Quantitative Chemical Analysis9th EditionDaniel C. Harris Publisher: W. H. FreemanISBN: 9781464135385
Quantitative Chemical Analysis
Solutions for Quantitative Chemical Analysis
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Chapter 0.2 - General Steps In A Chemical AnalysisChapter 1 - Chemical MeasurementsChapter 1.1 - Si UnitsChapter 1.2 - Chemical ConcentrationsChapter 1.3 - Preparing SolutionsChapter 1.4 - Stoichiometry Calculations For Gravimetric AnalysisChapter 2 - Tool Of The TradeChapter 2.6 - Pipets And SyringesChapter 3 - Experimental ErrorChapter 3.4 - Propagation Of Uncertainity From Random Error
Chapter 4 - StatisticsChapter 4.1 - Gaussian DistributionChapter 4.2 - Comparison Of Standard Deviations With The F TestChapter 4.3 - Confidence IntervalsChapter 4.4 - Comparison Of Means With Student's TChapter 4.7 - The Method Of Least SquaresChapter 4.8 - Calibration CurvesChapter 5 - Quality Assurance And Calibration MethodsChapter 5.1 - Basics Of Quality AssuranceChapter 5.2 - Method ValidationChapter 5.3 - Standard AdditionChapter 5.4 - Internal StandardsChapter 6 - Chemical EquilibriumChapter 6.1 - The Equilibrium ConstantChapter 6.4 - Complex FormationChapter 6.6 - PhChapter 6.7 - Strengths Of Acids And BasesChapter 7 - Let The Titrations BeginChapter 7.2 - Titration CalculationsChapter 7.3 - Precipitation Titration CurvesChapter 8 - Activity And The Systematic Treatment Of EquilibriumChapter 8.1 - The Effect Of Ionic Strength On Solubility Of SaltsChapter 8.2 - Activity CoefficientsChapter 8.3 - Ph RevisitedChapter 8.4 - Systematic Treatment Of EquilibriumChapter 9 - Monoprotic Acid-Base EquilibriaChapter 9.1 - Strong Acids And BasesChapter 9.4 - Weak Base EquilibriaChapter 9.5 - BuffersChapter 10 - Polyprotic Acid-Base EquilibriaChapter 10.1 - Diprotic Acids And BasesChapter 10.2 - Diprotic BuffersChapter 10.3 - Polyprotic Acids And BasesChapter 10.4 - Which Is The Principal Species?Chapter 11 - Acid-Base TitrationsChapter 11.3 - Titration Of Weak Base With Strong AcidChapter 11.5 - Finding The End Point With A Ph ElectrodeChapter 11.8 - Kjeldahl Nitrogen AnalysisChapter 12 - Edta TitrationsChapter 12.2 - EdtaChapter 12.5 - Auxillary Complexing AgentsChapter 12.7 - Edta Titration TechniquesChapter 13 - Advanced Topics In EquilibriumChapter 14 - Fundamentals Of ElectrochemistryChapter 14.1 - Basic ConceptsChapter 14.2 - Galvanic CellsChapter 14.4 - Nernst EquationChapter 14.5 - E° And The Equilibrium ConstantChapter 14.6 - Cells As Chemical ProbesChapter 14.7 - Biochemists Use E°'Chapter 15 - Electrodes And PotentiometryChapter 15.2 - Indicator ElectrodesChapter 15.3 - What Is A Junction PotentialChapter 15.6 - Ion-Selective ElectrodesChapter 15.7 - Using Ion-Selective ElectrodesChapter 16 - Redox TitrationsChapter 16.1 - The Shape Of A Redox Titration CurveChapter 17 - Electroanalytical TechniquesChapter 17.1 - Fundamentals Of ElectrolysisChapter 17.2 - Electrogravimetric AnalysisChapter 17.3 - CoulometryChapter 18 - Fundamentals Of SpectrophotometryChapter 18.2 - Adsorption Of LightChapter 18.4 - Beer's Law In Chemical AnalysisChapter 18.5 - Spectrophotometric TitrationsChapter 19 - Applications Of SpectrophotometryChapter 19.1 - Analysis Of A MixtureChapter 20 - SpectrophotometersChapter 20.2 - MonochromatorsChapter 20.4 - Optical SensorsChapter 21 - Atomic SpectroscopyChapter 22 - Mass SpectrometryChapter 22.1 - What Is Mass Spectrometry?Chapter 22.2 - Oh, Mass Spectrum, Speak To Me!Chapter 22.3 - Types Of Mass