Analytical Chemistry: Kenneth Soriano, Eit, Amicheme

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Analytical Chemistry Kenneth Soriano, EIT, AMIChemE

Course Outline • Solutions and Their Concentrations • Gravimetric Analysis • Acid – Base Equilibria • Volumetric Analysis

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Solutions and Their Concentrations

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OH

Aspirin O

Stearic Acid

OH

Molarity (M) • Molar concentration is the number of that species that is contained in 1L of the solution

Part per Million (ppm) • For very dilute solutions, parts per million (ppm) is a convenient way to express concentration

Parts per Million (ppm) • A handy rule in calculating parts per million is to remember that for dilute aqueous solutions whose densities are approx. 1.00 g/mL,

Parts per Billion (ppb)

Normality (N) • is the number of equivalents that is contained in 1L of the solution

Normality

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Gravimetric Analysis

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OH

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Progesterone H

H O

Testosterone

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O

Anthraquinone

Gravimetric Analysis • 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 • Sample is dissolved in an appropriate solvent. • A precipitant is used to convert the analyte into a sparingly soluble precipitate. • The precipitate is converted into a product of known composition by a suitable heat treatment. • The percentage of the analyte in the sample is calculated using the gravimetric factor (GF)

Gravimetric Analysis

WT. SAMPLE

Prepare Solution of the Sample • Choice of Solvent Water ( w/ or w/o heating) Non-Oxidizing Acids Oxidizing Acids (HNO3) Aqua Regia (3:1 HCl:HNO3) Fluxing Agents Acid: K2S2O7 Redox: Na2O2 Basic: Na2CO3

DISSOLVED SAMPLE

EXCESS PRECIPITATING AGENT

Precipitation of Analyte Elements / Ion Precipitated

Precipitating Agent

Final/Ignited Form

Cl, Br, I

AgNO3

AgCl, AgBr, AgI

Fe, Al ,Cr

NH3

Fe2O3 , Al2O3 , Cr2O3

Mg

NH4 HPO4

Mg2P2O7

K

H2PtCl6

K 2PtCl6

Ni

Dimethylglyoxime

Ni – DMG

(NH4)2C2O4

CaC2O4 - 110°C CaCO3 - 500°C CaO - 800°C

Ca

PRECIPITATION OF THE ANALYTE

FILTRATION AND IGNITION OF THE PRECIPITATE

Gravimetric Analysis

WT. FINAL FORM

Calculations

Example • A 0.5662-g sample of an ionic compound containing chloride ions and an unknown metal is dissolved in water and treated with an excess of AgNO3. If 1.0882 g of AgCl precipitate forms, what is the percent by mass of Cl in the original compound? ANS. 47.54%

Example 1 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%

Example 2 • A 250 mg sample containing 45% MgCO3 and 55% CaCO3 was ignited producing CO2 as decomposition product. Assuming that the decomposition reaction is complete, what is the change in weight of a NaOH solution used to absorb CO2 ANS. 119.21 mg

Example 3 • 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? ANS. 24.73 mg

Example 4 • 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. ANS. 49%

Example 5 • 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 Cl2 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-

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Acid – Base Equilibria

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OH

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Progesterone H

H O

Testosterone

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Anthraquinone

Arrhenius Acids and Bases • Arrhenius defined acids as substances that ionize in water to produce H+ ions and bases as substances that ionize in water to produce OH- ions. ARRHENIUS ACID

ARRHENIUS BASE

Produces H3O+ in sol’n

Produces OH- in sol’n

Arrhenius Acids and Bases • Monoprotic Acid - each unit of the acid yields one hydrogen ion upon ionization

Arrhenius Acids and Bases • Diprotic Acid - each unit of the acid gives up two H+ ions, in two separate steps

Arrhenius Acids and Bases • Triprotic Acid - which yield three H+ ions, are relatively few in number

Arrhenius Acids and Bases

Strength of Acids and Bases • Classification of Electrolytes STRONG 1. Inorganic acids such as HNO3, HClO4 , H2SO4 (only on first ionization) , HCl, HI, HBr, HClO3, HBrO3

WEAK 1. Many inorganic acids, including HF HCN, H2CO3, H3BO3, H3PO4, H2S, H2SO3

2. Alkali and Alkaline-Earth (Grp I and 2. Most organic acids ( ex. HC2H3O2 II) hydroxides - KOH, NaOH, acetic acid, HCOOH formic acid) LiOH, RbOH, Mg(OH)2, Ba(OH)2 , 3. Ammonia and most organic bases Ca(OH)2 3. Most Salts

4. Halides, cyanides, and thiocyanates of Hg, Zn, and Cd

Bronsted-Lowry Acids and Bases • Brønsted acid is a substance capable of donating a proton, and a Brønsted base is a substance that can accept a proton. BRONSTED ACID

BRONSTED BASE

Proton Donor

Proton Acceptor

Bronsted-Lowry

HCl : Proton Donor (H+) H2O: Proton Acceptor

Bronsted-Lowry • An extension of the Brønsted definition of acids and bases is the concept of the conjugate acid-base pair, which can be defined as an acid and its conjugate base or a base and its conjugate acid.

