material. How do we tell what it does? We need a new equation for this (Dahl refers to it as “big” because it is pretty big). [H+]=sqrt(K2*Ca/(1+Ca/K1)) Plugging in our values gives [H+]=0.000279 M and pH=3.55. Now that our H A i2 gone, we will be using K2 for our Ka value. K2 is small enough that we can use our regular equations to solve for pH, instead of the unsimplified ones. - 2- At 40 mL, do an ICE table, and you will get 1 mmol HA and 1.5 mmol A . Use Henderson-Hasselbalch. Use K2 for your Ka. pH=pKa+log(A/HA)=5.22+log(1.5/1)=5.4 At 50 mL, you only have A in solution. Use [OH-]=sqrt(KbCb). Your base, A , 2- contains no protons, so you will use Kb . Ka K1 =Ka 1b =2w, s2 fo1 Kb we 1 substitute Kw/Ka . D2n’t forget to adjust the concentration (it is now 0.025 M instead of 0.05 M) to make up for the volume added during the titration. [OH-]=6.45*10^-6, pH=8.8 At 60 mL, you have A and OH- in solution. The OH- is a much stronger base, so we do ICE to calculate the amount of OH- and then calculate pH=14+log[OH-]=11.96. Mixture of Strong and Weak Acids and Bases You mix 20 mL 0.2 M HCl, 5 mL 0.1 M Ba(OH) , 20 mL 0.22M H CO , and 5 mL 2.1 M3 Na 2O . 3hat is the resulting pH? First, convert everything to mmol. (For the sake of simplicity, we represent 2- carbonate as A .) We have 4 mmol H+, 1 mmol OH- (not 0.5 mmol, pay attention to the formula), 4 mmol H A, an2 0.5 mmol A . 2- From a kinetics perspective, the strong and weak acids are going to react first. 1 mmol of H+ will react with the 1 mmol OH- to give 1 mmol water. In solution, we 2- now have 3 mmol H+, 4 mmol H A, and 0.52mmol A . We still have strong acid left. It will react with the remaining base. 0.5 mmol of H+ 2- will react with 0.5 mmol A to form 0.5 mmol HA-. In solution, we now have 0.5 mmol HA-, 2.5 mmol H+, and 4 mmol H A. 2 We still have strong acid left. HA- is amphiprotic, but the strong acid forces it to act as a base. 0.5 mmol H+ will react with the 0.5 mmol HA- and convert it to 0.5 mmol H 2. In solution, we now have 2 mmol H+ and 4.5 mmol H A. 2 We don’t have any bases left, only acids, so the reaction is done. Ignore the pH contribution of the weak acid. The pH is calculated as pH=-log[H+]=-log[2/50]=1.4. (We get the 50 from adding up the volumes of the various components of the mixture.) To do the calculations for the lab: [H 3O ]4= (Eq point 2 – Eq point 1)*(molarity NaOH)/(10mL) [H 2O ]4= (Eq point 1 – (Eq point 2 – Eq point 1))*(molarity NaOH)/(10mL)/2 KHP = 204.23g/mol Choosing a pH indicator: color change happens at pH=pKa +/- 1. For instance, phenolphthalein has a Ka of about 10^-9, so it changes color from pH 8-10. You want to choose an indicator that will change colors at the pH at which the titration finishes. You saw this in the soda ash lab. Phenolphthalein was used for the first breakpoint, which occurred at a high (basic) pH. Bromophenol blue was used for the second breakpoint, which occurred at a lower (acidic) pH. Polyprotic acids often have multiple breakpoints, so you may use multiple indicators which change in the corresponding pH ranges. Back titrations There has been a drastic drop in the local bird population, and biologists think that thin eggshells may be the issue (eggs break in the nest and don’t last long enough to hatch). Your job: figure out the percent calcium in the eggshell (calcium is needed for eggshell strength). Eggshells consist primarily of calcium carbonate. Carbonate is a base, so you should be able to determine the amount of carbonate by titrating it with an acid. You wash and dry and crush the eggshell, you put it in your volumetric flash, you add DI water, and the eggshell won’t dissolve. Now what do you do? It turns out that calcium carbonate is not soluble in water; however, it is soluble in sufficiently concentrated acid (say HCl). You have no way of knowing how much acid is needed, so you have to add a fairly large amount until all of the eggshell dissolves. You will unavoidably add more acid than is needed to neutralize the base. So, how are you supposed to figure out how much of the acid