Analytical Chemistry Topics Lecture (Adamson University)

Gravimetric Method of Analysis – deals with the measurement of the mass of a substance that is chemically related to the analyte. Basic Steps on Precipitation Method 1. Sample is dissolved in an appropriate solvent. 2. A precipitant is used to convert the analyte into a sparingly soluble precipitate. 3. The precipitate is converted into a product of known composition by a suitable heat treatment. 4. The percentage of the analyte in the sample is calculated using the gravimetric factor (GF):

Ex. The phosphorus in a 0.1969 gram sample was precipitated as the slightly soluble (NH4)3PO4 12MoO3. This precipitate is filtered, washed and then redissolved in acid. Treatment of the resulting solution with an excess Pb+2 resulted in the formation of 0.2554 gram of PbMoO4 (367.1376). Express in the analysis of P2O5 (141.945). a. 1.045% b. 2.090% c. 4.180% d. 8.360% Ex. The aluminum in a 759.08 mg of impure aluminum sulfate sample was precipitated as Al(OH)3 and ignited at 1100°C to yield a precipitate of Al2O3 weighing 387.953 mg. Express the result of analysis in terms of %Al. a. 27.05% b. 13.53% c. 18.67% d. 23.29% Ex. What weight of an impure NaCl sample must be taken for analysis so that the weight of AgCl precipitate obtained in mg will be equal to the %Cl in the sample? a. 19.76 mg b. 12.66 mg c. 24.73 mg d. 4.04 mg Ex. A sample containing NaBr and KBr only weighs 253.02 mg. The sample was dissolved in water and treated with excess AgNO3. The precipitate formed was found to weigh 429.85 mg. Calculate the %NaBr in the sample. a. 45% b. 55% c. 49% d. 51% Ex. A 0.6407-g sample containing chloride and iodide ions gave a silver halide precipitate weighing 0.4430 g. This precipitate was then strongly heated in a stream of Cl 2 gas to convert the AgI to AgCl; on completion of this treatment, the precipitate weighed 0.3181 g. Calculate the percentage of chloride and iodide in the sample. Ans. 4.72% Cl- and 27.05% I-.

Volumetric Methods of Analysis – measures the volume of solution necessary to react completely with the analyte Standard Solution – solution of known concentration Standardization – process of determining the concentration of an unknown solution Primary standard – a substance of high purity Characteristics of a Good Primary Standard 1. 2. 3. 4.

High purity and high equivalent weight Stable towards air, high temperature and humidity Soluble in water Readily available and fairly inexpensive

Conditions for Volumetric Analysis 1. The reaction must be rapid and can be represented by a simple balanced equation. 2. The reaction is complete and no side reaction occurs. 3. An appropriate indicator must be available in order to detect the end point of the reaction Types of Titration 1. Direct Titration – the analyte reacts with the standard solution directly 2. Back Titration – an excess standard solution is added and the excess is determined by the addition of another standard solution 3. Replacement Titration – the analyte is converted to a product chemically related to it and the product of such reaction is titrated with a standard solution CLASSIFICATION OF VOLUMETRIC METHODS There are four general classes of volumetric or titrimetric methods. Acid-Base. Many compounds, both inorganic and organic, are either acids or bases and can be titrated with a standard solution of a strong base or a strong acid. The end points of these titrations are easy to detect, either by means of an indicator of by following the change in pH with a pH meter. The acidity and basicity of many inorganic acids and bases can be enhanced by titrating in a nonaqueous solvent. The result is a sharper end point, and weaker acids and bases can be titrated in this manner. Acid-Base Titration Concepts of Acids and Bases LEWIS BRONSTED-LOWRY ACID Electron-pair acceptor Proton donor BASE Electron-pair donor Proton acceptor *hydronium ion, protonated water or solvated proton