SpectrometersChapter 23 - Introduction To Analytical SeparationsChapter 23.1 - Solvent ExtractionChapter 23.3 - A Plumber's View Of ChromatographyChapter 23.4 - Efficiency Of SeparationChapter 24 - Gas ChromatographyChapter 24.1 - The Separation Process In Gas ChromatographyChapter 25 - High-Performance Liquid ChromatographyChapter 25.1 - The Chromatographic ProcessChapter 25.3 - Method Development For Reversed-Phase SeparationsChapter 26 - Chromatographic Methods And Capillary ElectrophoresisChapter 26.1 - Ion-Exchange ChromatographyChapter 26.7 - Conducting Capillary ElectrophoresisChapter 27 - Gravimetric And Combustion AnalysisChapter 27.1 - An Example Of Gravimetric AnalysisChapter 27.3 - Examples Of Gravimetric CalculationsChapter 27.4 - Combustion AnalysisChapter 28 - Sample PreparationChapter 28.1 - Statistics Of Sampling
Book Details
The gold standard in analytical chemistry, Dan Harris' Quantitative Chemical Analysis provides a sound physical understanding of the principles of analytical chemistry and their applications in the disciplines.
Sample Solutions for this Textbook
We offer sample solutions for Quantitative Chemical Analysis homework problems. See examples below:
Explanation: Calculate the number of moles of solute methanol to determine. Given that volume of...Explanation: Assume that volume of solution is 1.000 L which is equivalent to 1.50 kg as density of...Explanation: Each mol of Ca(NO3)2 furnishes one mol Ca2+ ion and two mol NO3− ions. Formula of NO3−...Explanation: The quantity 10−13 joules can be expressed as 100 fJ. fJ is femtoJoule and the prefix...Explanation: Formula mass of acetic acid, CH3COOH is calculated as – atomic mass of C = 12.0107...Explanation: To calculate true mass of water at 15°C Given, Mass of water in air (balance reading)...Explanation: To calculate the mass of Benzoic acid to prepare 250 mL of 100.0 mM aqueous solution....Explanation: To calculate the mass of 50.037 mL water at 20°C Given, Volume of water at 20°C in 50...Explanation: To give the equation predicts that the time at which 50% of micro pipets remain within...
Explanation: Given data: The weight of the empty crucible is 12.4372 g The weight of the same...Given data: [12.41(±0.09)÷4.16(±0.01)]×7.0682(±0.0004)=? Calculation of absolute and percent...Given data: 9.23(±0.03)+4.12(±0.02)−3.26(±0.06)=? Calculation of absolute and percent relative...Explanation: Given data: The given numbers are 116.0, 97.9, 114.2, 106.8 and 108.3 To Calculate: The...Explanation: Given data: The sediments in North Sea which are traces of toxic, man-made...Explanation: Given Interval: μ±σ Calculation of Gaussian Population: The given interval μ±σ...Explanation: Given data: The results for the measurement of nitrite (NO2−) using the given two...Explanation: Given data: Consider least-squares in Figure 4-11 A new single measurement gives a y...Explanation: Given, The low concentrations samples near the detection limit are....Explanation: Given, Initial volume ( Vi ) = 25.0 mL Final volume ( Vf ) =25.5mL Final concentration...Explanation: Given, An unknown sample of Cu2+ gave an absorbance of 0.262. 