Conjugate means “joined together”

Bronsted-Lowry • The conjugate base of a Brønsted acid is the species that remains when one proton has been removed from the acid. • Conversely, a conjugate acid results from the addition of a proton to a Brønsted base.

Bronsted-Lowry • Every Brønsted acid has a conjugate base, and every Brønsted base has a conjugate acid. For example, the chloride ion (Cl- ) is the conjugate base formed from the acid HCl, and H3O+ (hydronium ion) is the conjugate acid of the base H2O.

Bronsted-Lowry

Bronsted-Lowry

Bronsted-Lowry

Lewis Acids and Bases • Lewis base as a substance that can donate a pair of electrons. A Lewis acid is a substance that can accept a pair of electrons. LEWIS ACID

LEWIS BASE

Electron Pair Acceptor

Electron Pair Donor

Lewis

Autoprotolysis of Water

Autoprotolysis of Water

• In 1 L, or 1000 g, of water, there are 1000 g/(18.02 g/mol), or 55.5 moles, of water. Therefore, the “concentration” of water, or [H2O], is 55.5 mol/L, or 55.5 M. This is a large quantity compared to the concentrations of other species in solution (usually 1 M or smaller), and we can assume that it does not change appreciably during the course of a reaction.

Autoprotolysis of Water

Ion Product Constant

p Function

Whenever [H+] = [OH-], the aqueous solution is said to be neutral.

Example At 100°C, what is the neutral pH of water?

pH of Strong Acid

pH of Strong Acid

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B

pH of Strong Acid • Neglecting contribution of H2O to H3O + function………. C

pH of Strong Acid • pH of 1 x 10-7 M HCl • Using C…………... pH = ? • Using A and B…… pH =? • pH of 1 x 10-10 M HCl • Using A and B…….pH =?

pH of Strong Acid • Alternative Formula: D

E

pH of Strong Base F

G

pH of Strong Base • Calculate the pH of 1 x 10^-8 M NaOH • Calculate the pH of a 0.02 M Ba(OH)2 solution • The pH of a Ba(OH)2 solution is 10.0. What is the H+ concentration?

Weak Acid

Acid Ionization Constant (Ka)

Percent Ionization

Example Calculate the pH of a 0.036 M nitrous acid (HNO2) solution and calculate the % ionization. Ka = 4.5 x 10^-4 The pH of a 0.10 M solution of formic acid (HCOOH) is 2.39. What is the Ka of the acid?

Example • A 0.03 M HClO2 solution is 10.0% ionized. Calculate Ka • Ka for HC2H3O2 is 1.75 x 10-5. What is the pH of a 0.003 M solution?

Weak Base

Weak Base

Example • Calculate the pH of a 0.08 M NH3 solution. Kb = 1.8 x 10-5

Common-Ion Effect • 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

Example • 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?

An important relationship between the acid ionization constant and the ionization constant of its conjugate base can be derived as follows, using acetic acid as an example:

• The conjugate base, CH3COO-1, supplied by a sodium acetate (CH3COONa) solution, reacts with water according to the equation

Hydrolysis Reaction of Salts • Acidic Salt: NH4Cl NH4+ + H2O H3O+ + NH3 • Basic Salt: NaCN CN- + H2O HO- + HCN

KH = KW/KNH3

KH = KW/KHCN

pH of Salts Acidic Salt: when Csalt/KH >>> 1000 Basic Salt: when Csalt/KH >>> 1000

Example 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? What is the pH of 0.256 M NH4Cl? Kb of NH3 = 1.8 x 10-5?

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

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 g

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

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

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Volumetric Analysis

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O OH

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Progesterone H

H O

Testosterone

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Anthraquinone

Volumetric Methods of Analysis measures the volume of solution necessary to react completely with the analyte

Volumetric Analysis Standard Solution – solution of known concentration

NaOH PELLETS

Volumetric Analysis • Primary standard – a substance of high purity

Volumetric Analysis • Characteristics of a Good Primary Standard High purity and high molecular weight Stable towards air, high temperature and humidity Soluble in water Readily available and fairly inexpensive

Primary Standards for Bases • • • •

Benzoic Acid, C6H5COOH (1 OH-) Oxalic Acid, H2C2O4∙2H2O (2 OH-) Potassium Biiodate, KH(IO3)2 (1) Potassium Hydrogen Phthalate (KHP), C6H4(COOH)(COOK) (1) • Sulfamic Acid (HSO3NH2) (1)

Primary Standards for Acids • • • •

Calcium Carbonate, CaCO3 (2 H+) Mercuric oxide, HgO (2) Sodium Carbonate, Na2CO3 (2) Tris-hydroxymethylaminomethane (THAM), (CH2OH)3CNH2 (1)

Standardization – process of determining the concentration of an unknown solution

Volumetric Analysis • Conditions for Volumetric Analysis The reaction must be rapid and can be represented by a simple balanced equation. The reaction is complete and no side reaction occurs. An appropriate indicator must be available in order to detect the end point of the reaction

NaOH Solution V mL = ? DISTILLED WATER DISSOLVED KHP (known amount) INDICATOR (Ph) (1-2 drops)

Record the volume of NaOH consumed

END POINT Very Light Pink Solution

Example In a titration experiment, a student finds that 23.48 mL of a NaOH solution are needed to neutralize 0.5468 g of KHP. What is the concentration (in molarity) of the NaOH solution?