ARRHENIUS Produces *H3O+ in solution Produces **HO- in solution

**hydroxide ion Strength of Acids and Bases Ionization Reaction – reaction involving formation of ions Strong Acids/Bases – completely ionized in solution HCl + H2O H3O+(aq) + Cl-(aq) NaOH + H2O Na+(aq) + HO-(aq) Weak Acids/Bases – partially ionized in solution HF + H2O H3O+(aq) + F-(aq) NH3 + H2O NH4+(aq) + HO-(aq) Autoprotolysis of Water H2O + H2O H3O+(aq) + HO-(aq) KW = 1x10-14 at 25°C Strong Acids: HCl, HBr, HI, HClO4, HNO3, H2SO4 (only on the first ionization) Strong Bases: Bases of Group IA and 2A Weak Acids: HF, HCN, H2SO3, H3PO4 and organic acids Weak Bases: Ammonia and derivatives Calculation of pH Strong Acids: pH = -log[Cacid] Strong Bases: pH = 14 + log[nHO-Cbase] Weak Acids: pH = - ⁄ log[KaCacid] when Cacid/Ka >> 1000 Weak Bases: pH = 14 + ⁄ log[KbCbase] when Cbase/Kb >> 1000 At 25°C . . . pH + pOH = 14 Ex. Calculate the pH of 1 x 10-3 M HCl Ex. Calculate the pH of 1 x 10-8 M HCl Ex. A 0.03 M HClO2 solution is 10.0% ionized. Calculate Ka. Ex. Kb for NH3 is 1.8 x 10-5 .Calculate the % ionization and the [OH-] in a 0.05 M solution. Ex. What is the pH of a 0.003 M HNO2 solution that is 15% ionized? Ex. Ka for HC2H3O2 is 1.75 x 10-5. What is the pH of a 0.003 M solution? Ex. Calculate the pH of a 0.08 M NH3 solution. Kb = 1.8 x 10-5 Ex. Calculate the molarity of NaOH solution if 12.25 mL was used to titrate 0.2615 gram of primary standard KHP. a. 0.1045 M b. 0.1354 M c. 0.2509 M d. 0.1697 M Ex. In standardizing a solution of NaOH against 1.431 grams of KHP, the analyst uses 35.50 mL of the alkali and has to run back with 8.25 mL of acid ( 1mL=10.75 mg NaOH). What is the molarity of the NaOH solution? a. 0.2118 M b. 0.2044 M c. 0.7831 M d. 0.2598 M Common Ion Effect *is the shift of equilibrium caused by the addition of a compound having an ion in common with the dissolved substance *reduction in the ionization of the weak electrolyte Ex. Calculate the [H3O+] in a 0.005 M HC2H3O2 solution. Ka = 1.8 x 10-5 Ex. What is the [H3O+] in a 0.005 M HC2H3O2 which contains 0.001 M NaC2H3O2?

Hydrolysis of Salts Acid and base reacts to form salt and water. As a general rule, salts coming from weak acids or weak bases hydrolyze in water, that is, only the strong conjugate hydrolyzes in water. An acidic salt is formed from the reaction of a strong acid and weak base. A basic salt results from the reaction of a strong base and weak acid. Thus, a neutral salt is a product of the reaction between a strong acid and a strong base. Hydrolysis Reaction of Salts Acidic Salt: NH4Cl NH4+ + H2O H3O+ + NH3 Basic Salt: NaCN CN- + H2O HO- + HCN

KH = KW/ KH = KW/

pH of Salts Acidic Salt: pH = 7 Basic Salt: = 7 +

⁄

⁄

] when Csalt/KH >>> 1000

log[

log[

] when Csalt/KH >>> 1000

Ex. What is the pH of the resulting solution made by mixing 25 mL of 0.1 M HCl and 15 mL of 0.1 M NaOH? a. 1.40 b. 1.60 c. 1.00 d. 0.4 Ex. What is the pH of 0.256 M NH4Cl? Kb of NH3 = 1.8 x 10-5? a. 9.49 b. 11.0 c. 11.36 d. 4.92 Buffer Solutions Solutions that contains weak acid or weak base and its conjugate salt. These solutions tend to resist changes in pH. pH of a Buffer Solution [ [