1.00 mL Of a solution had...Explanation: Given, In chromatographic analysis, a solution has an analyte concentration of 3.47mM...Explanation: Given Ag++ Cl⇌ AgCl(aq) K =2.0×103 AgCl(s)⇌ Ag++ Cl- K = 1.8×10−10 By adding these two...Explanation: Given Given reaction is, BaCl2.H2O(s)⇌BaCl2(s)+ H2O(g) ΔH0= -63.11kJ/mol at 25°C ΔH0=...Given Given reaction is, H2(g)+Br2(g)⇌2HBr(g) K=7.2×10-4 ΔHo is positive. Amount of HBr in the...Explanation: Given Li++OH-⇌LiOH(aq) K1=0.83Na++OH-⇌NaOH(aq) K1=0.20 To prepare: table for initial...Explanation: Given reaction is H3N+CH2CH2N+H3+ H2O ⇌H3N+CH2CH2NH2+H3O+ Acid base conjugate base...Explanation: Concentration of the given solution is 0.010M The concentration of hydrogen ion is...Explanation: To calculate: The formula mass of ascorbic acid. The molecular formula of ascorbic acid...Explanation: To calculate: The pAg+ values for various volumes of silver ions of given titration....Explanation: To calculate: The pCa2+ value before equivalence point for the titration of Na2C2O 4 Vs...Explanation: To calculate: The pAg+ values of silver ions of given titration. Given, 0.0502M...Explanation: Given information, 0.2 mM solution of KNO3 Calculate the ionic strength by using the...Explanation: Given information, 0.10 M NaClO4 saturated with Mn(OH)2.Value of µ = 0.1.Size of MnOH+...Explanation: Given information, 0.01 M sodium acetate is given. Pertinent reactions are: A- +H2O ⇌Kb...Explanation: Pertinent reactions are: Ca(OH)2(S) ⇌KSP Ca2+ + 2OH- KSP = [Ca2+]γCa2+[OH-]2γ2OH- =...Explanation: Molarity of NaOH = 1.0×10-2 M [OH-] = 1.0×10-2 M γOH-=0.900 μ=0.0100 M pH can be...Explanation: Charge balance of HBr = [H+]=[OH-]+[Br-] Mass balance of HBr = [Br-]=1.0×10-8 M...Explanation: Mass of Glycine amide hydrochloride = 1.00 g Mass of Glycine amide = 1.00 g Volume of...Explanation: The chemical reaction with equilibrium constants Kb and Ka for Imidazole and Imidazole...Explanation: Mass balance and Charge balance is obtained if we dissolve B and BH+Br- . Mass balance...Explanation: Equilibria: β1=[AlOH2+][H+][Al3+] (1) β2=[Al(OH)2+][H+]2[Al3+] (2)...Explanation: Given, Dissociation of the given acid H2SO3⇌ HSO3-+H+ 0.050-x x x x20.050-x=K1=...EXPLANATION: Given, Dissociation of the given acid x20.100-x=K1 x= 3.11×10-3[H+]=[HA-] pH=...Explanation: Charge balance: [K+]+[H+]= [OH-]+[HP-]+2[P2-] (1) Mass balance: [K+]=[H2P]+ [HP-]+[P2-]...Explanation: Mass balance: F= [Na+]=[H2A]+ [HA-]+[A2-] (A) Charge balance: [Na+]=[H+]+ [HA-]+2[A2-]...Explanation: The expressions for KH, Ka1 and Ka2 as derive as equation C ,[CO32-]= Ka2[HCO3-][H+]...Explanation: Given that volume of acid, HCl and it is denoted by Va. Given the strength ( M1 ) and...Explanation: Given that volume of base, KOH and it is denoted by Vb. Given the strength ( M1 ) and...Explanation: Given that volume of acid, HNO3 and it is denoted by Va. Given the strength ( M1 ) and...Explanation: Given that volume of base, NaOH and it is denoted by Vb. Given the strength ( M1 ) and...Explanation: Given that volume of base, KOH and it is denoted by Vb. Given the strength ( M1 ) and...Explanation: Given that volume of acid, HCl and it is denoted by Va. Given the strength ( M1 ) and...Explanation: When NaCN and HClO4 are made into aqueous solution, they are completely dissociated...Explanation: Piperazine is a weak base and HCl is a strong acid. So the titration of piperazine with...Explanation: To calculate the concentration of K+ ion in the original sample EDTA Given, Volume of...Explanation: To determine the volume of solution at equivalent point in EDTA titration. Given,...Explanation: To determine the volume of solution at equivalent point in EDTA titration. Given,...Explanation: To calculate the pCu2+ at given volume of EDTA Given, 50 .0 mL of 0.001 M Cu2+ 0.0100 M...Explanation: To give the equations for formation