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Analytical Chemistry Dr. Manal A. Tooma First year 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, traditionally divide it into five fields: Organic, inorganic, physical, analytical, and biochemistry. Analytical chemistry is a field of science consisting of a set of powerful ideas and methods that are useful in all fields of science, analytical chemistry concemed with determinins the composition of substances. It comprises of two branches: Oualitative analvsis: deals with the determination of the identity of chemical constituents in samples. - Ouantitative analvsis: deals with the determination of the amounts (concentration or composition) of various substances in samples. There are many methods of determining these amounts, all based on chemical or physical properties of atoms, molecules and ions. In general chemical analysis methods can be classification to: . Gravimetric methods determine the mass of the analyte or some compound chemically related to it. This method is base on the measurement of the reaction product by precipitation. .Volumetric method. the method is based on determining the volume of a solution of known strength thatrrtiredreacting with specific amount of the , sample such as acid-base titratioSy' . ElectroanalVtical methods involve the measurement of such electrical properties as voltage, cunent, resistance, and quantity of electrical charge. . spectroscopic methods are based on measurement of the interaction between electromagnetic radiation and analyte atoms or molecules or on the production ofsuch radiation by analytes. Finally, there is a group of Analytical Chemistry Dr. Manal A. Tooma First year miscellaneous methods that includes the measurement of such quantities as mass-to-charge ratio of molecules by mass spectrometry, rate of radioactive decay, heat ofreaction, and rate of reaction, sample thermal conductivity, optical activity, and refractive index. Fundamental Concepts Atomic weieht of element: The mass of a single atom in grams is much too small a number for convenience, and chemists therefore use a unit called an atomic mass unit (amu) also known as a dalton (Da). One amu is defined as exactly one-twelfth the mass of carbon isotope '2C and equal to 1 .66054 x l0-24 g. Example:prove that carbon weighing 1.0 x10-3 g contains 5.01x 1.0 x10-3 e" *#m= G;#\ 1Ole carbon atom? 5'01 x 10re c atoms Molecular weight (Mwtl: is the average mass of a substance's molecules. Numerically, is equal to the sum of the atomic weights of all atoms in the molecule' molecular weight = sum of atomic weight M.wt of water (H2O):2(1.008) +l(16'00):18'02 M.wt of hydrogen (H2) : 2(1'008) :2'016 These weights are all relative to the mass of r2c atom as 12.000 with (SI) system of units' .Molecular weight is expressed by g/mole \.2 H.IY: Calculate the M.wt' of CzHr+Oz, CzHsNzSO+, NazCOr, PzO: Mole (moD: its Avogadro number (6.022137 x somethine consists of 6.022137 x 1023 1023) of species, so one mole of units of that substance' Analvtical Chemistrv Dr. Manal A. Tooma First year The mass of 1 mole of an element is equal to its atomic . no.of mole 1 mot mass of substance in g ------ M.*;-- 36ss ofHCI" 36.5 /mol g Examples: Calculate the number of moles in 200 mg of - in ,/ / -Nr P---e mass in srams D/ CaO? -s-' 10 g of P2O5? in 5 mg of NaCl? in 25 mg olH2Soa? in 400 mg of K2CrO3? L Equivalent weieht: is the molecular weight divided by the number of reacting units (no.of equivalent - valency) e quiv arent w e i g ht - mole cular w eig ht no.of equivalent For acids: the number of reacting units is the number of hydrogen ions that will fumish (acidic hydrogen). For bases: the number of reacting units is the hydrogen ions that will react with it' (Bases hydroxide) For salt: the number of positive parts in the salt Multiplied by its oxidation number' For redox reaction: the number ofreacting units is based on the number of electrons that the oxidizing or reducing agent will take on or supply' It can be computed also as the change in oxidation number (transfer electrons)' H.ftCalculate the Eq.wt. of HzSOr, HrPO+, HCI' CH3COOH, Na2CO3' NaHCO:, Ca(OH)2, NaOH, KOH, Ba(OH)2, K2CrOa, FeCr2Oa, CaClz,FezO:, (Cr*3- Cr*7), (KMnoa- Mnoz), (Mno4- -Mn*2),1K2Cr2or- 2Cr*3) AgNO3, HgCl2 ' -t Analvtical Chemistrv First year Dr. Manal A. Tooma C o ncentr atio No.of mole = ns express io ns mass of th substance in gm wt M.wt. M.wt. moles of solute Molarlty lMl = volum of solution in Liters wt M=-+M.wt. N omaltty IN tt- wt 1000 V meq.of solute volum of solution in liteTs s inml ., ',v wt = Eq.wt. -*- 1000 V i.n mI volume mI Density and SpeciJic Gravity Density (p): the mass per unit volume at the specified temperature. It is usually expressed in g/ml or g/Cm3 at 20 oC oC and O '99823 glml at 2Q'C Density of water: is taken as 1 .0 g/ml at 4 Specific gravity (sp.g.): is defined as the ratio between the densities of solution at2}octo that of water at 4oC, hence it is a dimensionless quantity' Numerical values of density and specific gravity are equal since d*u1., at 4 oC :1 .00 s.lml The wetght fractton -- r"# Z*ff)= Themole -n, fraction n =-= of solute x 100 mass of sample mass 1oo moles of solute total moles of samPle = lmobo/o- 1.00 I Analvtical Chemistrv First year Dr. Manal A. Tooma The volume fractton volume of solute o/o{ = * 100 vt total volume of sample - weight . ---:Lwt fracti"on -- o/o--voLume' V The Parts per of solute * 100 total volume of sample mass million (ppm): Used to express trace concentration of solute. It can be expressed either PPm /wt\ l-l \wt/ = of solute(g) * 106 mass of sample(g) mass mass of solute(mg) --------"------ rJUtrL - mass of sample(kg) mass of solute(g) PP /wt\ \v)=@*''u- PPm = ^A pg solute(mg) S -- or ---i or t*e s"rrrat"A mass ""t a of "f There are a number of expression in which the concentration of solution can be expressed Molaritv: It is defined Molarity [M] = as the number of moles moles of solute vol.of solution in liters of solute in each liter of solution Molarity tMl -#-;ffi Mpls.lifr;the solution which contains I mole of solute per kilogram of solvent. Lr Molali*, t*l t'Irrr - moles of solute 1000 gof solvent Analvtical Chemistrv First year Dr. Manal A. Tooma Normalitv: A one normal solution (1 N) is that contains one equivalent weight of solute per liter of solution. Normality [N] - no of equivalent of vol.of solution in solute liters The relation between Normality and ' Normality [N] = - -!- * +00, Eo.wt. V in ml Molarity: N: M (valency) H.W: 1) What volume of 3 .25 M sulfuric acid is needed to prepare 0.500 L of 0. I 30 M H2S04? ( i' Zl How do we prepare 75.0 ml of 0.96 M sulfuric acid from 18 M acid? ^r, 3) What volume of 6.39 M sodium chloride contains 51.2mmol sodium chloride? ({. 4) Rubbing alcohol is an aqueous solution containing 70% isopropyl alcohol by volume. How would you prepare 250 ml rubbing alcohol from pure isopropyl alcohol? 5) How many grams ofglucose and ofwater are in 500 gof a 5.3%oby-mass ^r/ glucose solution? Calculate the molar concentration (Molarity) of a solution contains I ppm of 6) lead (Pb), (l M.utt:207? 7) A sample with mass of 2.6gcontains 3.6 pg of zinc; calculate the \?' concentration of zinc in PPm? 8) 150mL of an aqueous sodium chloride solution contains 0'00459 ir' NaCl. Calculate the concentration of NaCl in parts per million (ppm)' 9) What mass in mg of potassium nitrate is present in 0.25kg of a 500ppm KNO31uql? . 10) { 1I A student is provided with 500m1 of 600ppm solution of sucrose. what volume of this solution in millilitres contains 0.159 of sucrose? ) Convert 78 ppm of Ca2* ions to mol/L? d" (t Analytical Chemistrv First year Dr. Manal A. Tooma Concepl of chemical equilibrium If two reactants A and B are mixed together a chemical reaction can take place to yield the product C&D The chemical reaction can be presented as follows: aA + bB -------> cC + 6P a, b, c, d: stoichiometric coefficients In many chemical reactions the reverse reaction (i.e.) reaction between c with D to form A&B the reaction is also possible hence the reaction is presented by: aA+bB <+------> cC+dD when forward and reverse reaction becomes equal, the system has reacted to the condition known as "chemical equilibrium" the reaction appear to stop when equilibrium condition is actually reached since there is no change in the concentration of all substances. The equilibrium constant the equilibrium condition is often expressed by a factor or known as equilibrium constant &: &: [B]b ku [C]" [D]o kr [A]" kr [A]" [B]b : rate of the forward reaction rate of the backward reaction ku [C]" [D]d when the system reach equilibrium K:# = ffis - When R; decreases with time, R6 increases with time and the two rate become equal at equilibrium. - Equilibrium constants are useful for calculating the concentration of varies ions in equilibrium Analvtical Chemistrv First year Dr. Manal A. Tooma Faclors affecting on chemical equilibrium Chemical equilibrium is a very delicate system that represents a perfect balance between forward and reverse reaction. A small disturb in the equilibrium may shift the equilibrium position either to right forming more products or to left forming more reactants. This reaction by the system is ofcourse temporary and eventually the system will come back to equilibrium. This phenomenon can be expressed in the form ofLe Chatelier's Principle (ifan external the system will tend to react in such way stress is applied to a system at equilibrium as to relieve the applied stress and tries to reestablish the equilibrium). In chemical reaction terminology, the "stress" means change in concentration, pressure, volume or temperature. Change in Concentration Consider the following equilibrium reaction Nz+3H2 +-> ZNH, If we add either Nz or H2, we increase the collisions between Nz and Hz thereby forming more product NHl. This is how the equilibrium counteracts the applied stress; we say the equilibrium shifts from left to right. If we add more NH1, we increase the concentration of NH3' As a result, some NH3 decomposes and forms more reactants. We say the equilibrium shifts from right to left' If we remove either Nz or H2, now there is less concentration of Nz and