] ]

If Kb is given . . . [ [

] ]

Ex. What mass in grams of NaC2H3O2 must be dissolved with 500 mL of 0.100 M acetic acid to make 2L of buffer solution of pH = 5? Ka = 1.8 x 10-5 a. 2.28 g b. 7.19 g c. 7.38 g d. 2.12

Ex. What is the pH of the resulting solution made by mixing 5 mL of 0.2178 M HCl and 15 mL of 0.1156 M NH3? Kb = 1.8 x 10-5? a. 9.49 b. 11.00 c. 9.02 d. 12.74 Ex. What volume of 0.200 M HCl must be added to 80 mL of 0.150 M NH3 to produce a 2L buffer solution with a pH of 8.00? Kb of NH3 = 1.8 x 10-5 a. 3.2 mL b. 9.6 mL c. 28.8 mL d. 56.8 mL Applications of Acid-Base Titration 1. Kjeldahl Method (Determination of Organic Nitrogen) Step 1. Digestion The sample is oxidized in hot, concentrated sulfuric acid, H2SO4 and turns black. . . Step 2. Distillation The oxidized solution is cooled and then treated with NaOH to liberate ammonia gas: NH4+ + HO- NH3(g) + H2O Step 3. Titration 1. Using an excess amount of HCl. . . NH3 + HCl NH4Cl The excess HCl is determined using a standard NaOH solution HCl + NaOH NaCl + H2O 2. Ammonia distilled is collected in a boric acid solution. . . NH3 + H3BO3 NH4+ + H2BO3-2 Titrate the H3BO3-NH3 solution with standard acid. . . H2BO3-2 + H3O+ H3BO3 + H2O Percentage Protein in the sample %protein =%N * f = 5.70 (cereals) = 6.25 (meat products) = 6.38 (dairy products) Ex. A 758-mg sample of full cream milk was analyzed by the Kjeldahl method; 38.61 mL of 0.1078 M HCl were required to titrate the liberated ammonia. Calculate the % N in the sample. a. 12.04% b. 7.69% c. 15.59% d. 10.93% Ex. A 5.8734-gram sample beef was analyzed for its N content and the liberated NH 3 was collected in a 50.00 mL of 0.4691 M HCl and a 12.55 mL back titration with 0.0256 M NaOH was required. Calculate the percentage protein in the beef sample. a. 17.32% b. 5.54% c. 34.46% d. 11.08% Ex. A 2060 mg sample of flour was taken through a Kjeldahl procedure and the ammonium produced was distilled into 100 mL of 0.1006 M H3BO3 solution. If this solution required 34.7 mL of 0.174 M HCl for titration to methyl red end point, what is the percentage of protein in flour? Use 5.70 for flour. Ans. 23.39%

