Solution Manual for Quantitative Chemic...9th EditionDaniel Harris Publisher: Palgrave Macmillan Higher EdISBN: 9781464175633
Solution Manual for Quantitative Chemic...
Solutions for Solution Manual 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
Sample Solutions for this Textbook
We offer sample solutions for Solution Manual 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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