H2' To offset the applied stress, the equilibrium shifts from right to left. If we remove some NH3, the equilibrium shifts from left to right to counteract the applied StTESS. lr t 2 '1\, ---d Chanses in Pressure and Volume Pressure does not have any effect on concentrations of species that are present in solid, liquid or solution form. On the other hand, the change in pressure affects the concentrations ofgases. According to ideal gas law, pressure and volume are inversely proportional to each other; the greater the pressure, the smaller the volume, and vice versa' Analvtical Chemistrv First year Dr. Manal A. Tooma Thus, it is just enough to understand the effect of change in pressure on the equilibrium system. Nz+3H2 <+ 2NH3 lmol+3mol 2mol vol + 3 vol 2vol 1 What will happen if we increase the pressure (decrease the volume) on the system at constant temperature? In order to understand the affect ofpressure, it is very important to understand the total volume of reactants and products. According to Avogadro's law, the number of moles is directory proportional to the volume of the gas. Think that one mole is one volume. Therefore, there are 4 volumes (l vol + 3 vol) on the reactant side and 2 volumes on the product side. The increase in pressure always affects the side that has more volume. Hence, increase in pressure shifts the equilibrium from left to right. The Dressure has no effect if the total volume of reactants is equal to the total volume of the products as in the following example. Hzs + Changes in Temperature: A Izs <+ ZHI, change in concentration, pressure or volume alters the position of the equilibrium but not the magnitude (value) of the equilibrium constant. However, the change in temperature changes the value of the equilibrium constant. To understand the effect of temperature, we must know whether the reaction is endothermic (absorption of heat) or exothermic (release ofheat). Let us consider the equilibrium reaction between dinitrogen tetroxide and nitrogen dioxide: NzO+ (g) ---_-=- 2NO2@) This reaction is endothermic in the forward direction and exothermic in the reverse direction (g) ------------r> 2NO2 (g) 2NO2 @) ----------> NzOq (g) NzO+ aH= AH = 58.0 kJ/mol -58'0 kJ/mol - Analytical Chemistry First year Dr. Manal A. Tooma To understand the effect oftemperature (heat), let us re-write the above equations treating the heat as chemical reagent. Thus heat+NrOo 2NO2 (g) ------------> 2NO2 (g) G) N2Oa (g) + heat Therefore, increase in temperature favors the endothermic reaction (forward -_-r> reaction, i.e. left to right) while decrease in temperature favors the exothermic - reaction (reverse reaction, i.e., right to left). What does it mean? It means that the value ofthe equilibrium constant increases when the heat is added (increase in temperature) and decreases when the heat is removed (cooling the system) that can be explained by the following equilibrium constant expression: . r<" = - [Nor]z fi, or) The Effect of a Catalvst The function of a catalyst is to speed up the reaction by lowering the activation energy. The catalyst lowers the activation energy of the forward reaction and reverse reaction to the same extent. Due to this, there is no shift in equilibrium or the change in the value ofthe equilibrium constant. Therefore, we conclude that the catalyst has no effect on the equilibrium system. (When the initial concentration is > 100k, we will neglect the small change in concentration. IJ not we can solve it by general quadratic formula) x - initial -ot'lo'-+ac 2a Type of equilibrium conslants There are several types ofequilibrium constants according to the type of reaction as below: - Dissociation constant (IQ) Formation constant (K1) Acid & base dissociation constant (K" , Ku) Ionization constant of water (K*) 10 Analytical Chemistrv First year Dr. Manal A. Tooma The following examples give an idea for calculations using equilibrium constants t( l. the chemical substances (A) and (B) are reacted to produce (C) and (D) according to the reaction A +B <--+ C +D if 0.2 mol of (A) and 0.5mol of (B) are dissolved in 1 liter solution and the reaction takes place with equilibrium constant K: 0.3 calculate the concentrations of reactants and products at equilibrium 2. Calculate the equilibrium concentration ofA and B in a 0.10 M solution weak electrolyte AB with an equilibrium constant of 3xl0'6 3. Calculate the equilibrium constant for a reaction Aaq + AB Baq <------+ a of A+B? 3Coo between 0.10 M A and 0.30 M B, given that the product, C has a concentration of 0.20 M and that these values are those when the reaction is in a state of eq ui I i brium. 