was actually used to neutralize the base, and how much acid was in excess? You will need to do a back titration. (You did this in Chem 223 with the nickel- ammonia complex.) You add an indicator (any indicator will do, since you are titrating a strong acid with a strong base so you will have a very large break). You then titrate your acidic solution with NaOH until you reach the equivalence point. You subtract moles of NaOH added from moles HCl added to get the amount of HCl that was used to neutralize the carbonate. The rest is stoichiometry. If you added 6 mmol HCl to dissolve the eggshell and you needed 1.2 mmol NaOH to remove the excess HCl, 4.8 mmol HCl was spent neutralizing the carbonate. Do the stoichiometry and you get 0.096 g calcium. Complexometric titrations Used when you have complex metal ions. Number of ligands a metal ion can have is twice the metal’s valency (silver has a valency of 1 and thus can have 2 ligands). The most common titrant used in complexometric titrations is EDTA, because it reacts with the sample you are titrating in a 1:1 mole ratio. When fully protonated, EDTA has a charge of +2. When fully deprotonated, EDTA has a charge of -4. If we symbolize the non-proton part of the molecule as Y, fully protonated EDTA is H Y 6 2+ 4- and fully deprotonated is Y . A Lewis acid accepts an electron pair, a Lewis base donates an electron pair. Photon emissions and such From longest wavelength to shortest wavelength: radio waves, microwaves, infrared waves, visible light, ultraviolet light, x-rays, gamma rays. Longer wavelengths have lower frequency and therefore lower energy. Imagine that your hand is holding a string that is tied to a hook on the wall. You move your hand up and down very fast, making waves on the string that have small wavelengths. If you keep up for a while, your hand will get tired (you run out of energy) and start moving more slowly and the waves on the string will be longer. Long wavelength, low energy. Energy is Planck’s constant times wave frequency, or Planck’s constant times the speed of light divided by the light’s wavelength. Speed is frequency times wavelength. Total energy of a molecule is the sum of its translational energy, rotational energy, vibrational energy, and electronic energy. Energized molecules emit radiation, and we can identify them from this. Atoms by themselves cannot rotate or vibrate as a whole molecule can, their emission is caused only by electrons changing energy states. This produces line spectra. Whole molecules, with more different types of radiation, produce band spectra. When something is heated and it gives off heat as a result, this produces continuous black body radiation. Multiplicity = 2s + 1, where s is the spin on an electron. M=1 is a singlet (+1/2 electrons are balanced by -1/2 electrons, thus no net s). M=2 is a doublet and it means that you have an unpaired electron. M=3 is a triplet, which means that one of the electrons in a pair has been excited so that it changes its spin (say instead of a +1/2 and -1/2 we have two +1/2 electrons, thus a net s of 1). A resonance emission is when the wavelength emitted by an electron is equal to the same wavelength absorbed by the electron. This does not normally happen; for instance, if you excite an organic compound with UV light, the compound emits light that you can see (visible light is lower energy than UV light, so lower-energy photons are being emitted than what was received, this is called fluorescence). This is because photons can lose energy in a variety of ways before they are re-emitted. Vibrational relaxation: excited electron “vibrates down” to a lower energy level before emitting a photon. Internal conversion: electron crosses from a higher energy level to a lower energy level, such as from s2 to s1. Inter system crossing: electron flips its spin. Vibration is preferred over inter system crossing when possible; therefore, glow in the dark toys work by making vibration impossible (the material is held in a matrix; think glow in the dark plastic Frisbee). A triplet-to-singlet