2. Double Indicator Method (Mixture of Bases) Ex. A sample that may contain NaOH, Na2CO3, NaHCO3, and inert matter alone or in compatible combination is titrated with 0.1000 N HCl with phenolphthalein as the indicator and the solution became colorless after the addition of 48.8 mL. Methyl orange is then added and 14.55 mL more of the acid are needed for the color change. If the sample weighs 2.345 grams, it contains a.5.842% NaHCO3 and 6.577% Na2CO3 c. 65.77% Na2CO3 and 5.842% NaHCO3 b.6.577% Na2CO3 and 5.842% NaOH d. 65.77% Na2CO3 and 5.842% NaOH Ex. A sample consisting of Na2CO3, NaHCO3 and inert matter weighs 1.179 grams. It is titrated with 0.100 N HCl with phenolphthalein as the indicator, and the solution became colorless after the addition of 24.00 mL. Another duplicate sample was titrated with HCl using methyl orange as indicator. It required 50.25 mL of the acid for the color change. What is the percentage of Na2CO3 in the sample? a. 18.70% b. 5.17% c. 12.56% d. 21.58% Precipitation. In the case of precipitation, the titrant forms an insoluble product with the analyte. An example is the titration of chloride ion with silver nitrate solution to form silver chloride precipitate. Again, indicators can be used to detect the end point, or potential of the solution can be monitored electrically. One of the oldest analytical techniques that started in the mid-1800’s. Silver nitrate (AgNO3) is commonly employed in such technique. Titration with AgNO3 is often termed as argentometric titration. Indicators in Precipitimetry The equivalence point can be observed by the following: a. Formation of a colored secondary precipitate Mohr Method (K.F. Mohr, Germany, 1865) Direct Method for halides and cyanides Titrant: Silver Nitrate, AgNO3 Titration Reaction: Ag+ + Cl- AgCl(s) white Indicator: sodium chromate, Na2CrO4 Indicator Reaction: 2Ag+ + CrO42- Ag2CrO4(s) red Primary Standard for AgNO3: NaCl Titration is carried out between pH of 7-10. Usually, a low concentration of chromate is desired to detect the endpoint clearly since a chromate ion imparts an intense yellow color. Ex. What is the molar concentration of AgNO3 solution standardized against 712 mg primary standard NaCl (58.45 g/mol) requiring 23.8 mL of the solution for titration?

a. 0.5027 M

b. 0.5118 M

c. 0.5263 M

d. 0.5329 M

Ex. A 1.500-gram sample of impure AlCl3 was dissolved in water and treated with 45.32 mL of 0.1000 M AgNO3 using Mohr method. Determine its purity as %AlCl3 (133.33) a. 40.28% b. 13.43% c. 4.48% d. 27.36% b. Formation of colored complexion Volhard Method (Jacob Volhard, Germany, 1874) Direct method for silver - Indirect method for halides Titrant: Potassium thiocyanate, KSCN Direct Titration Reaction: Ag+ + SCN-1 AgSCN(s) white Indirect Titration Reaction: Ag+(excess) + Cl-1 AgCl(s) white Ag+ + SCN-1 AgSCN(s) white Indicator: ferric alum Indicator Reaction: Fe+3 + SCN-1 Fe(SCN)+2 red Titration is carried out in acidic condition to hasten precipitation of ferric ion to its hydrated oxide form. Ex. Chloride in a brine solution is determined by the volhard method. A 10.00-mL aliquot of the solution is treated with 15.00 mL of standard 0.1182 M AgNO3 solution. The excess silver is titrated with standard 0.101 M KSCN solution, requiring 2.38 mL to reach the red Fe(SCN) 2+ end point. Calculate the concentration of chloride in the brine solution, in g/L. Ans. 5.434 g/L Ex. A mixture of LiBr and BaBr2 weighing 800 mg is treated 50.00 mL of 0.1879 M AgNO3 and the excess is found to require 8.76 mL of 0.3719 M KSCN for back titration, using ferric alum as indicator. What is the percentage of BaBr2 in the sample? ANS: 80.34% a. 67.95% b. 32.05% c. 35.62% d. 64.38% c. Formation of a colored adsorption complex Fajans Method (K. Fajans, Poland, 1874) Titrant: Silver nitrate, AgNO3 Titration Reaction: Ag+ + Cl-1 AgCl(s) white Indicator: dichlorofluorescein, best for determination of halides and cyanides End point: color change from yellow to pink Titration is carried out between pH of 4-7. Dextrin is added to prevent excessive coagulation of the AgCl precipitate. Complexometric. In complexometric titrations, the titrant is a reagent that forms a watersoluble complex with the analyte, a metal ion. The titrant is often a chelating agent (a type of complexing agent that contains two or more groups capable of complexing with a metal ion). The reverse titration may be carried out also. Ethylenediaminetetraacetic acid (EDTA) is one of the most useful chelating agents used for titration. It will react with a large number elements ,