4. (ans. 0.267 ) Calculate the equilibrium constant (K) for the following reaction: HrG)+Iz(e)+2HI(e) when the equilibrium concentrations at 25 "C were found to be: lHrl :0.050s M, [Ir] :0.0498 M, [HI] :0.389 M ?? ans:60.02 5. Calculate the equilibrium constant (K'o) for the following reaction: Nz(g)+3Hdg)+2NHr(g) when the equilibrium concentrations at 500 t3Ht :0.037 6. M, K were found to be: [N2] :0.03 M, [NH3] = 0.016 M ?? ans:0.017M Calculate the equilibrium constant (K"o) for the following reaction: 2SOz(g)+Oz(g)+2SO3@) when the equilibrium concentrations at 800 M, [SO2] :0.003 M, K were found to be: [O2] : 0.0035 [SO3] = 0.05 M ?? L1, Analytical Chemistry First year Dr. Manal A. Tooma Acids, Bases and buffers Acids - base Theories l- Arrhenius Theory (1884) Acid is any substance that ionized in water (partially or completely) to give hydrogen ions Base: is any substance that ionized in water to give hydroxyl ions According to this definition, HCI is an acid NaoH is a base as shown below: HCI rf + ct- _+ NaOH ----+Na* + 6p Acid chemically reacts with a base as follow: HCI + NaOH ------f Na* + 61- a g16 2- Brsnsted-Lowry Theory (1923) Acid: is any substance that can donate a proton. Base: is any substance that can accept a proton. 3- Lewis Theory (1923) Acids: is a substance which can accept an electron pair by combining with a second substance with al unshared pair of electrons. A base: Is a substance which shares it unshared electron - pair during chemical reaction BCl3 +:NH3 C6B:NH., --_' Conjugate Pairs: Acids: are base pairs in the ionization reaction are called pairs coniuqate base The conjugate base ofthe acid and acid is the conjugate base Acid = H'+ base 2H2O = H3O'+ Off 12 Analytical Chemistry First year Dr. Manal A. Tooma NH4*:11+ + NH: H2O: conjugated acid for OH- NH3: is conjugated base for (NI{4-) or Nlla": is conjugated acid for (NH3) base Acid- Base strength: when acid or base is dissolved in water it will dissociate or ionize. The degree ofionization depends on the strength ofthe acid A strong electrolyte is completely dissociated A weak electrolyte is partially dissociated weak acid has a relatively small dissociation constant (Ka) where's strong acid has a large dissociation constant. The strength ofan acid depends on its type and is not related to the concentration. (e.g.): HCI strong acid regardless of whether its concentration. 1M or l0-a M Strong base: is a base with relatively large K5 Weak base: is a base with relatively small K5 Acid bases equilibria in water Kw: tfl tOHl : 1.0 x 1O-ta(ionization trfl toHl : 1.0 x lo-'a Ffl = toHl: Conc. of 1.0 x content of water) 1o-7 [Ff] in a solution is often expressed as the pH of the solution which is the negative of the logarithm of [H-] pH:-log [Ff] pOH: - 1og [OFI] pKw:pH+pOH : l4 H.l/. : Calculate the pH of 2xl0-3 M of hydrochloric acid solution? In order to determine the pH of weak (partially ionized), strong (completely ionized) acids or bases and salts, different calculations must be followed as shown in the followine: Analytical Chemistry First year Dr. Manal A. Tooma 1- Strong acid / base Since the strong electrolye is completely dissociated for that the concentration of IFfl/ [OH-] is equal of the initial concentrarion. Exp: What is the pH of a 0.010 M HCI solution? Since HCI is a strong acid, the hydronium ion concentration will be equal to the HCI concentration: lH,o-l:0.010 M The pH can be found by taking the negative log ofthe hydronium ion concentration: pH 2- : : -log[H3O-] lVeak acid -log(0.01 0) : 2.00 / base It's important to realize that a weak acid is not the same thing as a dilute solution ofa strong acid. Whereas a strong acid is 100% dissociated in aqueous solution, a weak acid is only partially dissociated. It might therefore happen that the concentration from complete dissociation of a dilute strong acid is the same as that from partial dissociation of a more concentrated weak acid. Equilibrium exists between the weak acid, water, H:O', and the anion of the weak acid. The equilibrium lies to the left hand side ofthe equation, indicating that not much H3O* is being produced. The fact that very little H3O* is being produced is the definition of a weak acid. The Ku for a weak acid is small, usually a number less than 1' HA (ad + H2O 0) € t,^ nt,. -- Aiuql + H3O*1uq; lH3O+llAl IHA] L4 AnalWical Chemistrv First year Dr. Manal A. Tooma Note that water has been omitted from the equilibrium equation because its concentration in dilute solutions is essentially the same as that in pure water and pure liquids are always omitted from equilibrium equations Exp: The initial concentration of HNO2 is 0.45 HNO2luqy Init + Hrofl) 5 NO2 ("q) 0.45 0.45 Ku : 4.5 x 10-a H3O*1aqy 0 Change -x Equal. * M, *x +X -x Ka= [NO,-ILH3O-I li{No,l _ (x)(x) _,,,<,.,^.4 0.45 - x Simplifies to: since K" *100 < Ci x: [No2] : tH.l : "f 4 c' []rNot :0.45-0.014 roHt=#:# - :,l4sffix 0.436,lJ{ 7.7x'J.0-73 H.IY: Codeine (CrsHzrNO:), a 0.45 = 0.014 a drug used in painkillers and cough medicines, is naturally occurring amine that has kb :l .6 x 10-6. Calculate the pH and the concentrations of all species present in a 0.001 2 M solution of codeine. I) Analytical Chemistrv First year Dr. Manal A. Tooma We've seen that the strength of an acid can be expressed bv its valuc of K" and the strength of a base can be eipressed by its value of ks. For a lonjugate acid-base par, the hvo eq,rili!5ig6 constants are related in a simple way thar makes it possible to calculate either one from the other. L.et's consider the conjirgate acid-base pair NH.1' and NH.1, for example, $'here K., refers to proton transfer fiom the acid NHl - to n'ater and K6 refers to proton transfer from u'ater to the base NH3 Thc sum of the two reactions is simply the dissociation of water: N]+fldiil + H2O(/) lnH:trrl) + Net H2O1/,1 2H2O1/; -'. H,"O+(rr(i) + N!