crossing is called phosphorescence. Unlike fluorescence, phosphorescence can continue long after the source of excitation has been removed. Cool things that glow under UV light stop glowing when the light is turned off, but phosphorescing plastic Frisbees that have been left in the sun continue to glow after the sun goes down. The three types of electron pairs we will consider in this class are sigma, pi, and nonbonding. Bonding electrons can be present as bonds or antibonds, antibonds have higher energy. Antibond is symbolized with a star (π and π*, σ and σ*). Sigma antibonds have the highest energy, followed by pi antibonds, nonbonding electrons, pi bonds, and sigma bonds. A nonbonding pair, when energized, can jump to a pi antibond or sigma antibond. Pi bond can jump to pi antibond, and sigma antibond can jump to sigma antibond. For a compound to be analyzed, it must contain either lone pairs or pi bonds, because of the large energy required for σ/σ* transitions and vice versa. Calculating concentration from absorbance Calculating %T: (intensity of light that passed through cuvette)/(intensity of light before passing through cuvette) * 100% Converting from %T to absorbance: log(100%/%T) A=Ebc where E is molar absorptivity in units of L/(mol*cm), b is width of cuvette in cm, and c is concentration of solution in M. If using equation A=abc, a is absorptivity (not molar absorptivity) and things can be in any units. In the homework problems, b is in cm and c is in ppm. If you have a solution with an unknown concentration, you can determine its concentration by measuring the absorbance of the unknown, adding a known amount of a standard of the same analyte in solution, and re-measuring the absorbance. Use this equation: Absorbance −Absorbance standard unknown Volume unknown Concentrationof Unknown= Absorbance unknownoncentration standardlume standard ¿ For instance, you have two 50 mL samples of copper, unknown concentration. You put them in 100mL volumetric flasks. The first you dilute to 100mL without doing anything to it. The second, you add 4 mL of 2 ppm copper, and then you dilute to 100 mL. Absorbance of the first is .2, and of the second is .31. Using the equation above: Concentration = (0.2*2ppm*4mL)/(50mL(0.31-0.2))=0.29ppm. It is not necessary to use any particular unit for this calculation; just be sure that the volume units cancel (absorbance is unitless) and your concentration of your unknown will be in whatever units you used for the concentration of the standard (ppm, ppb, M, it doesn’t matter). We don’t have to convert mass to moles because our standard is the same solute as our unknown and thus has the same molar mass. Sample problem: you have a sample of iron(II) in solution, concentration unknown. You have a standard of iron(II), 0.0002 M. You are using absorbance to determine concentration, and you adjust the sizes of your cuvettes until both solutions have the same absorbance. When the standard cuvette has a width of 2 cm and the unknown cuvette has a width of 4.5 cm, absorbance is the same. Calculate the concentration of the unknown. Use A=Ebc. Absorbance was the same for both samples, so A=A. Both samples are of iron, and E is an intrinsic property of a material, so E=E. We are left with bunknownunknownb standastandardPlugging in numbers, we get 8.9*10^-5 M for the concentration of the unknown. Electrochemistry Standard Hydrogen electrode (SHE) has reduction potential of 0.00V by definition. Pressure is 1 atm and [H+]=1.00. SHE is theoretical, cannot actually be achieved because [H+]=1 cannot be achieved due to activities. A real life SHE must be corrected with the Nernst equation. Example: you have a hydrogen electrode at 720 torr and a pH of 4. What is the potential? The reduction of hydrogen is: 2H+ + 2e- = H . 2 Using Nernst equation: E = E – 0.0592/n*log(Q) = 0 – 0.0592/2*log((720/760)/(10^-4) ) = -0.236 V Remember to square the hydrogen concentration. Q expression is products over reactants raised to the balancing coefficients.


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