constants K1 and K2 Given, Figure 1 Fe3++T ⇌K1...Explanation: The given solution containing 0.010 mol hydroxybenzene (HA), 0.030 mol dimethyamine...Explanation: Given information, The given 1.00 L of solution containing 0.040 mol 2-aminobenzoic...Explanation: Derivation of equation 13-16 from mass balance 13-15: Equation 13-15 (mass balance for...Explanation: Given data: Eo=1.35 V To calculate: cell voltage The cell voltage = 1.35V. Because of...Explanation: Given data: The given reaction is I2(s) + 5Br2(aq) + 6H2O ⇌ 2IO3− +10Br−+12H+...Explanation: To determine: The cell voltage and direction flow of electrons. Right hand cell: Br(l)...To write: The Nernst equation for given cell. Right hand cell: H+ + e- ⇌ 0.5H+(g) E+0 = 0V - Left...Explanation: To write: The net reaction for given solutions. Elements with two oxidation states are...Explanation: The half-reaction and Nernst equation for each half-cell can be calculated using below...Explanation: M2++ 2e- ⇌ M(s) E+o= -0.266 V-2H+ + 2e- ⇌ H2 (0.50 bar) E-o= 0 V_H2(g)+ M2+ ⇌ 2H+ +...Explanation: Given: The titration reaction is a reduction reaction. Ag++e-⇌Ag(s) The cell voltage of...Explanation: Given: The given reactions are the small amount of HgY2- added to analyte equilibrates...Explanation: Cell Voltage: Ecell= E+- E- Given, E+ = 0.523 V E- = 0.00 V 0.523 V Versus S.H.E = x...Explanation: In phase α , 0.1 M H+ Has u=36.3×10-8m2s-1V-1 0.1 M Cl- Has u=7.91×10-8m2s-1V-1 In...Explanation: Given: pH is 6.865 at 250C for 0.0250mKH2PO4|0.0250mNa2HPO4 buffer. The change in...Given Amount of solution of Sn2+ = 20.0 mL Strength of Sn2+ solution = 0.00500 M Strength of Ce4+...Explanation: Figure 1 shows the titration of iron (II) with cerium (IV). 100.0 mL of 0.0500M Fe2+...Explanation: Given Scheme 1: MnO4-→Mn2+ 2[8H++MnO4-+ 5e−→Mn2++ 5H2O] +7 +2 5[ H2O2→O2+2e-+2H+] −1...Explanation: To calculate: Average oxidation number of Bi and copper Experiment A: Amount of initial...Explanation: To determine: The voltage required to electrolyze sodium sulfate with given current and...Explanation: To determine: The cathode potential at which Sb deposition will happen from SbO+ ....Explanation: To determine: The cathode potential required to reduce Co2+ to Co . The conversion of...Explanation: To determine: The voltage required to complete the given net reaction. When the...Explanation: To determine: The current density and overpotential of reduction of azobenzene. The...Given information: Transmittance = 45.0 %. Apply the Beer’s law to calculate the value of absorbance...Explanation: Given information, Concentration is 3.96 × 10-4 M.Absorbance (A) is 0.624 at 238 nm in...Explanation: Given information, Mass of NH4Cl is 1.00 × 10-2 g.Sample Absorbance (at 625 nm)Blank...Explanation: Given information, Vapor pressure is 30.3 µbar.transmittance is 24.4 % at 266...Explanation: From absorbance, the concentration can be calculated as,...Explanation: The quantity of HIn is small when compared to Aniline and Sulphanilic acid. The ICE...Explanation: A620=εHIn-620b[HIn-]+εIn2-620b[In2-]A434=εHIn-434b[HIn-]+εIn2-434b[In2-] The solution...Explanation: Given λ= 10.00 μmand 10.01 μm the value of λ= 10.00 μm and Δλ0.01 μm λΔλ= 10.000.01 =...Explanation: Given, Wavelengths is 512.23 and 512.26 nm According to resolution of grating:...Explanation: The limit for O2 is relative detection air from photoacoustic spectroscopy decreases...Explanation: The Li standard contained 1.62 μg Li/mL . Graph, that has intensity