*+*tnq)-t NHttr4i OH*(rr4) ::*:: Hro-(d(l) + OH-tnf) Kn * [H3O*][NH3] _ 5.6 x 10*ro lNHr*l [NH{,'][OH ] _ J{6 _ i.8 x 10 "5 INH:l K," = [H3O+][OH*] = 1.0 >( 10-14 The eqrrilibrium constant fnr the net reacfion equals the product of the equilibrium constants for the reactions added: Kn;\Kb= [{3C)-]tr{r{51 lNHr*l (5.5 x 10-10X1.8 k*: {N}*al[oH ] :[H:O'][OH]:K". {NHt ;< 10-5) = 1.0 x 10-14 ku. k5 ku: k*/ k6 or k6: kw/ka 3- pH ofSalt Salts such as NaCl that are derived from a strong base (NaOH) and a strong acid (HCl) yield neutral solutions because neither the cation nor the anion reacts appreciably with water to produce or ions. As the conjugate base of a strong acid, has no tendency to make the solution basic by picking up a proton from water. As the cation of a strong base, the hydrated ion has only a negligible tendency to make the solution acidic by transferring a proton to a solvent water molecule. The following ions do not react appreciably with water to produce either H3O- or OHions: I Analvtical Chemistrv First year Dr. Manal A. Tooma . Cations from strong bases: Alkali metal cations of group 1A: Li*, Na*, K* Alkaline earth cations of group 2A: Mg*2 , Ca*2, Sr*2, Ba*2 except for Be*2 . Anions from strong monoprotic acids: Cl-, Br-, I-, NO3-, and ClO3- Salts that contain only these ions give neutral solutions in pure water (pH: 7). _ Salts such as NlIaCl that are derived from a weak base (NH3) and a strong acid (HCl) produce acidic solutions. In such a case, the anion is neither an acid nor a base, but the cation is a weak acid NH4 *tuq) f HzOtrt <-----------+ H3O*1uq; -I- NH31uq; Finally, salt such as (NHa)2CO3 in which both the cation and the anion can undergo proton-transfer reactions. Because is a weak acid and is a weak base, the pH ofan (N}I4)2CO3 solution depends on the relative acid strength of the cation and base strength ofthe anion: We can distinguish three possible cases: . ku> ku If ku for the cation is greater than k5 for the anion, the solution will contain an excess of H3O' ions (PH< 7 ). . ku< ku If ku for the cation an excess of OH- ions (PH> . ku= is less than k5 for the anion, the solution will contain 7 )' kr If k" for the cation and k5 for the anion are comparable' the solution will contain approximately equal concentrations of H3O* and OH- ions (pH:7 )' - Calculate the pH of a 0.10 M solution of AlCl3; k" for Al(H2O;u*3 , is 1'4 x 10r' - calculate the pH of a 0.10 M solution of NacN; ku forHCN is 4.9 x10-r0 - Calculate the pH of 0.20 M NaNO2; ku for HNO2is 4'6x 1 0'-' T7 Analvtical Chemistrv First year Dr. Manal A. Tooma The human eye is not sensitive to color differences in solution containing a mixture of HIn and In- , particularly when the ratio [HIn]/ [In-] is greater than about 10 or smaller than 0.1 So we can write that the average indicator, HIn, exhibits its pure acid color . lHlnl 10 when t; and is base color when ffi luInl HrO' , H,O' - = K^= 'ltn-l = l0 lHtnl : K"= unj :0.1 K" K" [Htnl 1 - ffi =* for full acid for full base : PK" -1 pH = - log (0.1 K") : PKu +1 pH: - log (10 K") Indicator PH range : PKa +1 pH (acid- base): 2 --1 - 4-Reaction of strong acids and strong bases M HCI with 0' 1 M NaOH Plot titration curve for the titration of 50 ml of 0'05 (9, 10, 2j, 24,25,26,30, 4p ml. . 2- Calailarethe pH during the titration of 50 ml of 0'05M NaOH with 0'l ',. volume of reagent after the addition afteladding the following M HCI (24'5'25'25'5) ml? 3- when mixing 20 ml of 0'2 M HCI Calculate the pH of the solution that result with 25 ml of 0.132M NaOH, 0'323M NaOH? 19 Analytical Chemistry First year Dr. Manal A. Tooma 4- Plot titration curve for the titration of 50 ml of 0.1 M HCI with 0.2 M NaOH.(0. r0,20, 24, 25, 26, 30,40)ml. Titration I - for weak acids or bases Plot curve for the titration of 50 ml of 0.1M acetic acid with 0. 1 M NaOH (0. 10,25,40,50, 60, 70) ml. 2- Calculate the pH for the titration of 50 ml of 0.05 M NaCN with 0.1 M HCy(0, 10,25,26) ml. 3 - Calculate the pH of solution prepared by :a) dissolving 43 g of lactic acid in water and diluting to 500 ml Titrationfor weak acids and strong bases (buffer) If ions into a a strong base is added, the buffer converts the excess hydroxide base by reacting with the Brsnsted acid: s" o !Ho49! llofore titration tH1=JKaiH6'Ee1 With titration pH: At equivalent Point loHl:.p too.-l [oAc1 > looH After equivaience pKa * foe loAc-l ffi Henderson-Hasselbalch Equation [OHl=[excesslitrant] poiut excess tihant q for titration of 50m1 of calculate the pH and plot the resulting titration curve Ka = 0.1M HOAC by addition of 0, 10, 25' 50 and 60 ml of 0'lMNaOH' 1 .75x 10-s. 20 Analvtical Chemistrv Dr. Manal A. Tooma First year Titrationfor weak Ifa base and strong acid (buJfer) strong acid is added, the buffer converts the excess protons into an acid by reacting with a weak conjugate base Befbre titration IOHI: With titration pH At equivalent point [H*] = Alter cquivalcnce : pKa + loe INH"i fi;fr lNtr4l > loo # fH*J=[excesstitrant] '''--i e ltNH-r l([ q ':---" 100 rft[NH3] ,- ' -\ i . '. i" r r..