Example How many grams of KHP are needed to neutralize 18.64 mL of a 0.1004 M NaOH solution?

Types of Titration Direct Titration – the analyte reacts with the standard solution directly Back Titration – an excess standard solution is added and the excess is determined by the addition of another standard solution

Types of Titration • 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

Example In standardizing a solution of NaOH against 1.431 gram 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 c. 0.7831 b. 0.2044 d. 0.2598

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 Using an excess amount of HCl. . . NH3 + HCl NH4Cl The excess HCl is determined using a standard NaOH solution HCl + NaOH NaCl + H2O

Using Excess HCl – Back Titration (a) mmol excess reagent added - (b) mmol excess reagent reacted w/ back titrant (c) mmol excess reagent reacted w/ analyte

Ammonia distilled is collected in a boric acid solution. . . NH3 + H3BO3 NH4+ + H2BO3-1 Titrate the H3BO3-NH3 solution with standard acid. . . H2BO3-1 + H3O+ H3BO3 + H2O

Using excess H3BO3 acid – Replacement Titration *data about boric acid – IRRELEVANT

• Percentage Protein in the sample %protein =%N * f = 5.70 (cereals) = 6.25 (meat products) = 6.38 (dairy products)

A 5.8734-gram sample beef was analyzed for its N content and the liberated NH3 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. 34.64% d. 11.08%

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. 23.89%

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%

Double Indicator Method  Mixture of NaOH, Na2CO3 or NaHCO3

• Hargis

Double Indicator Method  Mixture of NaOH, Na2CO3 or NaHCO3

- no more than two of these three constituents can exist in appreciable amount in any solution

Double Indicator Method  Mixture of NaOH, Na2CO3 or NaHCO3

• Skoog

Na2CO3 and NaHCO3

Na2CO3 and NaOH

Double Indicator Method (Mixture of Bases)

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 b. 6.577% Na2CO3 and 5.842% NaOH c. 65.77% Na2CO3 and 5.842% NaHCO3 d. 65.77% Na2CO3 and 5.842% NaOH

• 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 NaHCO3 in the sample?

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. •

Precipitation • 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 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

MOHR METHOD 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.

MOHR METHOD 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

MOHR METHOD 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%

VOLHARD METHOD Formation of colored complexion Volhard Method (Jacob Volhard, Germany, 1874) Direct method for silver - Indirect method for halides Titrant: Potassium thiocyanate, KSCN Titration is carried out in acidic condition to hasten precipitation of ferric ion to its hydrated oxide form.

VOLHARD METHOD 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

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

FAJANS METHOD 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

FAJANS METHOD • Titration is carried out between pH of 4-7. Dextrin is added to prevent excessive coagulation of the AgCl precipitate.

COMPLEXOMETRIC 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.

COMPLEXOMETRIC 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.

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 wine red royal blue

Example What volume of 0.0305 M EDTA is needed to titrate the Ca in 178.56 mg of CaCO3? a.58.54 mL b. 29.27 mL c. 43.91 mL d. 14.64 mL

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%

• 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.

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

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)

Example 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%

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)

Example 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%

REDOX TITRATION • Reduction-Oxidation. These “redox” titrations involve the titration of an oxidizing agent with a reducing agent, or vice versa.

OXIDANT OXIDANT

Combining Ratio, also (f)

KMnO4 (a)

5

KMnO4 (b,n)

3

K2Cr2O7

6

I2

2

MnO2

2

REDUCTANT REDUCTANT

Na2C2O4 (C2O4-2) FeSO4 (Fe+2) Na2S2O3 KI Fe metal As2O3 H2O2 Cu+

Combining Ratio, also (f) 2 1 1 1 2 4 2 1

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

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%

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%

A sample of iron ore weighing 385.6 mg was dissolved in acid and passed through a Jones reductor. If the resulting solution required 52.36 mL of 0.01436 M K2Cr2O7 for titration, calculate % Fe3O4 (231.55 g/mol) in the ore sample. a. 15.05% b. 45.15% c. 90.30% d. 67.98%

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 MnO2 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 Mn3O4 (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%

A hydrogen peroxide solution is analyzed by adding a slight excess of standard KMnO4 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%

A sample of a pyrolusite weighs 0.5000 g. To this is added 0.6674 g of As2O3 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 MnO2. Ans. 78.19%