against the...Explanation: The metals that are present in the snow was identified by atomic fluorescence. The mean...Explanation: We know that, ΔE = hν = hcλ = (6.626 × 10−34 Js)(2.998 × 108 m/s)422.7 × 10−9 m = 4.699...Explanation: Calculation of Wavelength for Sodium; To calculate wavelength the formula that can be...Explanation: The width which is present at half-height is 0.60mz The given peak at m/z = 53 Now,...Explanation: Form the given intensity we can identified a compound for M+· =94 is phenol. The...Explanation: The given information in figure 22-9 is recorded as follows, The molecular formula for...Explanation: The expected intensities for the given species can be calculated as follows. The...Explanation: Let’s take the term AxCxmx, which applies to the unknown. AxCxmx =(μmol isotope Aμmol...Explanation: Given information, Partition coefficient value is 4.0. Extracted volume of phase 1 is...Explanation: Given information, The value of relative retention (α) is 1.068 . The value of...Explanation: Given information, Column length is 30.1 m.Thickness of the stationary phase is 3.1...Explanation: Given information, Column 1: N = 1000 ; k2 = 1.2 ; α = 1.16 ; resolution = 0.6Column 2:...Explanation: Given information, Figure 1 The relationship between N and w1/2 N = 5.55 tr2w1/22Where,...Explanation: Given information, Elution time for unretained solute is 1.06 min.Length of the column...Explanation: The retention index of the compounds is based on the overall polarity phases and the...Explanation: All analytes responds to thermal conductivity gas chromatography detector.Explanation: Given, tr = 8.4 min tm = 3.7 min The adjusted retention time is calculated as,...Explanation: Given, Boiling point of Octane = 126°C Boiling point = 126°C+273.15 Boiling point =...Explanation: As k→0 , Hminr=13Hminr=0.58 As k→∞ ,...Explanation: Peak area of A is 5.97 . Peak area of B is 6.38 . AreaA[A] = F(AreaB[B])10.86[1.03] =...Explanation: To find the retention time of R-isomer, tm is given as 1.00 min . Retention factor ( k...Explanation: To explain: The greatest and least retention of components in mobile phase Increase in...Explanation: For L-enantiomer: Retention time ( tr ) for L-enantiomer is given as 4.70 min ....Explanation: The given column length is 4400mm. The linear velocity is measured with unretained...Explanation: To determine the volume of 0.023 1 M NaOH is needed to titrate the eluate form...Explanation: To determine the mass percentages of given compounds. In ion-exchange chromatography,...Explanation: To calculate the elution volume of excluded molecules. Given, Column diameter is 7.8 mm...Explanation: To describe the working method of ion mobility spectrometry. The analyte are converted...Given Formula mass of an organic compound = 417g/mol Amount of the sample = 25.42 mg Amount of...Explanation: Given information: Oxidation state of Oxygen is -2. Oxidation state of Lanthanide is...Explanation: To calculate: The value of α when all the reactants are 100% deuterated during...Explanation: To calculate: The value of x when YBa2Cu3O7-x(s) undergoes thermogravimetric analysis....For experiment: 1 To find: mean for Cl− x¯ = 7.8+9.8+7.8+7.8+7.8+7.8+13.7+12.7+13.7+12.710 = 10.160...Explanation: The given box contains 12000 red marbles and 88000 yellow marbles. The given random...Explanation: The mass of single particle of Na2CO3 and K2CO3 can be determined by knowing its volume...
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