+.." Calculate the pH and plot the resulting curve for the titration of 20 ml of 0'1lM ammonia by the addition of 0,5, 11,15,20 ,22,25,3},35,and 40ml of 0'1M HCI Kb: 1.79x l0-5? 'ii\ ./ consider the titration of 100.0 ml of 0.016 M HOCI with 0.0400 M NaOH' How many milliliters of 0.0400 M NaOH are required to reach the equivalence M NaOH point? Calculate the pH: (a) After the addition of 10'0 ml of 0'0400 (b) Halfway to the equivalence point (c) At the equivalence point - acid titration, the Shows the pH titration curve for a typical weak base-strong l0-5 40.0 ml of 0.100 M NH3 with 0'100 M HCI' Kb 1'8 x : titration of H.t/: or [H3O-] 1- For each of the following solutions, calculate [OH] from [H:O-], from [OH-] Classifi each solution as acidic' basic' or neutral' [H:ot] : toHl :1.x10-r0M, 0.01M, 5.6x10-eM, 3.4 xl0-e M , 8.6 xl0-5 M, M, 2'5 xl0-4 M, 2'0M 1'5xl0'M, l'0xI0-7M 1.0 x10-7 2I Analvtical Chemistrv First year Dr. Manal A. Tooma 2- Water superheated under What is [H3O *]and pressure to 200 oC and 750 atm has Kw: I .5 Xl0-1' [OH-] at 200'C? Is the water acidic, basic, or neutral? 3- Water at 500 "C and 250 atm is a supercritical fluid. Under these conditions, Kw is approximately I .7 Xl0-re Estimate [HrO *] and [OH-] at 500 oC. Is the water acidic, basic. or neutral? 4- Calculate the pH to the correct number of significant figures for solutions of Question I? 5- Calculate the concentration to the correct number of [H3O "], [OH] significant figures for solutions with the following pH values: 4.1 , 10.82 ,0.00 , 14.25 , 5.238, -1.0,9,14.25 -0.3, 10.75 ,/6- A solution of NaOH has a pH of 10.50. How many grams of CaO should be dissolved in sufficient water to make 1.00 L of a solution having the same pH? 7- A solution of KOH has a pH of 10.00. How many grams of SrO should be dissolved in sufficient water to make 2.00 L of a solution having the same pH? 8- Calculate the pH ofsolutions prepared by: (a) Dissolving 4.8 g of lithium hydroxide in waterto give250 mL of solution (b) Dissolving 0.93 g of hydrogen chloride in water to give 0.40 L of solution (c) Diluting 50.0 mL of 0. I 0 M HCI ro a volume of 1 .00 L (d) Mixing 100.0 mL of HCI and 400.0 mL of HCIOa (Assume that volumes are additive.) 9- Calculate the pH ofsolutions prepared by: 22 Analvtical Chemistrv First year Dr. Manal A. Tooma (a) Dissolving 0.20 g of sodium oxide in water to give 100.0 mL of solution (b) Dissolving 1.26 g of pure nitric acid in water to give 0.500 L of solution (c) Diluting 40.0 mL of 0.075 M Ba(OH)2 to a volume of 300.0 mL (d) Mixing equal volumes of 0.20 M HCI and 0.50 M HNO3 (Assume that volumes are additive.) l0-Look up the values of k for C6H5OH, HNO3, CH3CO2H, and HOCI, and aftange these acids in order ofi a) Increasing acid strength Increasing conjugated acid strength 1 l- Look up the values of k for HCO2H, HCN, HCIO4, and HOBr, and arrange these acids in order of: a) Increasing acid strength, b) Increasing conjugated acid strength 12- The pH of 0.040 M hypobromous acid (HOBr) is 5.05. Set up the equilibrium equation for the dissociation of HOBr, and calculate the value of the aciddissociation constant. 13-Lactic acid (C3II6O3), which occurs in sour milk and foods such as sauerkraut, is a weak monoprotic acid. The pH of a 0.10 M solution of lactic acid is 2.43. What are the values of ka and oka for lactic acid? 14-Acrylic acid (HC3H3O2) is used in the manufacture of paints and plastics. The pka of acrylic acidis 4.25. (a) Calculate the pH and the concentrations of all species tH3Ol, C:H:O:-, HC3H3O2, and OH-in 0.150 M acrylic acid. (b) Calculate the percent dissociation in 0.0500 M acrylic acid. \\,, st year - Enviroffnental Chemistry Dr. Manal A. Tooma Photochemistry Photophysics and photochemistry both deal with the impact of energy in the form ), otphotons on materials. Photochemistry focuses on the chemistry involved as a material idimpacted by photons while photophysics deals with physical changes that result from theimpact of photons. we will focus on some tophotophysics and photochemistry followed by principleswill be related to of the basic principles related general examples. Finally these ways there is a great similarity between amaterial's the material is small ormacromolecular. to transferthe effects ofradiati Iex. Ph magnetic involves ofPlanck's thewavel icles; ejection of el particle behavior while the properties. The Grotthus-Draper law statE when a bal reactions onlyoccur .t -- photon of light is absorbed. for the First Law ofphoto_ chemistry, that is, only light that is absorbed can have aphotophysical/photochemical effect. '.tirTffiasis We can write this as follows. M + lig61-214* O.r, ,.