and the reactions can be controlled by adjustment of pH. Indicators can be used to form a highly colored complex with the metal ion. b. Titration with Ethylenediaminetetraacetic Acid (EDTA) The structure suggests six potential sites (hexadentate) for metal bonding: the four carboxyl groups and two amino groups. Commercially, the free acid and the dehydrate are available. Solutions of EDTA combines with any metal ions in a 1:1 ratio. The indicator used for titration is the Eriochrome Black T®. For metal ion in detections, it is necessary to adjust the pH to 7 or above so that the blue form predominates in the absence of a metal cation. Generally, metal complexes with EDTA are red as H2In-1. When an excess EDTA is added, the solution turns blue according to the reaction: MIn-1 + HY-3 HIn-2 + MY-2 red blue Ex. What volume of 0.0305 M EDTA is needed to titrate the Ca in 178.56 mg of CaCO 3? a. 58.54 mL b. 29.27 mL c. 43.91 mL d. 14.64 mL Ex. An EDTA solution prepared from its disodium salt was standardized using 506.3 mg of primary standard CaCO3 and consumed 28.50 mL of the solution. The standard solution was used to determine the hardness of a 2-L sample mineral water, which required 35.57 mL EDTA solution. Express the analysis in terms of ppm CaCO3. a. 89 ppm b. 316 ppm c. 158 ppm d. 269 ppm Ex. Aluminum is determined by titrating with EDTA: Al3+ + H2Y2- AlY- + 2H+ A 1.00 g sample requires 20.5 mL EDTA for titration. The EDTA was standardized by titrating 25.0 mL of a 0.100 M CaCl2 solution, requiring 30.0 mL EDTA. Calculate the percent Al2O3 in the sample. Ans. 8.71% Ex. Chromium(III) is slow to react with EDTA (H4Y) and is therefore determined by backtitration. A pharmaceutical preparation containing chromium (III) is analyzed by treating a 2.63g sample with 5.00 mL of 0.0103 M EDTA. Following reaction, the unreacted EDTA is backtitrated with 1.32 mL of 0.0122 M zinc solution. What is the percent chromium chloride in the pharmaceutical preparation. Ans. 0.213% A masking agent is a complexing agent that reacts selectively with a component in a solution to prevent that component from interfering in a determination. Ex. A 1.509-g sample of a Pb/Cd alloy was dissolved in acid and diluted to exactly 250.0 mL in a volumetric flask. A 50.00-mL aliquot of the diluted solution was brought to a pH of 10.0 with an NH4+/NH3 buffer; the subsequent titration involved both cations and required 28.89 mL of 0.06950 M EDTA. A second 50.00 mL aliquot was brought to a pH of 10.0 with an HCN/NaCN buffer, which also served to mask the Cd2+; 11.56 mL of the EDTA solution were needed to titrate the Pb2. Calculate the percent Pb and Cd in the sample. Ans. 55.16% Pb and 44.86% Cd