* - Environmental Chemistry Dr. Manal A. Tooma where M* is M after it has taken on some light energy that it has acquired energy duringa photochemical reaction. The asterisk is used to show that M is now in an excited state. Optical transmittance, T, is a measure of how much light that enters a sample is absorbed. T: I/IO If no light is absorbed then I = Io. Low transmittance values indicate that lots ofthe lisht has been absorbed. Most spectrophotometers give their results in optical absorbency, A, or opticaldensify (same) which is defined as: A= = log Transmittance T = Percent Transmittance %T = (1/T): -log T I Io L ylggoTo Io Absorbance A = -logT ,I = - loq'Io ,In = loq-I---:Beer's la ; is often bH absorbaqce of chromophoies, in rhores where k Beer's law €.% i_s a constant. A:kc rdlationship between I iJ : and -, known absorbance The optical path, ln looking at the color iri at the deep end because the optical path is Iaw wherek' is another empirical constant. A =k'l To the eye some colors appear similar but may differ in intensity when c and I arethe same. These solutions have a larger molar absorption coefficient, e, meaning theyadsorb more. The larger the adsorption coefficient the more the material adsorbs. The Beer-Lambert law combines the two laws giving -2- First year - Environmental Chemistry Dr. Manal A. Tooma A=elc The proportionality constant in the Lambert's law is e. The extinction coefficients of chromophores vary widely from less than 100 l/mol.cm,for a so-called forbidden transition, to greater than 105 l/mol.cm for fully allowedtransitions. We can redefine the elements of the Beer-Lambert law where I is the samplethickness and c is the molar concentration of chromophores. This can be rearranged todetermine the penetration depth of light into a material. Here I is defined as thepath length where 90Yo of the light of a particular wavelength is l (in Prm): es I giving. 104 e c ',ipsudaru"i4a*p,i In absorption, the light and the transmitted light angle, generally 90o, the resulting po energy and is sai ii it of cold body radiation. Otder TV principle tne.emlssron ot light occurs when i Iorescence and fluorescence! In a accel gunisitting behindJhe screen. In the el the cool. sets, isht when hit off light. The same of the screen is coated with tho green, and blue. The kinetic r andre-emitted as visible light to be seen by us. Jluorescence involves the morecurar uuroiiffiffi photon that triggers theemission of a photon oflonger wavelength (less energy). The energy differenceends up asrotational, vibrational, or heat energy loses. Here excitation is described as So Where So is * hv"* a Sland emission asSl a hv". + go the ground state, S1 is the first excited state. Firrt ye* - Enuironmental Chemistry Dr. Manal A. Tooma The excited state molecule can relax by a number of different, generallycompeting pathways. One of these pathways is conversion to a triplet state that cansubsequently relax through phosphorescence or some secondary non-radiative step. Relaxation ofthe excited state can also occur tlrough fluorescence quenching. Molecular oxygen is a particularly effrcient quenching molecule because of its unusualtriplet ground state. Watch hands that can be "seen in the dark" allow us to read the time in the dark. These watch hands typically are paintedyi paint. Like fluorescence, phosphorescence is the emisgi hit by electromagnetic radiation.Unlike afterglow for some time after the liCht emission is 10: The energy Jab 1935; a given in tn figure represents the electrohic ground J qr ! n u ! fl H M 4- h o r e's c en c e Inlerttal conversion Itt2s ll;?st - r,. ,rrr, ,.* - Environmental Chemistry Dr. Manal A. Tooma Classiftcafion of Analytical Methods Qualitative instrumental analysis is that measured property indicates pres ence of analyte in matrix Quantitative instrumental analysis is that magnitude of measured property is proportional to concentration of analyte in matrix Species of interest: All constituents including analyte and Matrix-analyte (concomitants) often need pretreatment - chemical extraction, distillation, separation, precipitation (A) Classical: Qualitative - identifi cation Quantitative - mass (B) Qualitative physical property ( qualitati Instrument to the concentration o than indicator to locate Spectro chemical tech niq ues Is one of the optical methods widery used for anarysis, i, ment of the wavelength frequency or energy of electromagnetic ij*:g the measure- radiation (EMR) that either absorbed or emitted by the sample-the EMR spectrometry is divided into severar energy regions as shown :_ -5- Dr. Manal A. Tooma c IArnt I I 10t" -: r, lo "r3 !- 1-0j rloa 4 Absorption Spectrum: D'-.- lo't -. a x_0' vlSbte Llght obs!€Es*F I lor lO' __ W'avclangth (m eters) velength) l, called absorption e can only In atoms Visible is Wand In M' with many AE: Fs: AE el For each Absorption spectra affected (l) -"t* a,' i ii.r Number of atoms in moleculeEEier (2) Solvent molecules blurred features -6- F-_.