b. Determination of Cyanide by the Liebig Method The titration is carried by the drop wise addition of AgNO3 in a solution of a cyanide forming a soluble cyanide complex of silver: 2CN- + Ag+ Ag(CN)2-1. The endpoint of the titration is the formation of a permanent faint turbidity: Ag(CN)2-1 + Ag+ Ag[Ag(CN2)](s) Ex. A 500-mg sample containing NaCN required 23.50 mL of 0.1255 M AgNO3 to obtain a permanent turbidity. Express the result of this analysis as % CN-. a. 15.34% b. 23.01% c. 17.25% d. 30.67% c. Determination of Nickel An ammoniacal solution of nickel is treated with a measured excess of standard cyanide solution and the excess of standard AgNO3 solution according to the reactions: Addition of Excess Cyanide: Ni(NH3)6+3 + 4CN-1 +6H2O Ni(CN)4-1 +6NH4OH Back Titration with Ag+: 2CN-1 + Ag+ Ag(CN)2-1 Endpoint: Ag(CN)2-1 + Ag+ Ag[Ag(CN)2](s) Ex. A 750.25-mg of alloy nickel was dissolved and treated to remove the impurities. The ammoniacal solution was treated with 50 mL of 0.1075 M KCN and the excess cyanide required 2.25 mL of 0.00925 M AgNO3. Determine %Ni in the alloy. a. 20.86% b. 37.69% c.10.43% d. 41.27% Reduction-Oxidation. These “redox” titrations involve the titration of an oxidizing agent with a reducing agent, or vice versa. An oxidizing agent gains electrons and a reducing agent loses electrons in a reaction between them. There must be a sufficiently large difference between the oxidizing and reducing capabilities of these agents for the reaction to go to completion and give a sharp end point; that is, one should be a fairly strong oxidizing agent (strong tendency to gain electrons) and the other a fairly strong reducing agent (strong tendency to lose electrons). You can use appropriate indicators for these titrations, or you may employ various electrometric means to detect the end point. Combining Ratio, also (f) OXIDANT KMnO4 (a) KMnO4 (b,n) K2Cr2O7 I2 MnO2 REDUCTANT Na2C2O4 (C2O4-2) FeSO4 (Fe+2) Na2S2O3 KI

5 3 6 2 2 2 1 1 1

Fe metal As2O3 H2O2 Cu+

2 4 2 1

Ex. What is the molarity of a KMnO4 solution standardized against 1.356 gram Na2C2O4 (134 g/mol) requiring 25.1 mL of the solution in acidic medium? a. 0.161 M b. 0.403 M c. 1.008 M d. 0.856 M Ex. The percentage of MnO2 in a 500 mg sample which after the addition of 80.00 mL of 0.1056 M FeSO4 solution required 8.50 mL of 0.0867 M K2Cr2O7 is a. 33.52% b. 35.00% c. 17.50% d. 67.04% Ex. A 240-mg sample of pyrolusite was treated with excess KI. The iodine liberated required 46.24 mL of 0.1105 M Na2S2O3 solution. Calculate the % MnO2 in the sample. a. 46.27% b. 30.85% c. 92.54% d. 76.12% Ex. A sample of iron ore weighing 385.6 mg was dissolved in acid and passed through a Jones redactor. If the resulting solution required 52.36 mL of 0.01436 M K 2Cr2O7 for titration, calculate % Fe3O4 (231.55 g/mol) in the ore sample. a. 15.05% b. 45.15% c. 90.30% d. 67.98% Ex. A 0.200-g sample of pyrolusite is analyzed for manganese content as follows. Add 50.0 mL of a 0.100 M solution of ferrous ammonium sulfate to reduce the MnO 2 to Mn+2. After reduction is complete, the excess ferrous ion is titrated in acid solution with 0.0200 M KMnO4, requiring 15.0 mL. Calculate the percent manganese in the sample as Mn 3O4 (only part or none of the manganese may exist in this form, but we can make the calculations on the assumption that it does). Ans. 66.74% Ex. A hydrogen peroxide solution is analyzed by adding a slight excess of standard KMnO 4 solution and back-titrating the unreacted KMnO4 with standard Fe+2 solution. A 0.587-g sample of the H2O2 solution is taken, 25.0 mL of 0.0215 M KMnO4 is added, and the back-titration requires 5.10 mL of 0.112 M Fe2+ solution. What is the percent H2O2 in the sample? Ans. 6.13% Ex. A sample of a pyrolusite weighs 0.5000 g. To this is added 0.6674 g of As 2O3 and dilute acid. After solvent action has ceased, the excess three-valent arsenic is titrated with 45.00 mL of 0.1000 N KMnO4. Calculate the oxidizing power of the pyrolusite in terms of percentage MnO 2. Ans. 78.30%