- E l-.,1 r ij R 1 :tt' First year - Environmental Chemistry U V U ltravio I et Dr. Manal A. Tooma . (',r so e ctro metm I {" when two atoms share to make new bond; there are electron in both atom will participate to form this bond the electrons occupied new orbital called molecular bonding orbital with low energy and antibonding orbital with high energy and the electron how don't share called nonbonding. when the molecular absorb energy in o-'rl*, rl-lt' uv range; electronic transfer will happened (o-o*, ,n- o', n- ,{, fi-o'1(200-380nm) -F - Itis the length ). Wav Vf/here:c is the speed - of Energy is related to wavelength and frequency by the fo'owing formulas: E - nt,- Il )," Were: h : Planck's constant, 6.6 x l0-3foules_sec - Note that energy is directly proportional to frequency and inversely proportional to wavelength. First year - Environmental Chemistry ' Dr. Manal A. Tooma IR is a device that measures the intensity of the electromagnetic spectrum which is between the visible and Microwave (400-4000 cm-';and expressed either transmittance or A: absorption: log (1/ T) An increase in wavenumber corresponds to an increase in energy, when the molecule absorb the energy will start to vibrate; stretching (changing the length ofthe bond or bending changing the angle ofthe bond) . we can calculate the L t:4 'IK K:5*1 l05dyne/cm for C: =3 x l01o i.1r..., & *ra -ts w.* asymrlrtricsl rtrE{c-lrhg 2925 cm-l r!'mmcticsl rarcaihilg 2850 (Jn-l rcisioridg, or bcndiog in-tl.'lc l465gtrl + q,P ,I / ".,. ourof-phna crn-l twirrioS, o. bcndhg l35Gl150 io-plaoe ?20erri rocting, or-bcDding wrggi||& or bending out-of_ptr Stretching ard bending vibntional modes for a CH2 group. -8- l35gllJ6gtrl r.-f= .]/ t:"t .# ;',# First year - Envirorunental Chemistry Dr. Manal A. Tooma con sample. element. flame into absorbed by ement interest is makes the method relati ofenergy at the characteristic the concentration of the element in the samold ost atomic absorption instruments are also equipped for Atomic absorption spectrophotometer consisting of a light source emitting the line spectrum of an element, a device for vaporising the sample, a means of isolating an absorption line and a photoelectric detector with its associated electronic amplifuing and measuring equipment . -9- of 1.. ,., -. First year - Environmental Chemistry Dr. Manal A. Tooma deute riurtr liBht sou'te msn*trrqmgtor trESOrUrnte d€lcctor hnlf rilvcrod mirrtrr cpilrry shsmbcr ofitrliftcr fucl ncbulizcr oxidant dfflxrrar S.' odo on them the relative strongly mobile the morQ the stationary Tvne ofchrometosraph Gas Jiquid moblle phose g3.s stadonarv ohase liquid Gas -solid gas solid Liquid-liquid liquid liquid Gel-permeation -10- liquid Gel First year - Environmental Chemistry Dr. Manal A. Tooma W(tp):isthetimerequiredforthemobilephasetoremovethecomponent from the stationary phase. series ofpeaks as shown Tx: can be computed from such plot as the time between sample injection and recording the peak. The time corresponds to the apffofsuch peak represents the tx ofthat component Concentration: they are area under the onal to theconcentration, hence the amount of sample can be Ifthe peak is Efftciency The (n) in the Theoreti the The by the Higher effrciency d be as short as possible to avoid a Two equations are used to the half peak width n= 76(3 w Or N = 5.55 peak width (w) and f-L)' \wrlz/ (p) must be accurately determined . (w): distance along the base line between extrapolated tangents. - rt- First year - Environmental Chemistry Dr. Manal A. Tooma wlthe width of the peak at half of the peak height pH-meter A pH Meter is a device used for potentiometrically (voltmeter which is calibrated to convert voltage to pH units) measuring the pH, which is either the concentration or the activity ofhydrogen ions, ofan aqueous solution. It usually has glass a electrode plus a calomel reference electrode, or a combination electrode. pH meters are usually used to measure the pH of liquids, though speci ggggglimes used to measure the pH of semi-solid substances. The electrode has a silver- inside a glass electrode, contained thin (typically electrode oride. The ,soit contains 're testing electrode the di potassium When hydrogen ions the metal ions insidd the unknown solution. to how a glass electrode works. the some of electrode into , and it's the key on the inside surface ofthe glass electrode from the potassium chloride solution. The two solutions on either side of the glass have different acidity, so a different amount of ion-swapping takes place on the two sides ofthe glass. This creates a different degree ofhydrogen_ion activity on the two surfaces of the glass, which means a different amount of electricar charge builds up on them' This charge difference means a tiny voltage (sometimes cared a potential difference, typically a few tens or hundreds of millivolts) appears between the two sides -t2- of First year - Environrnental Chemistry Dr. Manal A. Tooma the glass, which produces a difference in voltage between the silver electrode (5) and thereference electrode (8) that shows up as a measurement on the meter. (1)Solution tested; metal salts, inside which silver/silver glass. (7) H glass. (8) this 5srmed in. thc test sllutiorffi of the potassium chloride qolution inter i, rence in volhge betlveen the two rss and reading. (9) Rdference iilechode acts as or refe: of the 3 [,-_ $:z Glass pH relation: Constant k l) pH meter is adj 2) pH ofunknown --t5- iffinrs!